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Chemical Equilibrium (I): Types of Equilibrium. Equilibrium Constant & Calculations on Equilibrium Constant

2018-02-24T23:15:00.001+01:00

In this post, we shall be discussing the basic sub-topics on chemical equilibrium such as:

~Meaning of equilibrium
~Types of equilibrium
~Conditions for chemical equilibrium
~Equilibrium constant, K
~Calculations involving equilibrium constant

We will look at other areas such as Le Chatelier's Principle et al in our next/subsequent posts.

Meaning & Types of Equilibrium

Equilibrium is a state of rest of a body or system. It is the state when all the forces acting on a body counteract themselves, thereby making the net force on the body to be equal to zero. This is known as static equilibrium. Example of a body that exhibits static equilibrium is the seesaw, when two bodies of equal weights or masses sit on each of its arms.

Static Equilibrium

More importantly, equilibrium is the state in which there is no observable or detectable change in the properties of a system with respect to time. Though there are no detectable changes, there are still movements at the molecular level, which are not noticeable to the naked eyes. This is known as dynamic equilibrium, and can be subdivided into physical equilibrium (e.g. a saturated solution of sodium chloride) and chemical equilibrium (e.g. the thermal decomposition of calcium trioxocarbonate (IV) in a closed vessel).

In the saturated solution of sodium chloride, NaCl, the dissolved solutes are in equilibrium with undissolved solutes, as shown in the equation below:

NaCl(aq) <-----------------> NaCl(s)
dissolved solutes undissolved solutes

Let us assume that the solubility of NaCl at room temperature is 36 gdm^-3, i.e, the maximum amount of the salt that can dissolve in 1 dm^3 of water at 25°C is 36 g. If you add 50 g of the salt to 1 dm^3 of water at 25°C, 14 g will remain undissolved because the solution is saturated. At that point, those 14 g keep going into and coming out of solution. The fact is, as the excess salt particles above its solubility limit undergo dissolution, the same amount is being precipitated out of solution. Thereby making it look as if there is no change.

Dynamic Equilibrium

In the thermal decomposition of the calcium carbonate, if there is no space for the carbon (IV) oxide gas produced to escape, a point will be reached when the concentrations of the products and reactants will be at equilibrium with each other inside the reaction vessel.

CaCO3(s) + heat <-------------> CaO(s) + CO2(g)

For that to be achieved, certain conditions which are necessary for chemical equilibrium to occur must be met.

Conditions Necessary for Chemical Equilibrium

1. The reaction must occur in a closed vessel or it must be a closed system. Here, none of the products (especially gases) will be able to escape.

2. The reaction must be reversible. A reversible reaction is one that proceeds in both directions under appropriate conditions. The inability of the carbon (IV) oxide to escape will force it to combine with the calcium oxide to form the reactant again, thereby, making the reaction reversible.

3. The rate of the forward reaction must equal the rate of the backward reaction. As the products continue to combine to form the reactants and vice versa, a point will be reached when the concentration of the reactants and products will be constant, though not necessarily the same. This is the point of equilibrium at which the rate of the formation of products from reactants equals the rate of the decomposition of products back to reactants.

Equilibrium Constant, K

According to the law of mass action, the rate of a chemical reaction is directly proportional to the active masses of the reactants. The active mass of a reactant is its concentration raised to its stoichiometric value in a given equation.

Consider the hypothetical reaction stated below:

aA + bB <----------------> cC + dD

If the rate of the forward reaction (aA + bB ----------> cC + dD) is r1 ,

and the rate of the backward reaction (cC + dD -----------> aA + bB) is r2;

then, according to the law of mass action:

r1 & [A]^a × [B]^b ..................(i)
r2 & [C]^c × [D]^d ..................(ii)

where [A]^a, [B]^b, [C]^c and [D]^d are the active masses of A, B, C and D respectively
and, & = sign of proportionality.

From equations (i) and (ii),

r1 = k1 × [A]^a × [B]^b .................(iii)
r2 = k2 × [C]^c × [D]^d .................(iv)

where k1 and k2 are the velocity constants for the forward and backward reactions respectively.

At equilibrium, r1 = r2

=> k1 × [A]^a × [B]^b = k2 × [C]^c × [D]^d ..........................(v)

Rearranging equation (v) gives

k1/k2 = ([C]^c × [D]^d)/([A]^a × [B]^b) .....................(vi)

The ratio k1/k2 is equal to K, the equilibrium constant of the reaction. Since k1 and k2 are affected by temperature, it also implies that K will be affected by temperature.

Therefore, equation (vi) can be rewritten as

Kc = ([C]^c × [D]^d)/([A]^a × [B]^b) .........................(vii)

Equation (vii) gives the expression for the equilibrium constant of concentration, Kc of the reaction. Generally, it gives the relationship between the concentrations of the reactants and that of the products in an equilibrium mixture.

In other words, equilibrium constant of concentration, Kc of the reversible reaction

mReactants <--------------> nProducts

is given as:

Kc = [Products] ^n/[Reactants] ^m

Equilibrium Constant of a Gaseous System, Kp

The equilibrium constant of pressure, Kp, applies to a gaseous system only (while Kc applies to both aqueous and gaseous systems). To determine Kp, the partial pressure of the gases in the system is always used.

Hence, for a given gaseous system

aA(g) + bB(g) <------------------> cC(g) + dD(g)

Kp = ((PC)^c × (PD)^d)/((PA)^a × (PB)^b) ....................(viii)

where PA, PB, PC and PD are the partial pressures of gases A, B, C and D respectively in the system, while a, b, c and d are the their stoichiometric values in the equation.

Please refer to our post on Dalton's Law of Partial Pressure for the calculation of partial pressure.

Relationship between Kp and Kc

Recall from the ideal gas equation that

PV = nRT ............................(ix)

where P, V, T and n represent pressure, volume, temperature, number of moles respectively for a given gas. R is the molar gas constant.

Also, the relationship between number of moles, concentration and volume of a substance is given as

n = CV ..................................(x)

If we divide both sides of equation (ix) by V, we will obtain:

                                            P = nRT/V
which can be rewritten as
                                       P = (n/V) RT ..........................(xi)

From equation (x), it can be seen that n/V = C. Hence, substituting this into equation (xi), we will have

P = C RT ...............................(xii)

Equation (xii) gives us the relationship between the pressure P, concentration C, and temperature T, of a given gas.

Now, let us substitute equation (xii) into equation (viii) as follows:

PA = [A].RT; PB = [B].RT; PC = [C].RT; PD = [D].RT

where [A], [B], [C] and [D] are the concentrations of gases A, B, C and D respectively.

Therefore, equation (viii) becomes

Kp = (([C].RT)^c × ([D].RT)^d)/(([A].RT)^a × ([B].RT)^b)

Evaluating the above equation gives:

Kp = ([C]^c × [D]^d × (RT)^c × (RT)^d)/([A]^a × [B]^b × (RT)^a × (RT)^b)

If we rearrange the above equation and apply the laws of indices, then

Kp = (([C]^c × [D]^d)/([A]^a × [B]^b)) × (RT)^(c+d))/(RT)^(a+b)

Kp = ([C]^c × [D]^d)/([A]^a × [B]^b) × (RT)^(c+d)-(a+b)

If we substitute equation (vii) into the above equation, we will obtain

Kp = Kc × (RT)^(c+d)-(a+b) ...................(xiii)

but

(c+d) - (a+b) = ∆n (change in number of moles)

Hence, equation (xiii) becomes

Kp = Kc × RT^∆n ...........................(xiv)

Equation (xiv) gives us the relationship between the equilibrium constant of pressure Kp, equilibrium constant of concentration Kc, temperature T, and number of moles n, of a gaseous system at equilibrium.

Calculations on Equilibrium Constants

Example 1
Write the expressions and units for the Kc and Kp for the following reactions:

a) N2(g) + 3H2(g) <---------------> 2NH3(g)
b) 2SO2(g) + O2(g) <---------------> 2SO3(g)
c) CaCO3(s) <---------------> CaO(s) + CO2(g)

Answer
a) N2(g) + 3H2(g) <---------------> 2NH3(g)

i) Kc = [NH3]^2/([N2] × [H2]^3)

Since we are dealing with concentration, the unit will be moldm^-3. Therefore, the unit for Kc of the above reaction will be:

                                   Kc = (moldm^-3)^2/((moldm^-3) × (moldm^-3)^3)
                                         = (moldm^-3)^2/(moldm^-3)^4
                                         = (moldm^-3)^-2
                                         = dm^6 mol^-2

ii) Kp = (PNH3)^2/((PN2) × (PH2)^3)

The units of pressure include atm, Pa, Nm^-2, torr etc. For the sake of simplicity, we will use atm in this analysis. So, the unit of Kp for the above reaction is:

                                    Kp = atm^2/(atm × atm^3)
                                          = atm^2/atm^4
                                          = 1/atm^2
                                          = atm^-2

b) 2SO2(g) + O2(g) <---------------> 2SO3(g)

i) Kc = [SO3]^2/([SO2]^2 × [O2])

Unit
                                 Kc = (moldm^-3)^2/((moldm^-3)^2 × (moldm^-3))
                                       = (moldm^-3)^2/(moldm^-3)^3
                                       = (moldm^-3)^-1
                                       = dm^3 mol^-1

ii) Kp = (PSO3)^2/((PSO2)^2 × PO2)

Unit
                                  Kp = atm^2/(atm^2 × atm)
                                        = atm^2/atm^3
                                        = 1/atm
                                        = atm^-1

c) CaCO3(s) <---------------> CaO(s) + CO2(g)

i) Kc = [CaO] × [CO2]/[CaCO3]

The concentration of solids is always considered to be negligible (when compared to liquids or gases) and assigned a value of 1. Since CaCO3 and CaO are in the solid phase, their concentrations will reduce to 1.

Hence, the final expression for the Kc is:

Kc = [CO2]

Unit
Kc = moldm^-3

ii) Kp = PCO2

Since pressure has no effect on solid and liquid phases, CaCO3 and CaO exert no partial pressure on the system. This explains why CO2 alone appears in the Kp expression for the reaction.

Unit
Kp = atm

Example 2
Ethanol reacts with ethanoic acid to form ethyl ethanoate and water, according to the equation

CH3COOH(l) + C2H5OH(l) <-------------> CH3COOC2H5(l) + H2O(l)

250 cm^3 of the reaction mixture at equilibrium contained 0.135 mol of ethanoic acid and 0.0250 mol of ethanol together with 0.0820 mol of ethyl ethanoate and 0.0820 mol of water. Use this data to calculate a value of Kc for the reaction.

Answer
Here, the number of moles of the reactants and products and volume of mixture (or reaction vessel) are provided. From these, you can determine the equilibrium concentration of each species.

CH3COOH(l) + C2H5OH(l) <-------------> CH3COOC2H5(l) + H2O(l)

Volume of mixture, V = 250 cm^3 = 0.250 dm^3 (recall 1000 cm^3 = 1 dm^3)
Number of moles of CH3COOH, n = 0.135 mol
Concentration of CH3COOH at equilibrium, C = n/V
= 0.135/0.25
= 0.540 moldm^-3

Number of moles of C2H5OH, n = 0.0250 mol
Concentration of C2H5OH at equilibrium, C = n/V
= 0.0250/0.25
= 0.100 moldm^-3

Number of moles of CH3COOC2H5, n = 0.0820 mol
Concentration of CH3COOC2H5 at equilibrium, C = n/V
= 0.082/0.25
= 0.328 moldm^-3

Since the number of moles of ethyl ethanoate and water are equal, their equilibrium concentrations will also be equal.

Hence concentration of H2O at equilibrium = 0.328 moldm^-3

Therefore,

                                       Kc = [CH3COOC2H5][H2O]/[CH3COOH][C2H5OH]
                                            = (0.328 × 0.328)/(0.54 × 0.1)
                                             = 0.107584/0.054
                                             = 1.99

Since the stoichiometric values of the reactants and products are equal, the units will cancel out themselves. So, the value of the Kc for the reaction will have no unit.

Example 3
Propanone, CH3COCH3 reacts with hydrogen cyanide, HCN as follows:

CH3COCH3(l) + HCN(l) <----------------> CH3(OH)(CN)CH3(l)

A mixture of 0.0400 moldm^-3 propanone and 0.0400 moldm^-3 hydrogen cyanide is left to reach equilibrium at room temperature. At equilibrium, the concentration of the product is 0.0133 moldm^-3. Calculate Kc for this reaction.

Answer
Here, you are given the initial concentrations of the reactants and the equilibrium concentration of the product. From that, you are expected to calculate the equilibrium concentration of the reactants.

CH3COCH3 + HCN <----------------> CH3(OH)(CN)CH3

Initial concentrations of propanone and hydrogen cyanide = 0.0400 moldm^-3

If 0.0133 moldm^-3 of the product was formed at equilibrium, it means that same concentration of the reactants had decomposed.

Therefore, equilibrium concentration of propanone = 0.0400 - 0.0133 = 0.0267 moldm^-3

Similarly, the equilibrium concentration of hydrogen cyanide = 0.0267 moldm^-3

Hence,

                                          Kc = [CH3(OH)(CN)CH3]/[CH3COCH3][HCN]
                                                = 0.0133/(0.0267 × 0.0267)
                                               = 18.66 (moldm^-3) ^-1
                                                = 18.66 dm^3 mol^-1

Example 4
In the reaction

2SO2(g) + O2(g) <----------------> 2SO3(g)

The equilibrium partial pressures at constant temperature are 10 atm, 70 atm and 80 atm for SO2, O2 and SO3 respectively. What is the Kp for this reaction.

Answer
2SO2(g) + O2(g) <----------------> 2SO3(g)

Kp = (PSO3)^2/((PSO2)^2 × PO2)

where PSO3 = 80 atm, PSO2 = 10 atm and PO2 = 70 atm

Therefore, substituting the values into the above equation gives:

                                                 Kp = 80^2/(10^2 × 70)
                                                       = 6400/(100 × 70)
                                                       = 0.91 atm^-1 (check Example 1 to see how to determine the unit)

Example 5
Nitrogen reacts with hydrogen to form ammonia according to the equation:

N2(g) + 3H2(g) <---------------> 2NH3(g)

The pressure exerted by this mixture at constant temperature is 200 atm. Under these conditions, the partial pressure of nitrogen is 120 atm and that of hydrogen is 40 atm. Calculate the value of Kp and Kc for this reaction, assuming that it occurred at 25°C. (Take R = 0.082 atm dm^3 K^-1 mol ^-1)

Answer
Here, you are given the total pressure of the mixture of gases, and the partial pressure of the reactants. You are expected to determine the partial pressure of ammonia from the information provided.

N2(g) + 3H2(g) <---------------> 2NH3(g)

Total pressure of mixture = 200 atm
Partial pressure of nitrogen = 120 atm
Partial pressure of hydrogen = 40 atm
Partial pressure of ammonia = total pressure - (sum of partial pressures of nitrogen and hydrogen)
                                              = 200 - (120 + 40)
                                              = 200 - 160
                                              = 40 atm

                                               Kp = (PNH3)^2/((PN2) × (PH2)^3)
                                                     = 40^2/(120 × 40^3)
                                                     = 1/(120 × 40)
                                                     = 0.00021 atm^-2

but, Kp = Kc × (RT)^∆n

so, Kc = Kp/(RT)^∆n

where,                                     ∆n = n(products) - n(reactants)
                                                    = 2 - (3 + 1)
                                                     = 2 - 4
                                                    = -2

                              T = absolute temperature = 25°C + 273 = 298 K
                              R = molar gas constant = 0.082 atm dm^3 K^-1 mol^-1

Substituting these values into the last equation gives Kc as:

                                                 Kc = 0.00021/(0.082 × 298)^-2
                                                       = 0.00021/(24.436)^-2
                                                       = 0.00021 × 597.118
                                                       = 0.125 dm^6 mol^-2 (check Example 1a for the unit)

Do These

Question 1
Write the equilibrium expressions for the Kc and Kp of the following reactions and state their units:

a)                                      CO(g) + 2H2(g) <-------------> CH3OH(g)
b)                                      H2(g) + I2(g) <---------------> 2HI(g)
c)                                      N2(g) + O2(g) <----------------> 2NO(g)
d)                                4HCl(g) + O2(g) <---------------> 2H2O(g) + 2Cl2(g)

Question 2
Calculate the value of Kc for the following reaction given the information below:

H2(g) + CO2(g) <---------------> H2O(g) + CO(g)

Initial concentration of H2(g) = 5.00 moldm^-3
Initial concentration of CO2(g) = 5.00 moldm^-3
Equilibrium concentration of CO(g) = 3.74 moldm^-3

Question 3
The reaction below was carried out at a pressure of 5.00 × 10^4 Pa and at constant temperature

N2(g) + O2(g) <----------------> 2NO(g)

The partial pressures of nitrogen and oxygen are both 2.42 × 10^4 Pa. Calculate the partial pressure of nitrogen (II) oxide, NO(g) at equilibrium.

Question 4
The information provided below gives the data for the reaction of hydrogen with iodine at 500°C.

H2(g) + I2(g) <---------------> 2HI(g)

The initial partial pressures and the partial pressures at equilibrium of hydrogen, iodine and hydrogen iodide are:

Partial Pressure (atm)            H2                I2               HI
Initial                                     70                40                 -
At equilibrium                       34                  -                 77

Assuming that the total pressure remained constant throughout the experiment,

a) Deduce the partial pressure of iodine at equilibrium
b) Calculate the values of Kp and Kc for this reaction, including the units. (R = 0.082 atm dm^3 K^-1 mol^-1)

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Chemical Kinetics (Part I): Overview of Reaction Rates

2018-02-11T18:12:00.000+01:00

Chemical kinetics deals with two main aspects of a chemical reaction: the rate of reaction (i.e, the speed at which the reaction occurs), and the reaction mechanism (i.e, the details of all the steps involved in the reaction). In this post, our focus shall be on the rates of chemical reactions.
Rates of Reactions
Rate is the degree of change of the property of a substance with respect to time. A chemical reaction is a change which involves the conversion of reactants to products, as stated below:
       A ----> B
where A is the reactant and B is the product. In the above hypothetical equation, it can be said that at the beginning of the reaction at time, t = 0, the amount of A present will be 100%, while the amount of B will be 0%. After, a given time, say t1, the concentration of A will decrease, while that of B will be seen to increase. The speed at which these changes occur is said to be the rate of reaction.
Therefore, the rate of a chemical reaction is defined as the amount (concentration, mass or volume) of products formed per unit time. Also, in terms of the reactants, it is the amount (concentration, mass or volume) of reactants converted to products per unit time.
Mathematically, it can be expressed as:
rate of reaction = change in concentration of products formed/time taken, i.e,
rate = D[products]/Dt ………(i)
        = D[B]/Dt
(where D = delta (change), [ ] = concentration) or
rate of reaction = change in concentration of reactants converted/time taken, i.e,
rate = -D[reactants]/Dt………(ii)
        = -D[A]/Dt
The negative sign in equation (ii) indicates that the rate of reaction of the reactants decreases with respect to time, i.e,
rate of reaction & 1/time (where & is a sign of proportionality)
The concentration of the products or reactants is measured in mol dm^-3, while the time of reaction is measured in seconds. Therefore, the standard unit of rate of reaction is mol dm^-3 s^-1. However, if the mass or volume of the products or reactants is used, the unit will be expressed in g s^-1 or cm^3 s^-1. The rate of reaction can also be expressed in mol s^-1, if the mass of the product or reactant is converted to amount in moles.
As an illustration, let us consider the thermal decomposition of calcium trioxocarbonate (IV) to produce calcium oxide and carbon (IV) oxide according to the equation:
CaCO3(s) + heat ----> CaO(s) + CO2(g)
If, at the beginning of the reaction, the mass of the marble [CaCO3] was 100g, and after 100seconds, its mass was found to be 90g. Then, the rate of the reaction can be calculated thus:
rate of reaction = change in mass of CaCO3/time taken
                            = (final mass - initial mass)/time taken
                            = (90 - 100)/100
                            = -10/100
                            = -0.1 g s^-1
Similarly, the rate of reaction in mol s^-1 can be calculated by simply dividing the above result by 100 g mol^-1, the molar mass of CaCO3. Thus,
rate of reaction (mol s^-1) = rate of reaction (g s^-1)/molar mass
                                                = -0.1/100
                                                = -0.001 mol s^-1
Alternatively, the rate of reaction can be calculated in terms of the products. Let us assume that the 10g of CaCO3 were converted to CaO and CO2, then:
rate of reaction = change in mass of CaO + CO2 produced/time taken
                            = (final mass - initial mass)/time taken
                            = (10 - 0)/100
                            = 10/100
                            = 0.1 g s^-1
Since chemical reactions involve changes, there are certain properties of matter, which can indicate the possible occurrence of a reaction, and from which the rate of reaction can be determined. They include, but are not limited to:
a) change in mass
b) change in volume
c) change in concentration
d) change in colour intensity
e) degree of effervescence
Rate Curve
A plot of the concentration, mass or volume of the products against time gives a curve which starts from the origin with a steep slope and later flattens out. This curve is known as the rate curve and is used to describe the rate of a given reaction. The point at which the curve becomes parallel to the horizontal axis is the end-point of the reaction or the point at which the reaction is said to be complete. Click here to see the diagram of a rate curve.
Alternatively, the rate curve can be drawn by plotting the concentration, mass or volume of the reactants against time. This will give us a curve, with a negative gradient, that slopes from left to right. The negative gradient shows the inverse relationship between the concentration (mass or volume) of the reactants and time.
Click here to see an illustration, where the concentrations of the products and reactants are plotted on the same axis against time.
Instantaneous Rate: This is the rate of reaction at a given time, ti. It can be determined from the rate curve by calculating the gradient to the curve at a point on it, directly above ti. Click here to see the illustration. That is,
instantaneous rate = amount of product at time, ti /time, ti
Average Rate of Reaction: The rate curve can also be used to determine the average rate of a reaction by dividing the total amount of the products formed by the total time of reaction, bearing in mind, the end-point, i.e,
avg rate of reaction = total amount of products formed/total time of reaction
Collision Theory
Recall that a chemical reaction is the combination of the molecules or particles of the reactants to form products, as it involves bonds breaking and bonds formation. This implies that without the formation of products, a chemical reaction cannot occur. Therefore, to achieve this, there are two conditions that must be satisfied as explained by the collision theory. These are:
a) effective collisions (b) activation energy
Effective Collisions: During a reaction, millions of particles of the reactants undergo millions of collisions with one another to bring about the breaking of existing and subsequent formation of new intramolecular bonds. However, only a fraction of these collisions will actually result in the formation of products. This fraction is called effective collision, as illustrated in this chart.
Activation Energy: During a reaction, as the molecules of the reactants collide with one another, their potential energies are converted to kinetic energies. Hence, an increase in the frequency of collision leads to an increase in the average kinetic energy of the reactants. However, there is a minimum amount of energy, which the reactants must possess, without which no reaction can occur. This energy barrier is known as the activation energy, Ea.
The relationship between the rate of reaction, temperature and activation energy is given by the Arrhenius equation:
             Rate = Ae^(-Ea/RT) ...…(iii)
where,
A = a constant called the collision factor
Ea = activation energy (kJ mol^-1)
R = gas constant (8.314 J mol^-1 K^-1)
T = absolute temperature
The above equation is useful for determining the Ea of a reaction, through a graphical method or by studying the reaction at two different temperatures (T1 and T2), and evaluating the two equations simultaneously to obtain equation (vi) as shown:
Taking the natural log of both sides of equation (iii) under both conditions gives us,
                    ln Rate1 = ln A - Ea/RT1 ……………(iv)
                    ln Rate2 = ln A - Ea/RT2 ……………(v)
Subtract equation (iv) from (v)
ln Rate2 - ln Rate1 = -Ea/RT2 - (-Ea/RT1)
ln Rate2 - ln Rate1 = Ea/RT1 - Ea/RT2
      ln [Rate2/Rate1] = (Ea/R)[1/T1 - 1/T2] ………(vi)
Factors Affecting the Rates of Chemical Reactions
From the collision theory, it can be deduced that any factor that directly affects the frequency of collisions and the activation energy can influence the reaction rate. Such factors include:
a) effect of temperature
b) effect of concentration/pressure of the reactants
c) effect of nature of the reactants
d) effect of surface area of contact of reactants
e) effect of a catalyst
Temperature: According to the kinetic theory of matter, the average kinetic energy of a body is directly proportional to its absolute temperature. This implies that if the temperature of a reaction is increased, it will lead to a corresponding increase in the frequency of collisions of the particles of the reactants, which has a direct implication on the average kinetic energy. The increased collisions or kinetic energy causes more molecules to acquire enough energy to overcome the energy barrier and form products, as illustrated in these charts showing the relationship between temperature and activation energy.
This explains why a cube of sugar dissolves faster in hot water than cold water. This is also why the liberation of carbon (IV) oxide will be seen to be greater when marble is heated at a higher temperature (say 30°C), than when it is heated at a lower temperature (say 20°C). If the graphs of the mass of CO2 produced under the two conditions are plotted against time on the same axes, two rate curves A (high temp) and B (low temp) will be obtained.
Reaction A is seen to come to completion first before reaction B. Also, the slope of reaction A is steeper at the beginning than that of reaction B, which indicates that the rate of reaction A is higher or faster than that of reaction B. Generally, increasing the temperature of the reaction increases the rate of a chemical reaction.
Concentration/Pressure of Reactants: Concentration is a term used to describe the amount of molecules or particles of a substance present in a given volume, space or region. Pressure, on the other hand, is the force exerted per unit area by gaseous molecules. So, while concentration has an effect on the reaction rate of solid and aqueous systems, pressure affects the rate of reaction of gaseous systems.
The number of molecules available in a particular space will determine the frequency of collisions in the reaction vessel. If there are say, 10 molecules of a substance in about 1,000cm^3 of its solution, it will take a longer time for the molecules to collide with one another because the average distance between them is wide. This translates to a slow rate of reaction.
However, if the amount of molecules in the 1,000cm^3 is increased to about 1,000; then, the reacting molecules will take a shorter time to collide with one another, because the average distance between them has been reduced due to overcrowding (higher concentration). Since the frequency of collisions of the molecules per unit time is increased, the rate of reaction will also increase.
This also holds true for gaseous systems. A gaseous system is one, by which at least one of the reactants is in the gaseous phase. So, increasing the pressure of the system is tantamount to decreasing the volume and the available area for collisions. Hence, the frequency of collisions will be higher, because the gas molecules will collide more with one another within a short time than when the pressure is decreased (i.e, increased volume).
Generally, the higher the concentration or pressure of the reactants, the higher the rate of reaction.
Nature of Reactants: The ability of a substance to react with other substances can also determine how fast or far, a reaction can go. For instance, let us consider the reaction of sodium, iron and gold with water. If a pellet of sodium metal is dropped into a beaker of cold water, the reaction proceeds vigorously with the liberation of heat and hydrogen gas, according to the equation:
            2Na(s) + 2H2O(l) ----> 2NaOH(aq) + H2(g)
The iron, on the other hand, does not react with the cold water, but reacts moderately with steam at red heat to produce hydrogen gas.
           3Fe(s) + 4H2O(g) ----> Fe3O4(s) + 4H2(g)
The gold will not react with the water, irrespective of the conditions. This is because, it is relatively unreactive by nature.
So, based on their chemical reactivity, if the three metals are involved in the same type of reaction under the same conditions, the reaction of sodium will be expected to end first (fastest reaction rate), followed by iron, while that of gold will come last (slowest reaction rate).
Therefore, the more reactive the reactants are, the faster the rate of reaction will be.
Surface Area of Contact: This is the portion of the reactants exposed at a given time, for a reaction. Usually, it is analysed based on the fineness of the reactant particles, with the finest forms (powder, dust etc) having the largest surface area, and the coarsest forms (lumps, rods, pellets etc) having the smallest surface area. This is because in the powdery form, there are more molecules available for collision at a given time, than in the lumpy form.
Let us consider the dissolution of two forms of sugar of equal mass in the same volume of water. The granulated sugar will be seen to dissolve faster than the sugar cubes. This is because, all the particles of the granulated sugar are exposed to hydration by the water molecules at the same time; whereas, in the sugar cubes, only the particles at the surfaces are available for hydration at a given time, and until they are completely hydrated and the surface layers broken down, the inner particles cannot be hydrated. This delays the dissolution of the sugar.
Hence, the larger the surface area of contact of the reactants, the faster the rate of reaction.
Catalyst: A catalyst is a substance that alters the rate of a reaction. It can be positive or negative.
A positive catalyst increases the rate of a chemical reaction by lowering the energy barrier of the reaction, e.g, manganese (IV) oxide [MnO2] in the thermal decomposition of potassium trioxochlorate (V) [KClO3] to produce potassium chloride [KCl] and oxygen gas.
  2KClO3(s) + MnO2 + heat ---->   2KCl(s) + 3O2(g)

It achieves this by creating a new reaction pathway with a lower activation energy, by which the reactants can pass through to form the products. By reducing the energy barrier, more reacting molecules will possess the minimum energy required to form the products, as illustrated in this energy profile diagram.
A negative catalyst, on the other hand, decreases the rate of a chemical reaction by increasing the activation energy of the reaction.
There are certain reactions, which are catalysed by light. These reactions are known as photocatalytic reactions and examples include:
i) photosynthesis
ii) the chlorination of alkanes
iii) the reduction of silver ions to silver atoms in photography
iv) the reaction between hydrogen and halogens etc.
These reactions occur in the presence of sunlight or a source of UV light. The molecules of the reactants absorb the light energy and convert it to kinetic energy, which results in their excitement and increased frequency of collisions. This results in a higher percentage of effective collisions, and subsequent formation of products. The rates of the reactions vary depending on the intensity of the light and the nature of the reacting species. For instance, in the presence of sunlight, hydrogen reacts explosively with fluorine, vigorously with chlorine, moderately with bromine, and slowly with iodine. However, the reaction between hydrogen and chlorine will occur very slowly (almost negligibly) in the dark.
Collision Theory & Heat of Reaction
Recall that a chemical reaction involves bonds breaking and bonds formation. During this process, energy is evolved and absorbed in the form of bond energies, which are converted to heat energy of the system. Bond breaking occurs within the reactants, while bond formation occurs in products. It is important to note that when a bond is broken, heat is evolved, and when a bond is formed, heat is absorbed.
The heat content of a substance is also known as its enthalpy, and is represented by H. In a reaction, the enthalpy of reactants, Hr, is the total amount of heat energy evolved when the intramolecular and intermolecular bonds within the reactants are broken, while the enthalpy of products, Hp, is the total amount of heat energy absorbed when these bonds are formed within the products. The difference between Hp and Hr, is the enthalpy of reaction, also known as change in enthalpy, DH.
Hreaction = Hproducts - Hreactants
           DH = Hp - Hr
Exothermic and Endothermic Reactions
Depending on the nature of reaction, the heat content of the products can either be greater or less than the heat content of the reactants. If the heat content of the products is less than that of the reactants, the reaction is said to be exothermic, and DH will be negative. On the other hand, if the heat content of the products is greater than that of the reactants, the reaction is said to be endothermic, and DH will be positive, i.e:
Exothermic: Hp < Hr; DH < 0 (-ve)
Endothermic: Hp > Hr; DH > 0 (+ve)
Click here for the energy profile diagrams of exothermic and endothermic reactions.
In an exothermic reaction, the excess heat of the reactants is given off to the surrounding, and the reaction vessel is always hot to touch. Examples of exothermic reactions are the dissolution of sodium hydroxide pellets, dissolution of concentrated tetraoxosulphate (VI) acid. However, in an endothermic reaction, the inadequate heat energy which the reactants could not provide is absorbed from the surrounding, and this explains why the reaction vessel is always cold when touched. Example of an endothermic reaction is the dissolution of ammonium chloride [NH4Cl]. Click here to see the illustrations.
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The Kinetic Theory of Matter & Gases

2016-07-29T15:22:00.001+01:00

The Kinetic Theory of Matter postulates that matter is made up of tiny particles that are continuously in motion and so possess kinetic energy. It is also known as the Molecular Theory. The particles may be atoms, molecules or ions. Recall that matter exists in three common states, namely solid, liquid and gaseous states; although, we also have plasma and Einstein-Bosé condensate as additional states of matter. However, these are beyond our scope of discussion. The major differences in the properties of solid, liquid and gas are the degrees of intermolecular forces of attraction and average kinetic energy of the particles.

While the intermolecular force is strongest in the solid state, which confers upon a solid its rigidity in shape and form, it is weakest in the gaseous state, which explains why a gas is formless and takes the shape and volume of the containing vessel. The intermolecular force of attraction is however, mild (not very strong and not very weak) in the liquid state. This is why a liquid has a definite volume, but takes after the shape of the containing vessel.

Conversely, the average kinetic energy of the particles is lowest in the solid state and highest in the gaseous state. This is why a solid always tends to remain where it is, except it is acted upon by a force which results in its motion. It also explains why a gas can readily spread across a large room within seconds, and why a liquid tends to flow slowly when poured on a surface, because of the moderate average kinetic energy of its particles.

Change of State

It is possible for a substance to exist in the three different states of matter discussed above with the absorption or evolution of heat. This is known as change of state of matter and it can be explained using the Molecular Theory. The heat involved in a change of state is known as latent (hidden) heat because it cannot be detected by the thermometer. The various changes of state of matter include:

a) melting - solid to liquid
b) freezing/solidification - liquid to solid
c) vaporization - liquid to gas
d) condensation/liquefaction - gas to liquid
e) sublimation - solid to gas

Phase Diagram for Change of State of Water

Melting
This is the conversion of a solid to liquid through the application of heat. Given that the temperature of a substance is a measure of the average kinetic energy of its molecules, when the solid is heated, the particles at the point where the heat is applied, acquire kinetic energy in the form of heat energy and start vibrating about their axes. They translate this energy to the nearby particles. This continues until all the particles begin to vibrate about their axes.

When a solid (say ice) is heated, there are two forces contending with each other - the vibrational force driven by the applied heat and the intermolecular force of attraction, which holds the particles together. As the temperature increases (as recorded by the thermometer), the force of vibration increases, until a point is reached, when the force of vibration of two adjacent particles is greater than their intermolecular force of attraction. This will cause the latter to give way, allowing the particles to move about 'freely' as water molecules. At this juncture, melting is said to begin. No further heat applied will be recorded by the thermometer, until all the intermolecular forces of attraction holding the solid particles together are broken or melting is completed. The heat absorbed during melting is known as the latent heat of fusion, and the fixed temperature at which the process occurs is the melting point of the substance.

Freezing
This is the opposite of melting as it is the change in state of a liquid (like water) to solid (say ice). It is also known as solidification. During freezing, there is a loss of heat, which is marked by a drop in temperature as recorded by the thermometer. This decrease in temperature leads to a decrease in the average kinetic energy of the liquid, and this in turn increases the intermolecular forces of attraction existing between the molecules. At this point, the particles which were farther apart start getting closer until a bond is formed between any two adjacent particles. Immediately this occurs, no temperature change is further detected by the thermometer until all the molecules are linked together by the intermolecular forces of attraction peculiar to solids. During freezing, the latent heat of fusion is evolved, and the fixed temperature at which this occurs is the freezing point of the substance. It is important to note that the melting and freezing point of a pure sample of a substance are always the same. While the former is associated with the solid state, the latter is with the liquid state.

Vaporization
This is change in state of a substance from its liquid to gaseous state with the application of heat. Using water as an example, it is the stage at which the water molecules become water vapour (steam) molecules. It can also be said to be gasification. There are two different forms of vaporization - evaporation and boiling, and we shall look at each in details.

Evaporation: When the temperature of the liquid increases due to the applied heat, different particles of the liquid acquire different amounts of kinetic energy. This rise in temperature is recorded by the thermometer. As the temperature of the system keeps increasing, many particles acquire more kinetic energy and rise to the liquid surface. On getting there, they are prevented from breaking free by the surface tension of the liquid and the intermolecular attraction of its neighbouring particles. This continues until the molecules acquire sufficient energy to overcome the surface tension and the intermolecular force of attraction to exist as free gaseous molecules in the space above the liquid surface. At this point, evaporation is said to occur.

Factors Affecting the Rate of Evaporation
Before we discuss the factors that affect the rate of evaporation, it is important to know that evaporation occurs at every temperature, and only at the surface of the liquid.

a) For evaporation to occur, the liquid must be in an open vessel. The openness of the containing vessel will determine how rapid the particles at the surface will transit from the liquid to the gaseous phase.

b) The rate of evaporation is affected by humidity. Humidity is a measure of the amount of water vapour present in the atmosphere at a given time. A large amount of water vapour in the atmosphere around a liquid will reduce the kinetic energy of the liquid molecules, thereby reducing the rate of their vaporization. Hence, the higher the humidity, the lower the rate of evaporation and vice versa.

c) The windiness of the environment also affects the rate of evaporation. Evaporation is faster on a very windy day, than a less windy one. This is because as long as the space above the liquid surface is not saturated by vapour, there will always be room for other molecules of the liquid to break free and exist as gaseous molecules. The space above the liquid surface can only remain unsaturated, if there is sufficient wind to blow away the existing vapour molecules.

d) Evaporation is also affected by surface area. The larger the surface area of the material, the higher the rate of evaporation will be. This is because with a larger surface area, more particles of liquid are exposed to vaporization at the surface.

The above factors explain why the oceans cannot boil, why clothes dry faster on a dry and windy day and why a completely spread towel dries faster than a folded one.

Boiling: This is an advanced form of evaporation, except that it occurs at a fixed temperature and throughout the liquid. As the molecules rise to the surface of the liquid and break free as gaseous particles, some of the vapour molecules still return into the liquid. This continues as the temperature increases until an equilibrium is reached, when the number of molecules leaving the liquid as gaseous molecules equals the number of gaseous particles returning into the liquid. At this point, the space above the liquid surface is said to be saturated with vapour. The saturated vapour starts exerting pressure on the liquid surface, and this is known as the saturated vapour pressure (s.v.p) of the liquid. It varies with temperature.

The saturated vapour pressure of the liquid increases as the temperature increases, until a temperature is reached when its s.v.p equals the atmospheric pressure. At that point, an air bubble rises from the bottom of the liquid to its surface, and this marks the beginning of the boiling process. No further temperature rise will be detected by the thermometer until all the intermolecular forces of attraction are broken. Just like evaporation, the latent heat of vaporization is absorbed during boiling, but unlike it, the whole liquid is seen to be bubbling. Also, the temperature at which this occurs is the boiling point of the liquid. In other words, the boiling point of a liquid is the temperature at which its saturated vapour pressure equals the atmospheric pressure.

Condensation
This is the opposite of vaporization. It is the change in state of a gas to liquid, and involves the evolution of the latent heat of vaporization. It is also known as liquefaction. The loss in heat is accompanied by a decrease in temperature, which also leads to a decrease in the average kinetic energy of the gas molecules, and a subsequent decrease in their randomness. As they get closer to one another, the effect of the intermolecular forces of attraction begin to increase, until a bond is formed between two adjacent gaseous particles. At this point, the thermometer stops recording any further change in temperature until all the intermolecular bonds (forces of attraction) are formed. These bonds restrict the excessively random nature of the particles, and as a a result, they become liquid. The constant temperature at which this change occurs is called the condensation point of the gas.

The Heating Curve

The above graph is the heating of curve of water. It is a plot of the changes water undergoes when heated from the solid state (ice) to the gaseous state (steam). Between the solid and liquid phases is the melting point, where you have the solid-liquid (ice-water) equilibrium. Similarly, at the boiliing point, the water exists in a liquid-gas (water-steam) equilibrium. As seen from the diagram, these two temperatures are constant over periods of time, and the temperature changes only occur after the melting and boiling processes had been completed.

The fixed melting and boiling points are typical properties of crystalline substances such as water. Non-crystalline or amorphous substances like wax, melt and boiling over a range of temperatures, and as such their heating curve will not be as sharp as that of crystalline substances.

The Kinetic Theory of Gases

This is the extension of the kinetic theory of matter aimed at describing the behaviour of an ideal gas. It is the basis for the study of the Gas Laws, and it states the following:

1. The gas molecules move randomly in a straight line colliding with themselves and with the walls of the containing vessel. The collisions of the gas molecules on the walls of the vessel account for the gas pressure exerted in the container.

2. The volume of each gas molecule is negligible compared to the volume of the containing vessel.

3. The forces of cohesion existing between the gas molecules is negligible. This explains the wide distances between the gas molecules, which makes it easy for a gas to be compressed. It also explains why a gas opened from a container rapidly spreads to fill a room.

4. The collisions of the gas molecules between themselves and the walls of the containing vessel is perfectly elastic. In other words, when two molecules collide, they rebounce like elastic balls because their total kinetic energies remain the same, as there is no loss of energy or conversion of kinetic energy to heat energy.

5. The temperature of the gas is a measure of the average kinetic energy of the molecules. This holds true not only for gases, but for all states of matter. An increase in the temperature of a body leads to a corresponding increase in its average kinetic energy.

Do These

Question 1
Using water as an example, distinguish between the three common states of water.

Question 2
Using a heating curve, differentiate with explanation, the following changes in state of matter - melting and freezing, boiling and liquefaction.

Question 3
a) Differentiate between boiling and evaporation.
b) What do you understand by the terms, vapour pressure and saturated vapour pressure?

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Gas Laws (Part I): Boyle's Law. Charles' Law. Dalton's Law of Partial Pressure

2016-07-23T11:45:00.002+01:00

Gas Laws

These are empirical laws that are used to explain the behaviour of ideal gases. An ideal gas is a perfect gas. It is a gas that does not exist, because there is no perfect situation in the real world. The Gas laws are grouped into two major categories - the physical and chemical laws. The former seek to explain the behaviour of gases based on their physical properties such as volume, temperature and pressure, while the latter do so by using their chemical properties like the number of molecules in the gases. We had looked at the chemical laws in our post - Gas Laws (Part II): Gay Lussac's & Avogadro's Laws, so this post will be devoted to the discussion on the physical laws.

Boyle's Law

This law states that the volume, V, of a given mass of gas is inversely proportional to its pressure, P, provided the temperature remains constant. This implies that if the volume of a gas is doubled, it will lead to a decrease in the gas pressure by half. Similarly, if the volume of the gas is decreased by half, the pressure exerted by its molecules on the walls of the containing vessel will increase. This is because the area of the containing vessel available for collision has been reduced, thereby increasing the force per unit area.

Mathematically, Boyle's Law can be expressed as:

V & 1/P (at constant T)
V = k/P ....................(i)
k = PV .....................(ii)

From equation (i), a plot of V against 1/P will give a straight line graph which passes through the origin. Click here to see all the graph variants of Boyle's Law.

According to equation (ii), the product of the volume and pressure of a gas at constant temperature will always give a constant. Hence, for a given mass of gas whose volume is V1 at pressure, P1, and V2 at pressure, P2, we can conclude that:

P1V1 = P2V2 ............................(iii)

Note that the volume can be in m^3, cm^3 or dm^3, while the pressure can be in mm Hg (millimetres mercury), atm (atmospheres) or Nm^-2 (Newton per metre). The most important thing is the consistency of units. The following conversion factors can be of help:

1 dm^3 = 1000 cm^3 = 10^-3 m^3
1 atm = 760 mm Hg = 101325 Nm^-2

Example 1
50 cm^3 of oxygen at 20 atmospheres pressure is placed in a 30 cm^3 container. Calculate the new pressure if temperature is kept constant.

Answer
This question involves changes in volume and pressure only at constant temperature; therefore, we will apply the Boyle's law:

P1V1 = P2V2,

where

P1 - initial pressure = 20 atm,
V1 - initial volume = 50 cm^3,
V2 - final volume = 30 cm^3,
P2 - final pressure = ?

Making P2 subject of the formula gives:

P2 = P1V1/V2
= 20 × 50/30
= 33.33 atm

Example 2
A given mass of a gas occupies a volume of 250 cm^3 at a pressure of 740 mm Hg. What will be its volume at atmospheric pressure?

Answer
The question also involves changes in pressure and volume only. So, using Boyle's law:

P1V1 = P2V2,

where,

P1 - initial pressure = 740 mm Hg,
V1 - initial volume = 250 cm^3,
P2 - final pressure = 760 mm Hg, (atmospheric pressure)
V2 - final volume = ?

V2 = P1V1/P2
= 740 × 250/760
= 243. 42 cm^3

Charles' Law

This law states that the volume, V, of a fixed mass of gas is directly proportional to its absolute temperature, T, provided the pressure remains constant. This implies that an increase in the temperature of a gas will lead to a corresponding increase in the volume of the gas, and vice versa.

According to the kinetic theory of gases, the temperature of a gas is a measure of the average kinetic energy of the gas molecules. That is to say, an increase in the temperature of a gas will increase the kinetic energies of its molecules, thereby making them to collide more often, and increasing their force of collision, F, with the walls of the containing vessel. In order for the pressure to remain constant, there must be an expansion in the surface area of the containing vessel, so that F/A = constant. This expansion is in the form of an increase in height of the surface area, which translates to an increase in volume. ( Given that volume = area × height).

Mathematically, Charles' Law can be expressed as:

V & T (at constant P)
V = kT .............................(iv)
k = V/T .............................(v)

From equation (v), the volume - temperature ratio of a gas at constant pressure gives a constant. Therefore, for a given mass of gas whose volume is V1 at temperature, T1, and V2 at temperature, T2, then:

V1/T1 = V2/T2 ............................(vi)

Absolute Temperature & Absolute Zero
Absolute temperature is the temperature value based on the Kelvin scale. Hence, every temperature involved in the Gas Laws calculations MUST always be converted to the Kelvin temperature by adding 273 to the Celsius temperature because 0°C = 273K.

K = °C + 273; similarly, °C = K - 273

On the other hand, absolute zero, i.e,0K or -273°C, is the temperature at which the volume of a gas is considered to be 0 cm^3. According to Charles' Law, at that temperature, the gas is expected to varnish, since its volume is directly proportional to the temperature as depicted in the graph of V against T(°C).

However, studies have shown that no existing gas varnishes at extremely low temperatures, rather all known gases liquefy at temperatures below 273K, because they are not ideal gases. Therefore, Charles' & Boyle's Laws are only true for ideal gases, but fail for real gases.

Example 3
If the volume of a given mass of gas at 0°C is 30.7 cm^3, what will be the volume of the gas at 20°C, pressure remaining constant?

Answer

This is a question that involves changes in volume and temperature at constant pressure. Hence, Charles' Law is applicable:

V1/T1 = V2/T2

where,

V1 - initial volume = 30.7 cm^3,
T1 - initial temperature =0°C + 273
= 273K,
T2 - final temperature = 20°C + 273
= 293K,
V2 - final volume = ?

V2 = V1T2/T1
= 30.7 × 293/273
= 32.95 cm^3

Example 4
A given quantity of gas occupies 5.0 dm^3 at 250K. At what temperature will its volume be doubled keeping the pressure constant?

Answer
Applying Charles' Law:

V1/T1 = V2/T2, where

V1 - initial volume = 5.0 dm^3,
T1 - initial temperature = 250K,
V2 - final volume = 2 × V1
= 2 × 5
= 10.0 dm^3
T2 - final temperature = ?

T2 = V2T1/V1
= 10 × 250/5
= 500K

General Gas Equation

The general gas equation is a combination of Boyle's and Charles' Laws as shown below:

V & T (Charles)
V & 1/P (Boyle)

Combining both using joint variation,

V & T/P
V = KT/P ....................................(vii)
PV/T = k .....................................(viii)

Analysing a given mass of gas, under two different sets of conditions based on equation (viii), we will obtain:

P1V1/T1 = P2V2/T2 ....................(ix)

Standard Temperature & Pressure (s.t.p.)
Since the volume of a gas varies with temperature, and the room temperature varies from one location or country to another due to climatic variations; there was a need by scientists all over the world, to establish standard conditions under which experimental studies involving gases can be carried out in order to obtain consistent results, irrespective of the location. These conditions were a temperature of 273K and a pressure of 1 atm/760 mm Hg/101325 Nm^-2 and are considered to be the standard temperature & pressure (s.t.p.).

Example 5
12.0 dm^3 of a gas at s.t.p was subjected by means of a movable piston to two times the original pressure with the temperature being now kept at 384K. What is the volume of the gas in dm^3 at this pressure?

Answer
Since the question involves changes in volume, temperature and pressure, then, we will make use of the general gas equation:

P1V1/T1 = P2V2/T2

where,

P1 - initial pressure = 760 mm Hg (standard pressure),
V1 - initial volume = 12.0 dm^3,
T1 - initial temperature = 273K (standard temperature),
P2 - final pressure = 2 × P1
= 2 × 760 = 1520 mm Hg,
T2 - final temperature = 384K,
V2 - final volume = ?

V2 = P1V1T2/P2T1
= (760 × 12 × 384)/(1520 × 273)
= 8.44 dm^3

Example 6
If a gas occupies a container of volume 165 cm^3 at 27°C and 0.984 atm, calculate its volume in cm^3 at s.t.p.

Answer
Applying the general gas equation:

P1V1/T1 = P2V2/T2

where,

P1 - initial pressure = 0.984 atm,
V1 - initial volume = 165 cm^3,
T1 - initial temperature = 27°C + 273
= 300K,
P2 - final pressure = 1 atm,
T2 - final temperature = 273K,
V2 - final volume = ?

V2 = P1V1T2/P2T1
= (0.984 × 165 × 273)/(1 × 300)
= 147.75 cm^3

Example 7
At 25°C and 1 atm, a gas occupies a volume of 1.50 dm^3. What volume will it occupy at 100°C and 5 atm?

Answer
Using the general gas equation:

P1V1/T1 = P2V2/T2

where,

P1 - initial pressure = 1 atm,
V1 - initial volume = 1.5 dm^3,
T1 - initial temperature = 25°C + 273
= 298K,
P2 - final pressure = 5 atm,
T2 - final temperature = 100°C + 273
= 373K,
V2 - final volume = ?

V2 = P1V1T2/P2T1
= (1 × 1.5 × 373)/(5 × 298)
= 0.376 dm^3

Ideal Gas Equation

Since the ratio PV/T equals to a constant, according to the general gas equation (equation (viii)), then the ideal gas equation can be given as:

PV = nRT ...................................(x)

where,

P = pressure in atm
V = volume in dm^3
T = temperature in K
n = number of moles
R = molar gas constant (0.082 atm dm^3 mol^-1 K^-1)

For 1 mole of a gas at s.t.p,
standard pressure = 1atm
standard temperature = 273K
volume occupied by 1 mole of a gas = 22.4 dm^3

Therefore,

R = PV/nT
    = (1 × 22.4)/(1 × 273)
    = 22.4/273
    = 0.082 atm dm^3 mol^-1 K^-1

Alternatively, if standard pressure = 101325 Nm^-2, and volume = 0.0224 m^3

then,

R = (101325 × 22.4)/(1 × 273)
= 2269.98/273
= 8.314 J mol^-1 K^-1

If we divide both sides of equation (x) with V, we will obtain:

PV/V = nRT/V
P = (n/V) RT ........................(xi)

Recall, concentration in mol dm^-3, C = n/V ......(xii)

Substituting equation (xii) into (xi) gives us:

P = CRT ......................................(xiii)

Equation (x) gives the ideal gas equation in terms of pressure, volume, temperature and no. of moles, while equation (xiii) gives the equation in terms of pressure, concentration and temperature.

The ideal gas equation holds true for all gases because, according to the Avogadro's Law, equal volume of gases contain equal number of molecules at the same temperature and pressure.

Example 8
5.0 moles of an ideal gas are at temperature of –33°C and a pressure of 350 torrs. What volume in dm^3 will the gas occupy at that temperature? (R = 0.082 atm dm^3 mol^-1 K^-1, 1 torr = 1 atm)

Answer
Since the question involves moles, pressure, temperature and volume, then, the right equation to apply is the ideal gas equation.

PV = nRT

where,

P - pressure = 350 atm
T - temperature = –33°C + 273 = 240K
n - number of moles = 5.0 moles
R - molar gas constant = 0.082 atm dm^3 mol^-1 K^-1
V - volume = ?

V = nRT/P
= 5 × 0.082 × 240/350
= 0.28 dm^3

Example 9
Calculate the number of moles of gas, which occupies 500 dm^3 at temperature of 77°C and pressure of 2.52 × 10^5 Nm^-2.(R = 8.314 J mol^-1 K^-1)

Answer
Using the ideal gas equation:

PV = nRT

where,

P - pressure = 2.52 × 10^5 Nm^-2
T - temperature = 77°C + 273 = 350K
R - molar gas constant = 8.314 J mol^-1 K^-1
V - volume = 500 dm^3 = 500 × 10^-3 m^3 (since pressure is in Nm^-2, volume must be in m^3)
n - number of moles = ?

n = PV/RT
    = 2.52 × 10^5 × 500 × 10^-3/8.314 × 350
    = 252000 × 0.5/8.314 × 350
    = 43.30 moles

Dalton's Law of Partial Pressure

This law states that in a mixture of gases that are not chemically combined, the total pressure of the gas mixture is the sum of the partial pressure of the individual gases that make up the mixture. If Ptotal is the total pressure of a mixture of gases A, B, C...n, and Pa, Pb, Pc....Pn are the partial pressures due to gases A, B, C....n, then,

Ptotal = Pa + Pb + Pc + .... + Pn ........(xiv)

but,

Pa = na/ntotal × Ptotal
Pb = nb/ntotal × Ptotal
Pc = nc/ntotal × Ptotal

where,

na, nb, and nc = number of moles (molecules) of gases A, B and C in the mixture; and na/ntotal, nb/ntotal, nc/ntotal are the mole fractions of gases A, B and C respectively.

ntotal = total number of moles (molecules) of gases in the mixture (na + nb + nc....)

For instance, the total pressure exerted by air (made up of nitrogen, oxygen, carbon (IV) oxide, noble gases, water vapour and others) in a container, is the sum of partial pressure exerted by each of the gases listed above, assuming each of them was to act alone in the containing vessel.

Usually, most gases prepared in the laboratory are collected over water. If the pressure of the collected gas is measured at a given temperature, it would be observed that it contains the saturated vapour pressure (s.v.p) of water at that temperature, and as such, the measured pressure is not the true pressure of the dry gas. Therefore, to get the actual pressure of the dry gas (partial pressure), the s.v.p of water must be subtracted from the pressure of the wet gas (total pressure); i.e,

Pressure of Dry Gas = Pressure of Wet Gas - Saturated Vapour Pressure of H2O .............(xv)

Example 10
What is the partial pressure of hydrogen gas collected over water at standard atmospheric pressure and 35°C if the saturated vapour pressure of water is 30 mm Hg at that temperature?

Answer
Partial Pressure of H2 gas = Total (Atmospheric) Pressure, Ptotal - S.V.P of H20 at 35°C

partial pressure of H2 = 760 - 30
= 730 mm Hg

Example 11
A gaseous mixture of 40.0g of oxygen and 28.0g of nitrogen has a total pressure of 1.8 atm. Calculate the partial pressure of oxygen in the mixture (O = 16, N = 14)

Answer
Mass of O2 in gaseous mixture = 40g
Moles of O2 = Mass of O2/Molecular mass of O2
= 40/(16 × 2) = 40/32
= 1.25 moles

Similarly,

Mass of N2 in gaseous mixture = 28g
Moles of N2 = Mass of N2/Molecular mass of N2
= 28/(14 × 2) = 28/28
= 1.00 moles

Total number of moles of gaseous mixture = no. of moles of O2 + no. of moles of N2
= 1.25 + 1.00 = 2.25 moles

Total pressure of mixture = 1.8 atm

Therefore,

Partial pressure of O2 in mixture = mole fraction of O2 × total pressure
= (1.25/2.25) × 1.8
= 1.0 atm

Example 12
50 cm^3 of hydrogen was collected over water at 25°C and 790 mm Hg. If the vapour pressure of water at the temperature of the experiment was 15 mm Hg, calculate the volume of the gas at 750 mm Hg and 15°C.

Answer
This question is a combination of the general gas equation and Dalton's Law of Partial Pressure. The first thing to do is to determine the partial pressure of the hydrogen gas collected over water, and then use the result as the initial pressure in the general gas equation.

Given:
Total pressure = 790 mm Hg
Vapour pressure of water = 15 mm Hg
Initial volume, V1 = 50 cm^3
Initial temperature, T1 = 25°C + 273 = 298K
Final pressure, P2 = 750 mm Hg
Final temperature, T2 = 15°C + 273 = 288K
Final volume, V2 = ?

but,

Initial pressure, P1 = partial pressure of hydrogen
= total pressure - vapour pressure of water
= 790 - 15 = 775 mm Hg

Applying the general gas equation and substituting the values of the parameters into the equation gives:

P1V1/T1 = P2V2/T2
V2 = P1V1T2/P2T1
= (775 × 50 × 288)/(750 × 298)
= 49.93 cm^3

Do These

Question 1
To what temperature must a gas at 273K be heated in order to double both its volume and pressure?

Question 2
The ratio of the initial to the final pressure of a given mass of gas is 1:1.5. Calculate the final volume of the gas if the initial volume was 450 cm^3 at the same temperature.

Question 3
The pressure of 100 cm^3 of oxygen at 35°C is 750 mm Hg. What will be the volume of the gas if the pressure is reduced to 100 mm Hg at 25°C?

Question 4
A mixture of 0.40 mole of argon, 0.30 mole of helium and 0.50 mole of nitrogen exerts a total pressure of 3.5 atm. What is the partial pressure of nitrogen in the mixture?

Question 5
The partial pressure of oxygen in a sample of air is 350 mm Hg and the total pressure is 785 mm Hg. What is the mole fraction of oxygen?

Question 6
50cm^3 of a gas at 45°C is warmed to 85°C at a fixed pressure, what is the fractional increase in volume of the gas?

Question 7
75 cm^3 of a gas was collected over water at 20°C and 785 mm Hg. Calculate the volume of the gas at s.t.p, if the saturated vapour pressure of water at 20°C is 25 mm Hg.

Question 8
300 cm^3 of a gas at 25°C exerts a pressure of 750 mm Hg. Calculate its pressure if its volume increases to 450 cm^3 at 75°C.

Question 9
Find the volume occupied by 16.0g of oxygen under a pressure of 3.5 × 10^5 Nm^-2 at a temperature of 27°C. (Molar gas constant = 8.314 J mol^-1 K^-1, O = 16)

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Hydrolysis of Salts

2016-06-09T19:02:00.000+01:00

pH Chart

In our last post: Types, Preparation & Uses of Salts, we studied that salts are the products of the neutralization reactions between acids and bases. Following this fact, all salts are expected to have a neutral pH of 7 in solution. Interestingly, some salts produce solutions with pH > 7 (basic solutions), while others produce acidic solutions with pH < 7. Such salts, which are the products of either weak acids and strong bases or strong acids and weak bases, decompose in water to form two products. This phenomenon is known as hydrolysis.

Recall that from our post on pH in Acids & Bases (Part III), we learnt that water contains equal concentration of hydrogen and hydroxide ions. Therefore, the dissolution of a salt in water is likely to upset this ionic equilibrium based on the salt's composition. This is the underlying principle of hydrolysis of salts. So, in this post, we shall be looking at the hydrolysis of different types of salts based on their formation. To fully understand this topic, you may also need to read our posts on Acids & Bases (Part I) and Acids & Bases (Part II).

Hydrolysis of Salts of Weak Acids & Strong Bases

An example of a salt that falls into this category is sodium ethanoate, CH3COONa, formed from ethanoic acid and sodium hydroxide. In solution, sodium ethanoate ionizes completely to produce ethanoate and sodium ions.

CH3COONa(aq) ----------> CH3COO–(aq) + Na+(aq)

Being a weak acid/base, the water molecules undergo partial dissociation as discussed in Acids & Bases (Part I ).

H2O(l) <------------____> H+(aq) + OH–(aq)

Note that the reversible arrows points longer to the undissociated molecules, H2O, showing that they are the favoured species in the dissociation process.

The four ions in solution undergo interaction with one another, and being the conjugate base of a weak acid, the ethanoate ions from the salt, combine with the hydrogen ions from the water to form undissociated ethanoic acid molecules.

CH3COO–(aq) + H+(aq) <-----_______> CH3COOH(aq)

According to Le Chatelier's Principle, this will distort the chemical equilibrium of the system. Consequently, in order to replace the lost hydrogen ions, the undissociated water molecules must undergo further dissociation to produce hydrogen and hydroxide ions. This process continues until all the hydrogen and ethanoate ions are removed from the solution, leaving only sodium and hydroxide ions, which are products of the ionization of sodium hydroxide - a strong base.

NaOH(aq) ---------> Na+(aq) + OH-(aq)

These ions remain in solution, and account for the pH of the resulting solution of sodium ethanoate being basic (pH > 7). Other examples of salts that fall into this category (solution pH > 7), and their constituent acids and bases are:

1. Sodium trioxocarbonate (IV), Na2CO3 - a product of trioxocarbonate (IV) acid, H2CO3, and sodium hydroxide, NaOH.

2. Potassium ethanoate, CH3COOK - a product of ethanoic acid, CH3COOH, and potassium hydroxide, KOH.

3. Sodium trioxosulphate (IV), Na2SO3 - a product of trioxosulphate (IV) acid, H2SO3, and sodium hydroxide, NaOH.

In general, salts of weak acids and strong bases undergo hydrolysis to give basic or alkaline solutions.

Hydrolysis of Salts of Strong Acids & Weak Bases

An example of a salt of a strong acid and a weak base is ammonium chloride, NH4Cl, which is a product of aqueous ammonia and hydrochloric acid and ionizes in solution to produce ammonium and chloride ions.

NH4Cl(aq) ----------> NH4+(aq) + Cl-(aq)

Accordingly, there are also hydrogen and hydroxide ions from the partial dissociation of water.

H2O(l) <------------____> H+(aq) + OH–(aq)

The hydroxide ions combine with the ammonium ions to form ammonium hydroxide molecules. Being a weak base, the formation of the molecules is favoured over the formation of dissociated ions.

NH4+(aq) + OH–(aq) <-----_______> NH3.H2O(aq)

Expectedly, this action destabilizes the equilibrium of the system, and causes more water molecules to dissociate to replace the hydroxide ions, which had been removed as aqueous ammonia molecules. As this dissociation occurs, more hydrogen ions are also introduced into the system. This process occurs, until all the hydroxide and ammonium ions are eventually removed from the solution, leaving behind hydrogen and chloride ions. Being products of a strong acid, they remain in solution as ions, and this accounts for the acidic nature of the of the ammonium chloride solution (pH < 7).

HCl(aq) -------------> H+(aq) + Cl-(aq)

Other salts, whose solutions will produce pH < 7, and their constituent acids and bases include:

1. Aluminium chloride, AlCl3 - a product of aluminium hydroxide, Al(OH)3 and hydrochloric acid, HCl.

2. Copper (II) tetraoxosulphate (VI), CuSO4 - a product of copper hydroxide and tetraoxosulphate (VI) acid, H2SO4.

3. Zinc trioxonitrate (V), Zn(NO3)2 - a product of zinc hydroxide, Zn(OH)2 and trioxonitrate (V) acid, HNO3.

Generally, salts of strong acids and weak bases undergo hydrolysis to produce acidic solutions.

Hydrolysis of Salts of Strong Acids & Strong Bases

A typical example of a salt of this category is sodium chloride, NaCl. It ionizes in water to produce sodium and chloride ions.

NaCl(aq) ---------------> Na+(aq) + Cl–(aq)

Ordinarily, one would expect the hydroxide ions from the water molecules to be removed as sodium hydroxide, and the hydrogen ions as hydrochloric acid. However, being a product of a strong acid and a strong base, the four ions remain in solution due to the complete ionization of acid and the base.

HCl(aq) ----------------> H+(aq) + Cl–(aq)

NaOH(aq) ---------------> Na+(aq) + OH–(aq)

Consequently, the ionic equilibrium of water is not disturbed as there is no hydrolysis. Therefore, giving a neutral solution with a pH = 7.Other salts with solutions of neutral pH are:

1. Potassium chloride, KCl - a product of potassium hydroxide, KOH, and hydrochloric acid, HCl.

2. Sodium tetraoxosulphate (VI), Na2SO4 - a product of sodium hydroxide, NaOH, and tetraoxosulphate (VI) acid, H2SO4.

3. Potassium trioxonitrate (V), KNO3 - a product of potassium hydroxide, KOH, and trioxonitrate (V) acid, HNO3.

Generally, salts of strong acids and strong bases do not undergo hydrolysis, and are the only category of salts that exhibit this property.

Note that if any sodium salt exhibits a particular property, the chances that the corresponding potassium salt will behave similarly are very high. E.g, NaCl and KCl, Na2SO4 and K2SO4, Na2CO3 and K2CO3 etc.

Hydrolysis of Weak Acids & Weak Bases

An example of a salt formed from a weak acid and a weak base is ammonium ethanoate, CH3COONH4 - a product of aqueous ammonia, NH3.H2O, and ethanoic acid, CH3COOH. It ionizes in solution to form ammonium and ethanoate ions.

CH3COONH4(aq) ----------> CH3COO–(aq) + NH4+(aq)

Expectedly, the water molecules in the solution undergo partial dissociation to give hydrogen and hydroxide ions.

H2O(l) <------------____> H+(aq) + OH–(aq)

On reaction, the ammonium and hydroxide ions combine to produce aqueous ammonia molecules, while the ethanoate ions combine with the hydrogen ions to form ethanoic acid molecules.

NH4+(aq) + OH–(aq) <----_______> NH3.H2O(aq)

CH3COO–(aq) + H+(aq) <-----_______> CH3COOH(aq)

The formation of these molecules lead to the removal of the hydrogen and hydroxide ions, and creates an imbalance in the ionic equilibrium of water. In response to this, more water molecules undergo dissociation to replace the lost ions. This process continues until all the ammonium and ethanoate ions are converted to undissociated molecules. Although the salt is strongly hydrolysed in solution, the pH of the solution remains almost neutral (if not neutral).

Other examples of salts of weak acids and weak bases, and their constituent acids and bases are:

1. Ammonium trioxocarbonate (IV), (NH4)2CO3 - a product of aqueous ammonia, NH3.H2O, and trioxocarbonate (IV) acid, H2CO3.

2. Magnesium trioxosulphate (IV), MgSO3 - a product of magnesium hydroxide, Mg(OH)2, and trioxosulphate (IV) acid, H2SO3.

3. Calcium tetraoxophosphate (V), Ca3(PO4)2 - a product of calcium hydroxide, Ca(OH)2, and tetraoxophosphate (V) acid, H3PO4.

Do These

Question 1
Classify these salts according to the strength of the acids and bases they are formed from. Briefly discuss whether hydrolysis occurs in each case:

a) Potassium ethanoate
b) Ammonium trioxocarbonate (IV)
c) Potassium chloride
d) Ammonium tetraoxosulphate (VI)

Question 2
Consider the following salts:
KNO3, CH3COONa, NH4Cl
Which one of them would form:
a) an acidic solution
b) a basic solution
c) a neutral solution

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Salts: Types, Preparation & Uses

2016-05-16T10:38:00.001+01:00

Introduction

Before they are introduced to acids and bases, young chemistry students always think that sodium chloride (common salt) is everything there is to know about salts. However, from their knowledge of acids and bases, they also get to know about other substances, such as copper (II) tetraoxosulphate (VI), potassium trioxocarbonate (IV), ammonium chloride, calcium trioxonitrate (V) etc, which are classified as salts. These substances are the outcomes of the Arrhenius acid-base reactions. So, what is a salt?

Definitions

We will define a salt in terms of basicity (replaceable hydrogen ions) and neutralization.

I) A salt is a substance formed when all or part of the replaceable hydrogen ions in an acid, are replaced by metallic ions (Na+, K+, Mg2+, Ca2+, Cu2+ etc) or ammonium ions (NH4+). This implies that every acid has its corresponding salts. The list below shows examples of some salts and their parent acids.

1. Acid: Hydrochloric acid (HCl)
Salts: Sodium chloride (NaCl), Potassium chloride (KCl), Calcium chloride (CaCl2), Copper (II) chloride (CuCl2) etc

2. Acid: Trioxonitrate (V) acid (HNO3)
Salts: Magnesium trioxonitrate (V) (Mg(NO3)2), Lead trioxonitrate (V) (Pb(NO3)2), sodium trioxonitrate (V) (NaNO3), Silver trioxonitrate (V) etc

3. Acid: Tetraoxosulphate (VI) acid (H2SO4)
Salts: Potassium tetraoxosulphate (VI) (K2SO4), Aluminium tetraoxosulphate (VI) (Al2(SO4)3), Ammonium tetraoxosulphate (VI) ((NH4)2SO4), Copper (II) tetraoxosulphate (VI) (CuSO4) etc

4. Acid: Trioxocarbonate (IV) acid (H2CO3)
Salts: Sodium trioxocarbonate (IV) (Na2CO3), Calcium hydrogen trioxocarbonate (IV), (Ca(HCO3)2), Zinc trioxocarbonate (IV), (ZnCO3), Potassium trioxocarbonate (IV) (K2CO3) etc

II) A salt is a product of the neutralization reaction between an acid and a base.

Examples

a) Hydrochloric acid and sodium hydroxide react to form sodium chloride and water.

HCl(aq) + NaOH(aq) -------> NaCl(aq) + H2O(l)

b) Calcium hydroxide neutralizes tetraoxosulphate (VI) acid to produce calcium tetraoxosulphate (VI) and water.

Ca(OH)2(aq) + H2SO4 (aq) -------> CaSO4(aq) + 2H2O(l)

From the above reactions, we can see that the salts are formed from the exchange of the metallic and nonmetallic radicals of the reactants. In other words, the acids donate the nonmetallic radicals of the salts, while the bases donate the metallic radical components.

Types of Salts

There are five major types of salts, namely: normal, acid, basic, double and complex salts. Let's take a look at each category in details.

Normal Salts: These are salts formed as the products of the complete neutralization of acids by bases. The replaceable hydrogen ions in the acids are completely replaced by metallic or ammonium ions. Examples include sodium chloride, potassium trioxocarbonate (IV), calcium tetraoxosulphate (VI), magnesium trioxonitrate (V), ammonium chloride etc.

Acid Salts: These are salts in which the replaceable hydrogen ions in the acid are partially replaced by metallic or ammonium ions. They are formed when there is an insufficient supply of the base to completely neutralize the acid.

Examples

a) Tetraoxosulphate (VI) acid undergoes incomplete neutralization with potassium hydroxide to form potassium hydrogen tetraoxosulphate (VI) salt.

H2SO4(aq) + KOH(aq) -------> KHSO4(aq) + H2O(l)

On complete neutralization, the product will be potassium tetraoxosulphate (VI) salt

H2SO4(aq) + 2KOH(aq) -------> K2SO4(aq) + 2H2O(l)

From the above equations, it can be observed that for the tetraoxosulphate (VI) acid to be completely neutralized, we will need two moles of potassium hydroxide. Where that is not available as in the first equation, then the reaction will be incomplete and the product formed will be an acid salt. The normal and acid salts ionize in water to produce the tetraoxosulphate (VI) ions and hydrogen tetraoxosulphate (VI) ions respectively as the negative radicals as shown below:

K2SO4(aq) -------> 2K+(aq) + SO4––(aq)

KHSO4(aq) -------> K+(aq) + HSO4–(aq)

b) Sodium hydroxide undergoes incomplete neutralization with trioxocarbonate (IV) acid to form sodium hydrogen trioxocarbonate (IV) salt.

H2CO3(aq) + NaOH(aq) -------> NaHCO3(aq) + H2O(l)

Whereas, complete neutralization produces sodium trioxocarbonate (IV) salt.

H2CO3(aq) + 2NaOH(aq) -------> Na2CO3(aq) + 2H2O(l)

The normal and acid salts ionize in water to produce the trioxocarbonate (IV) ions and the hydrogen trioxocarbonate (IV) ions respectively:

Na2CO3(aq) -------> 2Na+(aq) + CO3––(aq)

NaHCO3(aq) -------> Na+(aq) + HCO3–(aq)

c) Similarly, calcium hydroxide can undergo incomplete neutralization with tetraoxophosphate (V) acid to form more than one acid salts, namely calcium dihydrogen tetraoxophosphate (V) (Ca(H2PO4)2), and calcium hydrogen tetraoxophosphate (V) (CaHPO4) because it is tribasic.

2H3PO4(aq) + Ca(OH)2(aq) -------> Ca(H2PO4)2(aq) + 2H2O(l)

H3PO4(aq) + Ca(OH)2(aq) -------> CaHPO4(aq) + 2H2O(l)

However, on complete neutralization, the reaction gives a normal salt, calcium tetraoxophosphate (V) salt.

2H3PO4(aq) + 3Ca(OH)2(aq) -------> Ca3(PO4)2(aq) + 6H2O(l)

The two acid salts dissociate in water to produce dihydrogen tetraoxophosphate (V) ions and hydrogen tetraoxophosphate (V) ions respectively.

Ca(H2PO4)2(aq) -------> Ca2+(aq) + 2H2PO4–(aq)

CaHPO4(aq) -------> Ca2+(aq) + HPO4––(aq)

It is important to note that only dibasic and tribasic acids form acid salts because they have more than one replaceable hydrogen ions. This implies that they have the ability to form both normal and acid salts as illustrated in the examples above. Also, the easiest way to theoretically identify acid salts is to look out for the replaceable hydrogen ion(s) in their molecules.

Basic Salts: Just like the acid salt, a basic salt is produced when there is an insufficient supply of an acid to completely neutralize the base. For a basic salt to be formed, the base must possess at least two hydroxide ions in its molecule.

Examples

a) When an insufficient supply of hydrochloric acid reacts with calcium hydroxide, it forms calcium hydroxide chloride salt.

Ca(OH)2(aq) + HCl(aq) -------> Ca(OH)Cl(aq) + H2O(l)

On complete neutralization, calcium chloride is formed.

Ca(OH)2(aq) + 2HCl(aq) -------> CaCl2(aq) + 2H2O(l)

b) Similarly, trioxonitrate (V) acid can undergo incomplete neutralization with magnesium hydroxide to form magnesium hydroxide trioxonitrate (V) salt.

Mg(OH)2(s) + HNO3(aq) -------> Mg(OH)NO3(s) + H2O(l)

While the complete neutralization reaction produces magnesium trioxonitrate (V)

Mg(OH)2(s) + 2HNO3(aq) -------> Mg(NO3)2(aq) + 2H2O(l)

c) Tetraoxosulphate (VI) acid can undergo incomplete neutralization with aluminium hydroxide to form aluminium hydroxide tetraoxosulphate (VI) salt.

Al(OH)3(s) + H2SO4(aq) -------> Al(OH)SO4(s) + 2H2O(l)

The complete neutralization reaction will produce aluminium tetraoxosulphate (VI) salt.

2Al(OH)3(s) + 3H2SO4(aq) -------> Al2(SO4)3(s) + 6H2O(l)

Double Salts: A salt is said to be a double salt when it ionizes in water to produce three different ions - two positively charged metallic or ammonium ions and one negatively charged nonmetallic ion. They are the products of the crystallization of a mixture of two identical salts in solution, and possess the general formula M+M3+X2-.xH2O because they are hydrated salts.

Examples

a) Aluminium potassium tetraoxosulphate (VI)-dodecahydrate (KAl(SO4)2.12H2O) also known as potash alum, formed from the crystallization of a mixture of equimolar solutions (solutions of equal concentration) of potassium tetraoxosulphate (VI) and aluminium tetraoxosulphate (VI). The ions present in potash alum are K+, Al3+ and SO4––.

b) Chromium (III) potassium tetraoxosulphate (VI)-dodecahydrate (KCr(SO4)2.12H2O), otherwise called chrome alum, is a product of the crystallization of a mixture of potassium tetraoxosulphate (VI) and chromium (III) tetraoxosulphate (VI) (Cr2(SO4)3). The three ions present in chrome alum are K+, Cr3+ and SO4––.

c) Ammonium iron (II) tetraoxosulphate (VI)-hexahydrate ((NH4)2Fe(SO4)2.6H2O), which is a product of the crystallization of a mixture of equimolar solutions of ammonium tetraoxosulphate (VI) and iron (II) tetraoxosulphate (VI). The ions present in ferrous alum are NH4+, Fe2+ and SO4––

The properties of double salts are the aggregate of those of the two salts they contain.

Complex Salts: These are salts that contain complex ions. Complex ions are those that contain positively charged central ions coordinately bonded to other ions or neutral molecules known as ligands. Examples of complex ions are:

a) tetrahydroxozincate (II) ion [Zn(OH)4]2-, found in sodium tetrahydroxozincate (II) salt (Na2[Zn(OH)4])

Na2[Zn(OH)4](aq) -------> 2Na+(aq) + [Zn(OH)4]2-(aq)

b) hexacyanoferrate (III) ion [Fe(CN)6]3-, present in potassium hexacyanoferrate (III) salt (K3[Fe(CN)6])

K3[Fe(CN)6](aq) -------> 3K+(aq) + [Fe(CN)6]3-(aq)

c) tetrahydroxoaluminate (III) ion [Al(OH)4]-, found in sodium tetrahydroxoaluminate (III) salt (Na[Al(OH)4])

Na[Al(OH)4](aq) -------> Na+(aq) + [Al(OH)4]-(aq)

General Methods of Salts Preparation

The method employed in the preparation of a particular salt in the laboratory depends on whether it is a soluble or an insoluble salt, because of the need to recover it from its solution or the reaction mixture.

I. Preparation of Soluble Salts: The following are the three ways soluble salts can be prepared. These methods are simply the chemical properties of acids, which we studied in Acids & Bases (Part I).

a) Action of Dilute Acids on Active Metals:

acid + active metal --------> salt + hydrogen gas

Examples

Reaction between hydrochloric acid and magnesium ribbons to produce magnesium chloride and hydrogen gas.

2HCl(aq) + Mg(s) -------> MgCl2(aq) + H2(g)

Action of tetraoxosulphate (VI) acid on zinc granules

H2SO4(aq) + Zn(s) -------> ZnSO4(aq) + H2(g)

b) Action of Acids on Bases: Acids undergo neutralization reaction with alkalis and insoluble bases to form salts and water.

acid + base -------> salt + water

Examples

Dilute trioxonitrate (V) reacts with potassium hydroxide to form potassium trioxonitrate (V) salt and water.

HNO3(aq) + KOH(aq) -------> KNO3(aq) + H2O(l)

Sodium ethanoate and water are produced when ethanoic acid is neutralized by sodium hydroxide.

CH3COOH(aq) + NaOH(aq) -------> CH3COONa(aq) + H2O(l)

Hydrochloric acid reacts with aqueous ammonia to form ammonium chloride and water.

HCl(aq) + NH4OH(aq) -------> NH4Cl(aq) + H2O(l)

When preparing salts, the reaction of acids with insoluble bases is quite different from that with alkalis. The insoluble base is always added to the acid in a beaker and heated with stirring until the acid reacts completely with it. At that point, the base will no longer dissolve in the acid, and the undissolved portion will be filtered off.

Examples

Heating copper (II) oxide in tetraoxosulphate (VI) acid to produce copper (II) tetraoxosulphate (VI) salt and water.

H2SO4(aq) + CuO(s) -------> CuSO4(aq) + H2O(l)

Heating zinc oxide in hydrochloric acid to form zinc chloride and water.

HCl(aq) + ZnO(s) -------> ZnCl2(aq) + H2O(l)

c) Action of Acids on Trioxocarbonate (IV) Salts: Acids react with trioxocarbonate (IV) salts, to form salts (other than trioxocarbonate (IV)) and water.

acid + trioxocarbonate (IV) salt -------> salt + water + carbon (IV) oxide

Examples

Dilute trioxonitrate (V) acid reacts with copper (II) trioxocarbonate (IV) to form copper (II) trioxonitrate (V) salt.

HNO3(aq) + CuCO3(s) -------> Cu(NO3)2(aq) + H2O(l) + CO2(g)

Tetraoxosulphate (VI) acid reacts with calcium trioxocarbonate (IV) to produce calcium tetraoxosulphate (VI) salt.

H2SO4(aq) + CaCO3(s) -------> CaSO4(aq) + H2O(l) + CO2(g)

Recovery of Soluble Salts from Solutions: Depending on their stabilty to heat, soluble salts are recovered from their solutions by evaporation to dryness (e.g, NaCl, K2CO3) or crystallization process (e.g, CuSO4)

II. Preparation of Insoluble Salts: Insoluble salts are salts that do not dissolve in water, whether at room temperature or when heated. They can be prepared in the laboratory by the following methods:

a) Double Decomposition: This is the process of using two soluble substances to produce a soluble and a volatile or gaseous or an insoluble substance. In the case of salts, it involves the use of two soluble salts to produce soluble and insoluble salts. The insoluble product is separated or recovered from the mixture by filtration.

The choice of the reagents used depends on the insoluble salt you want to prepare. They are selected in a way that each of them will provide one of the radicals needed in the insoluble product. A typical example is the preparation of silver chloride from sodium chloride and silver trioxonitrate (V) salts. The sodium chloride provides the chloride ions, while the silver trioxonitrate (V) provide the silver ions. These two, then, combine to form the silver chloride.

NaCl(aq) + AgNO3(aq) -------> NaNO3(aq) + AgCl(s)

Na+(aq).Cl–(aq) + Ag+(aq).NO3–(aq) ------> Na+(aq).NO3–(aq) + AgCl(s)

Cancel out terms that appear on both sides of the equation

Cl–(aq) + Ag+(aq) -------> AgCl(s)

The silver chloride precipitate is filtered off and dried.

b) Direct Combination of Constituent Elements: Some binary salts are produced by directly combining their constituent elements under appropriate conditions. Example is the combination of metallic iron and sulphur to form iron (II) sulphide.

Fe(s) + S(s) -------> FeS(s)

The following is a list of some soluble and insoluble salts:

I. Soluble Salts

a) All salts of sodium, potassium and ammonium.

b) All trioxonitrate (V) salts.

c) All chloride salts except silver chloride and lead (II) chloride, PbCl2. Lead (II) chloride is however soluble when warmed.

d) All tetraoxosulphate (VI) salts except barium tetraoxosulphate (VI), BaSO4, lead (II) tetraoxosulphate (VI), PbSO4. Calcium tetraoxosulphate (VI) is sparingly soluble.

II. Insoluble Salts

a) All trioxocarbonate (IV) salts, except those of sodium, potassium and ammonium.

b) Silver chloride, barium tetraoxosulphate (VI), lead (II) chloride, and lead (II) tetraoxosulphate (VI).

Uses of Salts

Salts have wide applications in different areas of life such as medicine, agriculture, building & construction, chemical industries etc. Some of them are:

1. Magnesium tetraoxosulphate (VI), also known as Epsom salt, MgSO4.7H2O, is used as laxative.

2. Calcium trioxocarbonate (IV), CaCO3, is used in the manufacture of cement.

3. Calcium tetraoxosulphate (VI)-dihydrate, otherwise called gypsum, CaSO4.2H2O, is used for making Plaster of Paris (POP).

4. Aluminium potassium tetraoxosulphate (VI)-dodecahydrate, called potash alum, KAl(SO4)2.12H2O, is used as a coagulant in water treatment, and in dyeing as a mordant.

5. Sodium chloride, NaCl, is used for cooking, food preservation and glazing of pottery.

6. Potassium trioxonitrate (V), KNO3, is used for making explosives and fertilizers.

Interaction of Substances with Air & Moisture (Efflorescence, Deliquescence & Hygroscopy)

Different substances behave differently when they come in contact with air and moisture.

Efflorescence: This is a phenomenon whereby substances, which possess water of crystallization, lose part of or all their molecules of water of crystallization when exposed to the atmosphere to become anhydrous. It is mainly hydrated salts that undergo efflorescence. Examples are washing soda, Na2CO3.10H2O, blue vitriol, CuSO4.5H2O. These substances are known as efflorescent substances.

Deliquescence: Deliquescent substances are those substances that absorb moisture from the atmosphere and dissolve to become solutions or dilute, if liquid. This phenomenon is known as deliquescence. Examples are sodium hydroxide, potassium hydroxide, calcium chloride, concentrated tetraoxosulphate (VI) acid. They are mostly used as drying agents in the laboratory.

Hygroscopy: This is a phenomenon in which substances absorb moisture from the atmosphere and become sticky, when exposed to air. Such substances as known as hygroscopic substances. Examples include copper (II) oxide, calcium oxide, iron (II) chloride, silica gel etc. They are also used as drying agents in the laboratory.

Do These

Question 1
a) Give two definitions of a salt.
b) Chrome alum is a double salt. Explain what this statement means using the chemical formula of the salt.
c) Give three examples of complex salts with their chemical formulae.

Question 2
a) State four methods of preparing salt with appropriate equations.
b) The basicity of tetraoxophosphate (V) acid is 3. Explain this statement with an equation showing the reaction between the acid and calcium hydroxide.
c) Explain why tetraoxosulphate (VI) acid can form two types of salts, but trioxonitrate (V) acid can form only one type of salt.

Question 3
a) Explain the following with one example each:
i) Deliquescence
ii) Hygroscopy
iii) Efflorescence
b) What substances are used in desiccators?
c) How would you determine experimentally, the water of crystallization of a sample of hydrated copper (II) tetraoxosulphate (VI)? Write the calculation? (To answer this, you may need to read our post on Water of Crystallization)

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Acids & Bases (Part III): pH, Indicators & Buffers

2016-05-10T12:59:00.001+01:00

pH & pH Scale

The pH (hydrogen ions potential) is a measure of the acidity or alkalinity of a solution. The concept of pH was introduced by Sörensén in 1909 to bring about the convenience of working with very dilute solutions. To this effect, he developed a scale consisting of fifteen numbers (0 - 14), which is used in pH meter, for measuring the relative acidity or alkalinity in solutions. This scale is known as the pH scale. The numbers in the pH scale are the values of the negative logarithms of the hydrogen ions concentrations in such solutions.

From the above, we can define pH as the negative logarithm of the hydrogen ions concentration [H+] to base 10. Mathematically, this is given as:

pH = - log [H+] ..........................(i)

Alternatively, the above equation can be expressed as

pH = log 1/[H+] .........................(ii)

From equation (ii), we can also define pH as the logarithm of the reciprocal of the hydrogen ions concentration to base 10.

The pH scale is interpreted as follows:

    0 - 2 = strongly acidic
    3 - 4 = moderately acidic
    5 - 6 = weakly acidic
    7 - 7 = neutral
    8 - 9 = weakly alkaline
10 - 11 = moderately alkaline
12 - 14 = strongly alkaline

An analysis of the pH scale shows that acidity decreases with increase in pH value, and alkalinity increases as the pH value increases.

Examples of some substances with their pH values are

pOH

Just like pH, pOH (hydroxide ions potential) is the negative logarithm of the hydroxide ions concentration [OH-] to base 10. It is mathematically expressed as:

pOH = - log [OH-] ........................(iii)

Alternatively, it can be defined as the logarithm of the reciprocal of the hydroxide ions concentration [OH-] to base 10, and expressed as:

pOH = log 1/[OH-] ........................(iv)

Calculations Involving pH & pOH

Dissociation of Water

The pH of pure or distilled water shows that it is neutral. This neutrality indicates that as water molecules dissociate into ions, they produce equal amounts of hydrogen and hydroxide ions, according to the equation:

H2O(l) <-----------> H+(aq) + OH-(aq)

Studies have shown that the dissociation constant of water, Kw at 25°C equals 1.0 × 10^-14, and since water contains equal amounts of [H+] and [OH-], it implies that

[H+][OH-] = 1.0 × 10^-14 mol^2 dm^-6 .......................(v)

Hence,

[H+] = [OH-] = 1.0 × 10^-7 mol dm^-3

Taking the logarithm of both sides of equation (v) gives:

log [H+][OH-] = log 10^-14

Applying the laws of logarithm, we will obtain:

log [H+] + log [OH-] = - 14 log10 (but, log 10 = 1)

log [H+] + log [OH-] = - 14

Divide through by - 1

- log [H+] + (- log [OH-]) = 14 ................(vi)

Substituting equations (i) and (iii) into equation (vi) gives:

pH + pOH = 14 .............................(vii)

Using equations (i), (iii), (v) and (vii); we can calculate the pH, pOH, [H+] and [OH-] of any given solution.

Example 1

Calculate the pH of the following solutions:

(a) 0.005M H2SO4
(b) 0.015M aqueous NH3
(c) M/25 HCl

Answer

(a) Tetraoxosulphate (VI) acid ionizes according to the equation:

H2SO4 (aq) --------> 2H+(aq) + SO4––(aq)

The given concentration of the acid is the hydrogen ion concentration, i.e, [H+] = 0.005M or mol dm^-3

But, H2SO4 contains 2 moles of H+
Therefore, the total concentration of hydrogen ions in the acid, [H+] = 2 × 0.005 = 0.01M

Hence, applying equation (i) or (ii)

pH = - log [H+]
      = - log 0.01 = - log 10^-2
      = - (-2 log 10) = - (-2)
      = 2

(b) Aqueous ammonia dissociates weakly as shown in the equation:

NH4OH(aq) <-------> NH4+(aq) + OH-(aq)

Being a base, the 0.015M concentration in the question is that of the hydroxide ion, and not hydrogen ion; i.e, [OH-] = 0.015M or mol dm^-3.

Therefore, applying equation (iii) or (iv)

pOH = - log [OH-]
         = - log 0.015 = - log 15 × 10^-3
         = - (log 15 + log 10^-3)
         = - (1.1761 - 3) = - (-1.8239)
         = 1.8239

but

pH + pOH = 14
            pH = 14 - pOH
                  = 14 - 1.8239
                  = 12.1

HCl(aq) -------> H+(aq) + Cl-(aq)

The concentration, M/25 = 1/25 M = 0.04M. Therefore, [H+] = 0.04M or mol dm^-3.

Using equation (i) or (ii)

pH = - log [H+]
      = - log 0.04 = - log 4 × 10^-2
      = - (log 4 + log 10^-2)
      = - (0.6021 - 2) = - (-1.3979)
      = 1.4

Example 2

Calculate:

(a) the hydroxide ion concentration in 80.01M tetraoxosulphate (VI) acid.
(b) hydrogen ion concentration in 0.003M sodium hydroxide.

Answer

(a) We know that the ionization of tetraoxosulphate (VI) acid is given as:

H2SO4 (aq) --------> 2H+(aq) + SO4––(aq)

and the given concentration of the acid is the hydrogen ion concentration, i.e, [H+] = 80.01M or mol dm^-3;

but H2SO4 contains 2 moles of H+
Therefore, the total concentration of hydrogen ions in the acid, [H+] = 2 × 80.01 = 160.02M = 1.60 × 10^2 M

Applying equation (v) gives

[H+][OH-] = 1.0 × 10^-14 M^2
       [OH-] = 1.0 × 10^-14 M^2/[H+], but [H+] = 1.6 × 10^2 M
Therefore,
       [OH-] = (1.0 × 10^-14)/(1.6 × 10^2)
       [OH-] = 6.25 × 10^-17 M

(b) Sodium hydroxide ionizes as shown:

NaOH(aq) --------> Na+(aq) + OH-(aq)

The given concentration of 0.003 M is the hydroxide ion concentration. Therefore, [OH-] = 3.0 × 10^-3 M

Applying equation (v) gives

[H+][OH-] = 1.0 × 10^-14 M^2
        [H+] = 1.0 × 10^-14 M^2/[OH-], but [OH-] = 3.0 × 10^-3 M
Therefore,
        [H+] = (1.0 × 10^-14)/(3.0 × 10^-3)
        [H+] = 3.33 × 10^-12 M

The above solutions to Examples 2(a) & (b) show that irrespective of the concentration of a substance, it contains both hydrogen and hydroxide ions.

Example 3

If the dissociation constant, Kw, for water is 1.0 × 10^-14 at 25°C, calculate:

(a) the hydrogen ion concentration
(b) pH value of water

Answer

(a) Water dissociates according to the equation:

H2O(l) <---------------> H+(aq) + OH-(aq)

The dissociation constant, Kw, is given as:

[H+][OH-] = Kw; at 25°C, Kw = 1.0 × 10^-14 M^2 and [H+] = [OH-]
Therefore,
     [H+]^2 = 1.0 × 10^-14 M^2
         [H+] = √1.0 × 10^-14 M^2
                 = 1.0 × 10^-7 M

(b) pH = - log [H+]
            = - log 10^-7
            = - (-7 log 10)
            = - (-7) = 7

Therefore, the hydrogen ion concentration and pH value of water at 25°C are 1.0 × 10^-7 M and 7 respectively. We can infer from the above that if the hydrogen ion concentration of a substance, [H+] = 10^-x, then pH = x

Example 4

(a) The dissociation constant of ethanoic acid at 25°C is 1.8 × 10^-5. In a buffer solution, the concentration of CH3COO- and CH3COOH are 0.025 M and 0.05 M respectively. What is the pH of buffer solution?

Answer

The dissociation of ethanoic acid is as shown:

CH3COOH(aq) <----------------> CH3COO-(aq) + H+(aq)

Its dissociation constant at 25°C, Ka, is given as:

Ka = [CH3COO-][H+]/[CH3COOH]
but Ka = 1.8 × 10^-5; [CH3COO-] = 0.025 M; [CH3COOH] = 0.05 M

1.8 × 10^-5 = 0.025 × [H+]/0.05
           [H+] = 0.05 × 1.8 × 10^-5/0.025
                   = (9.0 × 10^-7)/(2.5 × 10^-2)
                   = 3.6 × 10^-5 M

pH = - log [H+]
      = - log (3.6 × 10^-5)
      = - (log 3.6 + log 10^-5)
      = - (log 3.6 - 5 log 10)
      = - (0.5563 - 5)
      = - (-4.4437) = +4.4437
      = 4.4

(b) Calculate the hydrogen ion concentration and pH of a buffer solution, in which the concentration of Ca2+ is 0.02M and that of Ca(OH)2 in water is 0.04M, if the dissociation constant of calcium hydroxide, Kb = 5.5 × 10^-6 at 25°C.

Answer

The dissociation of calcium hydroxide in water is as follows:

Ca(OH)2(aq) <-------> Ca2+(aq) + 2OH-(aq)

Its dissociation constant at 25°C, Kb, is given as:

Kb = [Ca2+][OH-]^2/[Ca(OH)2]
but Kb = 5.5 × 10^-6; [Ca2+] = 0.02 M; [Ca(OH)2] = 0.04 M

5.5 × 10^-6 = 0.02 × [OH-]^2/0.04
      [OH-]^2 = 0.04 × 5.5 × 10^-6/0.02
                   = (2.2 × 10^-7)/(2.0 × 10^-2)
                   = 1.1 × 10^-5 M^2
          [OH-] = √1.1 × 10^-5 M^2
                   = 3.3 × 10^-3 M

But

[H+][OH-] = 1.0 × 10^-14
         [H+] = 1.0 × 10^-14/[OH-]
                 = (1.0 × 10^-14)/(3.3 × 10^-3)
                 = 3.03 × 10^-12 M

pH = - log [H+]
      = - log (3.03 × 10^-12)
      = - (log 3.03 + log 10^-12)
      = - (log 3.03 - 12 log 10)
      = - (0.4814 - 12)
      = - (-11.5186) = +11.5186
      = 11.5

Therefore, the pH of the buffer solution is 11.5

Importance of pH

We cannot overstate the significance of pH in different areas of our lives such as agriculture, medicine, pharmacy, water purification and sewage treatment. In agriculture, a farmer must ensure the soil he intends to cultivate has the correct pH requirements for the crops he intends to plant on it. Studies have shown that soil pH values vary from 4 - 9, depending on the location, and most plants thrive in soils of pH values between 7 - 8. The pH of soils with excess acidity is usually corrected by using calcium hydroxide (slaked lime) to neutralize them.

In medicine and pharmacy, for our bodies to function properly, the body fluids must be maintained at the correct pH values, as enzymes, which aid in the metabolic processes in our bodies, are pH dependent. So, deviations from the normal pH values of our body systems are signs of ill health. When this occurs, the doctors prescribe drugs, which can be used to correct the anomaly. Meanwhile, when producing the drugs, the pharmacists ensure that they are produced under the right pH conditions, without which their efficacy of the active ingredients cannot be guaranteed.

In water purification, it is important to ensure that the water distributed for town supply has a neutral pH to prevent the growth of microbes within a very short period. Similarly, in sewage treatment, the waste products must be made harmless and of neutral pH before they are finally released into the environment, especially, water bodies. If this is not done, it will foster the growth of bacteria and other micro-organisms, which compete for oxygen with the fishes and other aquatic animals in the water bodies. This leads to low oxygen level in the water bodies and eventual death of the water organisms.

Indicators

Indicators are weak organic acids and bases that ionize slightly in solution to produce different colours at different pH of the solution.

During ionization of indicators, there is always an equilibrium mixture of undissociated molecules and dissociated ions, as represented in the equation:

HIn(aq) <------------------> H+(aq) + In-(aq)

The undissociated molecules, HIn, always produce different colours from the dissociated ions. Therefore, the colour of the indicator in a given solution will depend on the ratio of the undissociated molecules and dissociated ions present. Let's take a look at how methyl orange and phenolphthalein behave in acidic and alkaline solutions.

Methyl orange: Let us represent the unionized molecules of methyl orange as HMe, and the dissociated ions as Me–. In solution, HMe is red, while Me– is yellow, and dissociates as follows:

HMe(aq) <------------------> H+(aq) + Me–(aq)

When an acid is added to a solution of methyl orange, the red colour becomes prominent because of the introduction of more hydrogen ions from the acid. This destabilizes the equilibrium of the system, which is forced to adjust according to Le Chatelier's Principle. The excess hydrogen ions combine with the dissociated ions to form the unionized molecules, and this causes the equilibrium position of the reaction to shift to the left. This shift is indicated by the increased intensity of the red colour of the solution.

However, when an alkaline is added to the methyl orange solution, the yellow colour becomes dominant. This is because the hydroxide ions from the alkaline combine with the hydrogen ions from the indicator to form water, thereby, removing them from the equilibrium mixture. This causes the equilibrium position to shift to the right, by forcing the dissociation of more undissociated molecules to replace the lost hydrogen ions. This also increases the concentration of dissociated ions in the solution, which is marked by the yellow colour.

Phenolphthalein: Let the unionized molecules of phenolphthalein be HPh, and the dissociated ions be Ph–. In solution, HPh is colourless, while Ph– is pink, and dissociates as follows:

HPh(aq) <------------------> H+(aq) + Ph–(aq)

In an acidic solution, phenolphthalein is colourless because of the introduction of more hydrogen ions from the acid. The excess hydrogen ions combine with the dissociated ions to form the undissociated molecules, and this causes the equilibrium position to shift to the left. This change is accompanied by the colourless nature of the solution. However, in an alkaline solution, the pink colour is predominant, and the reason is the same as that for methyl orange above.

Indicators & Acid-Base Titration

Indicators are very useful in acid-base titrations to determine the endpoints of the reactions. An endpoint is a stage in titration in which a change in colour of the indicator shows that no more titrant should be added. This is because of the colourless nature of the acid and base solutions. Different indicators give different colours at endpoints depending on the pH of the mixture, which is somewhat dependent on the types of acids and bases used. The list below shows the colour changes of some common indicators in acidic, basic and neutral solutions and their suitability in a given acid-base titration.

Indicator: Methyl orange
Colour in Solutions: Acid - red (pH: 1.0-3.2); Neutral - orange (pH: 3.2-6.1); Basic - yellow (pH: 6.1-14.0)
Titration Suitability: Strong Acid + Weak Base; Strong Acid + Strong Base

Indicator: Litmus
Colour in Solutions: Acid - red (pH: 1.0-6.0); Neutral - purple (pH: 6.0-8.5); Basic - blue (pH: 8.5-14.0)

Indicator: Phenolphthalein
Colour in Solutions: Acid - colourless (pH: 1.0-8.9); Neutral - pale pink (pH: 8.9-11.0); Basic - pink (pH: 11.0-14.0)
Titration Suitability: Weak Acid + Strong Base; Strong Acid + Strong Base

The pH of a solution can also be measured by using a universal indicator. Though not accurate, it can be used to detect values within the range of 2 - 12, through a series of colour changes. It is obtained by mixing different indicators together. These indicators work at different pH ranges. The pH values obtained are easily determined by comparing the colours obtained with the given standards.

Buffers

A buffer is a solution that shows resistance to changes in pH on dilution or when a little quantity (drops) of acid or base is added to it. It is made up of a mixture of a weak acid and its salt or a weak base and its salt. Examples of buffers are mixtures of:

a) ethanoic acid and potassium ethanoate (CH3COOH/CH3COOK)

b) aqueous ammonia and ammonium chloride (NH4OH/NH4Cl)

c) trioxocarbonate (IV) acid and sodium hydrogen trioxocarbonate (IV) (H2CO3/NaHCO3)

d) calcium hydroxide and calcium chloride (Ca(OH)2/CaCl2)

The concept of buffers and pH is very important in biochemical activities, because most metabolic processes in the human body are pH specific. This implies that a change in the pH of the system by about +/-0.5 may lead to a fatal outcome, if not carefully handled.

For example, the pH of the blood is about 7.4, and if an injection is to be administered to a patient intravenously (through the veins), such injection must be buffered to prevent it from altering the pH of the blood, since it is going directly into the bloodstream. Otherwise, the effect can be fatal.

Similarly, most of the processed foods and drinks we take are buffered. Different parts of the human digestive system have different pH. The mouth, where the digestion of food starts, is alkaline, the stomach is acidic and the small intestine, where the digestion ends, is also alkaline. Therefore, it is necessary that the foods and drinks we take are buffered to prevent unnecessary changes in their pH, as they pass through our alimentary canal. This also prevents complications that may arise due to indigestion.

Do These

Question 1
(a) Define pH in terms of hydrogen ions concentration and give two uses of pH.
(b) What is a buffer solution? Give two examples.

Question 2
The concentration of hydrogen ions in three solutions are:
(a) 2 × 10^-3 mol dm^-3
(b) 7 × 10^-9 mol dm^-3
(c) 5 X 10^-8 mol dm^-3
What is the pH of each solution?

Question 3
(a) A solution has a pH of 4.5. What is its hydrogen ions concentration?
(b) The pH values of two solutions are 5.0 and 10.5. What will be the colours of methyl orange, litmus and phenolphthalein indicators in these solutions?

Question 4
Which of the following pairs is/are buffer solution(s)?
(a) Hydrochloric acid and sodium chloride
(b) Ethanoic acid and sodium ethanoate
(c) Calcium chloride and potassium trioxocarbonate (IV)
(d) Potassium hydroxide and potassium chloride
(e) Aqueous ammonia and ammonium tetraoxosulphate (VI)

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Acids & Bases (Part II): Preparation, Properties & Uses of Bases

2016-04-25T16:39:00.001+01:00

In a layman's term, a base is the opposite of an acid. In other words, a base is everything an acid is not. Recall that in the Part I of this series, we looked at the definition of an acid using three different concepts. Similarly, we are going to define a base using the same concepts.

Definition

Lewis Bases
According to G. N. Lewis, a base is any species that can readily donate a pair of electrons. The availability of lone pair(s) of electrons increases a substance's ability to behave as a base. Examples include H2Ö, ÑH3, Cl-, F- etc. They are also considered to be nucleophiles. Any species with an electron-rich centre is said to be a nucleophile.

Brønsted-Lowry Bases
According to Brønsted and Lowry, an acid is a proton donor, while a base is a proton acceptor. In other words, any substance that has the ability to accept a proton (hydrogen ion, H+) by donating a pair of electrons to it, is said to be a base. Examples are H2Ö, ÑH3, Br- etc.

H2Ö(l) + H+(aq) <----> H3Ö+(aq)

ÑH3(l) + H+(aq) ----> NH4+(aq)

Cl-(aq) + H+(aq) ----> HCl(aq)

Arrhenius Bases
Recall that an Arrhenius acid produces hydrogen ion as the only positive ion, when dissolved in water. Similarly, a base is any substance, which produces hydroxide ion, OH- when dissolved in water. For example, sodium hydroxide dissolves in water to produce sodium ion, Na+ and OH- as follows:

NaOH(aq) ----> Na+(aq) + OH-(aq)

Also, potassium hydroxide dissolves in water to produce potassium ion, K+ and OH-.

KOH(aq) ----> K+(aq) + OH-(aq)

A base can also be defined as any substance that neutralizes an acid, or a substance that reacts with an acid to form salt and water. This is because not all bases produce hydroxide ions in water, as some of them are oxides and hydroxides of certain metals, which are insoluble in water.

Classification of Bases

From the aforementioned, bases can be classified into two major categories based on their solubility in water. These are the soluble and insoluble bases.

Soluble Bases: These comprise of oxides of Group I metals and calcium; and hydroxides of sodium, potassium, calcium and ammonium. These soluble hydroxides are known as alkalis.

Insoluble Bases: These are the oxides and hydroxides of other metals not mentioned above.

Examples of alkalis and insoluble bases are:

Classification of Alkalis

Alkalis are classified based on their strength, which is their degree of ionization in water. These are strong and weak alkalis.

Strong Alkalis: These are those, which ionize completely when dissolved in water. Examples are sodium hydroxide and potassium hydroxide.

NaOH(aq) ----> Na(s) + OH-(aq)

KOH(aq) ----> K+(aq) + OH-(aq)

Weak Alkalis: These are those hydroxides which undergo incomplete dissociation in water. Examples are calcium hydroxide and liquid ammonia.

Ca(OH)2(aq) <----> Ca2+(aq) + 2OH-(aq)

NH4OH(aq) <----> NH4+(aq) + OH-(aq)

NH4OH was was initially known as ammonium hydroxide, but was later changed to liquid ammonia (NH3.H2O), because unlike other hydroxides, it decomposes in the presence of heat to liberate ammonia gas.

NH4OH(aq) + heat ----> NH3(g) + H2O(l)

Physical Properties of Alkalis

1. They possess a bitter taste.
2. They are slippery to touch because of their soapy feel.
3. They turn red litmus paper blue.
4. They change the colour of phenolphthalein paper from colourless to pink.
5. Their concentrated forms are corrosive.
6. They are electrolytes.

Chemical Properties of Bases

1. Neutralization Reactions: All bases react with acids to form salts and water. This is known as neutralization reaction.

base + acid ----> salt + water

Examples
(a) Potassium hydroxide reacts with hydrochloric acid to form potassium chloride and water.

KOH(aq) + HCl(aq) ----> KCl(aq) + H2O(l)

Ionically,

K+(aq).OH-(aq) + H+(aq).Cl-(aq) ----> K+(aq).Cl-(aq) + H2O(l)

H+(aq) + OH-(aq) ----> H2O(l)

(b) Copper (II) oxide reacts with tetraoxosulphate (VI) acid to produce copper (II) tetraoxosulphate (VI) and water.

CuO(s) + H2SO4(aq) ----> CuSO4(aq) + H2O(l)

(c) Insoluble magnesium hydroxide reacts with dilute trioxonitrate (V) acid to form magnesium trioxonitrate (V) and water.

Mg(OH)2(s) + 2HNO3(aq) ----> Mg(NO3)2(aq) + 2H2O(l)

(d) Sodium hydroxide combines with ethanoic acid to produce sodium ethanoate and water.

NaOH(aq) + CH3COOH(aq) ----> CH3COONa(aq) + H2O(l)

Ionically,

Na+(aq).OH-(aq) + CH3COO-(aq).H+(aq) ----> CH3COO-(aq).Na+(aq) + H2O(l)

H+(aq) + OH-(aq) ----> H2O(l)

From (a) and (d) above, we can see that the actual species that 'partake' in a neutralization reaction are hydrogen/hydroxonium ions, and hydroxide ions.

Hence a neutralization reaction can be redefined as the reaction between hydrogen/hydroxonium ions and hydroxide ions to produce water molecules.

H+(aq) + OH-(aq) ----> H2O(l)

H3O+(aq) + OH-(aq) ----> 2H2O(l)

2. Reaction with Ammonium Salts: All alkalis, except liquid ammonia, react with ammonium salts to liberate ammonia gas when heated. This reaction is used in qualitative analysis for the identification of salts that contain ammonium ions.

NH4+(aq) + NaOH(aq) ----> NH4OH(aq) + Na+(aq)

NH4OH(aq) + heat ----> NH3(g) + H2O(l)

Methods of Preparation of Bases

The following are some of the ways bases can be prepared in the laboratory:

1. Burning metals in air. Metals burn in air to form basic oxides.

4Na(s) + O2(g) ----> 2Na2O(s)

2Ca(s) + O2(g) ----> 2CaO(s)

2Fe(s) + O2(g) ----> 2FeO(s)

2. Dissolving soluble basic oxides in water. Some metallic oxides dissolve in water to form alkalis. For instance, sodium oxide and potassium oxide dissolve in water to form sodium hydroxide and potassium hydroxide respectively.

Na2O(s) + H2O(l) ----> 2NaOH(aq)

K2O(s) + H2O(l) ----> 2KOH(aq)

Calcium oxide dissolves sparingly in water to form calcium hydroxide (lime water).

CaO(s) + H2O(l) ----> Ca(OH)2(aq)

3. Dissolving some metals in water. The very reactive Group 1 metals like potassium and sodium, and a few Group 2 metals such as calcium and barium react with water to form alkalis with the liberation of hydrogen.

2K(s) + 2H2O(l) ----> 2KOH(aq) + H2(g)

2Na(s) + 2H2O(l) ----> 2NaOH(aq) + H2(g)

Ca(s) + 2H2O(l) ----> Ca(OH)2(aq) + H2(g)

4. Dissolving some metallic hydrides in water. Sodium, potassium and calcium hydrides dissolve in water to form alkalis, with the evolution of hydrogen gas.

NaH(s) + H2O(l) ----> NaOH(aq) + H2(g)

KH(s) + H2O(l) ----> KOH(aq) + H2(g)

CaH2(s) + 2H2O(l) ----> Ca(OH)2(aq) + 2H2(g)

5. Thermal decomposition of some trioxocarbonate (IV) and trioxonitrate (V) salts. Heating some trioxocarbonate (IV) and trioxonitrate (V) salts will produce the oxides of the metallic cations in the salt.

Examples
(a) Copper (II) trioxocarbonate (IV) salt undergoes decomposition when heated to form copper (II) oxide, with the liberation of carbon (IV) oxide.

CuCO3(s) + heat ----> CuO(s) + CO2(g)

Other examples are:

ZnCO3(s) + heat ----> ZnO(s) + CO2(g)

CaCO3(s) + heat ----> CaO(s) + CO2(g)

(b) Calcium trioxonitrate (V) salt decomposes on heating to form calcium oxide. Oxygen and brown fumes of nitrogen (IV) oxide are also given off.

2Ca(NO3)2(s) + heat ----> 2CaO(s) + 4NO2(g) + O2(g)

Other examples include:

2Pb(NO3)2(s) + heat ----> 2PbO(s) + 4NO2(g) + O2(g)

2Zn(NO3)2(s) + heat ----> 2ZnO(s) + 4NO2(g) + O2(g)

6. Double decomposition. Some insoluble hydroxides can be prepared through the process of double decomposition by reacting a soluble salt containing the metallic cation of the intended hydroxide with an alkali.

For instance, copper (II) hydroxide can be prepared by reacting copper (II) trioxonitrate (V) solution with sodium hydroxide.

Cu(NO3)2(aq) + 2NaOH(aq) ----> Cu(OH)2(s) + 2NaNO3(aq)

Uses of Bases

The following are some of the important uses of bases:

1. They are used for making soaps. E.g. NaOH, KOH

2. They are used in the manufacture of fertilizer. E.g. NaOH, liquid ammonia.

3. Calcium hydroxide is used for reducing the level of acidity in acidic soils, and for the production of cement, mortar and Plaster of Paris (POP) etc.

4. Magnesium hydroxide is used for making antacid (milk of magnesia), laxatives and toothpaste.

5. Liquid ammonia is used in laundries as a solvent for removing grease and oil stains.

6. They are used in the petroleum industry for the refining of some petroleum products. E.g. NaOH

7. They are used as dyes in tanning industry. E.g. KOH

8. Alkalis like KOH, NaOH and liquid ammonia are used in water treatment, and in qualitative analysis for the identification of some cations.

Do These:

Question 1
(a) Elaborate on the statement: "All alkalis are bases, but all bases are not alkalis".
(b) Using appropriate examples, state five methods of preparing bases in the laboratory.
(c) List four uses of bases

Question 2
(a) What is a base?
(b) What is a neutralization reaction?
(c) Aqueous solution of trioxonitrate (V) acid neutralizes a solution of sodium hydroxide. Write the ionic equation for this reaction.
(d) In what other way, can it be shown that potassium hydroxide is a base without using litmus or phenolphthalein paper?

Question 3
Answer True (T) or False (F)
A base
(a) is a proton donor according to Brønsted and Lowry.
(b) produces hydroxide ions in water.
(c) accepts a pair of electrons based on Lewis Theory.
(d) is an electrolyte.
(e) does not have the ability to change the colour of an indicator.
(f) forms an aqueous solution with pH6.
(g) is neutralized by a salt.
(h) reacts with ammonium tetraoxosulphate (VI) to liberate ammonia when heated.

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Acids & Bases (Part I): Preparation, Properties & Uses of Acids

2016-04-13T19:46:00.001+01:00

Introduction

Growing up, we used to think that any substance that burns is an acid, until we learnt about acids, bases and salts in our foundational chemistry class. It was then clear to us that there is more to acids than corrosivity, and not every substance that is corrosive, is an acid.

Definition

The definitions of an acid is based on three different concepts of acid-base reactions, namely the Lewis, Brønsted-Lowry and Arrhenius concepts.

Lewis Acids
According to G. N. Lewis, in acid-base reactions, the reactants undergo co-ordinate covalent bonding, in which one reacting species has the ability to accept a lone pair of electrons, while the other can readily donate a lone pair of electrons. A Lewis acid is an electron pair acceptor, because it has an empty orbital. Examples are H+, H3O+, Cu2+, Fe3+.

Brønsted-Lowry Acids
According to J. H. Brønsted and M. Lowry, an acid-base reaction involves the transfer of proton from one of the reactants to the other. A Brønsted-Lowry acid is a proton donor. For instance,

HA ----> H+ + A-
HNO3 ----> H+ + NO3-

Our focus on this topic will be the Arrhenius concept of acid-base reactions, which gives the basic definition of acids.

Arrhenius Acids
According to Arrhenius, an acid is any substance that produces hydrogen ion, H+ or hydroxonium ion, H3O+ as the only positive ion, when dissolved in water.

Example, hydrogen chloride gas dissolves in water to form hydroxonium and chloride ions.

HCl(g) + H2O(l) ----> H3O+(aq) + Cl-(aq)

Other examples include:

H2SO4(l) + 2H2O(l) ----> 2H3O+(aq) + SO4--(aq)

HNO3(l) + H2O(l) ----> H3O+(aq) + NO3-(aq)

H3PO4(l) + 3H2O(l) ----> 3H3O+(aq) + PO4---(aq)

The above equations can be rewritten in the abbreviated form as follows:

HCl(aq) ----> H+(aq) + Cl-(aq)

H2SO4(aq) ----> 2H+(aq) + SO4--(aq)

HNO3(aq) ----> H+(aq) + NO3-(aq)

H3PO4(aq) <----> 3H+(aq) + PO4---(aq)

The presence of the H3O+ or the H+ in the above equations is what accounts for the acidity of the substances.

Examples of some acids are:

Basicity of Acids

The basicity of an acid is the number of replaceable hydrogen or hydroxonium ions present in one molecule of the acid when it dissociates in water.

The following equations show the dissociation of some acids in solution, and the accompanying table shows a summary of their basicity.

HCl(aq) ----> H+(aq) + Cl-(aq)

HCN(aq) ----> H+(aq) + CN-(aq)

HNO3(aq) ----> H+(aq) + NO3-(aq)

HClO3(aq) ----> H+(aq) + ClO3-(aq)

CH3COOH(aq) <----> CH3COO-(aq) + H+(aq)

H2SO4(aq) ----> 2H+(aq) + SO4--(aq)

H2SO3(aq) <----> 2H+(aq) + SO3--(aq)

H2CO3(aq) <----> 2H+(aq) + CO3--(aq)

H2C2O4(aq) <----> 2H+(aq) + C2O4--(aq)

H3PO4(aq) <----> 3H+(aq) + PO4---(aq)

Basicity (No of H+/molecule) |                Acids
            1 (monobasic)               |  HCl, HCN, HNO3, CH3COOH
            2 (dibasic)                      |  H2SO4, H2SO3, H2CO3, H2C2O4
            3 (tribasic)                      |  H3PO4

Types of Acids

Acids can be classified based on two major categories - their strength and nature.

Classification by Strength

There are two types of acids based on their strength. These are strong acids and weak acids.

Strong Acids: These are acids that dissociate or ionize completely when dissolved in water. Some examples of strong acids are hydrochloric acid, trioxonitrate (V) acid and tetraoxosulphate (VI) acid

HCl(aq) ----> H+(aq) + Cl-(aq)

HNO3(aq) ----> H+(aq) + NO3-(aq)

H2SO4(aq) ----> 2H+(aq) + SO4--(aq)

Weak Acids: These are acids that undergo partial or incomplete ionization in water. Examples include H2CO3, H2SO3, CH3COOH, H3PO4.

CH3COOH(aq) <----> CH3COO-(aq) + H+(aq)

H2CO3(aq) <----> 2H+(aq) + CO3--(aq)

H2SO3(aq) <----> 2H+(aq) + SO3--(aq)

H3PO4(aq) <----> 3H+(aq) + PO4---(aq)

The backward and forward arrows in the equations indicate an incomplete ionization. Actually, only 4 out of every 1000 molecules of ethanoic acid in solution are ionized. Generally, all organic acids like ethanoic acid and ethanedioic acids are weak acids.

Classification by Nature

There are also two types of acids by nature. These are the organic and inorganic acids.

Organic Acids: These are acids, which occur naturally in plant and animal matters. Some examples of organic acids are ethanoic (acetic) acid from vinegar, amino acids from protein, lactic acid from milk, ascorbic acid (Vit. C) from oranges, citric acid from lime and lemon, palmitic acid from palm oil, ethanedioic (oxalic) acid, etc.

Inorganic Acids: These are acids synthesized from mineral elements. They are also known as mineral acids. Examples include hydrochloric acid, trioxocarbonate (IV) acid, tetraoxosulphate (VI) acid, trioxonitrate (V) acid etc.

Concentration of Acids

The concentration of an acid is the amount of the acid in moles present in a given volume of water. If a little volume of water is added to a large amount of acid, then the solution is said to be a concentrated acid solution. Conversely, when a large volume of water is added to a small amount of acid, a dilute acid solution will be obtained.

For instance, given two solutions A and B of H2SO4, of concentrations 10 mol dm^-3 and 0.1 mol dm^-3 respectively; the former contains 10 moles or 980g of the acid in 1dm^3 (1000 cm^3) of distilled water, while the latter contains 0.1 mole or 9.8g of the acid in 1 dm^3 of distilled water. Hence, solution A is a concentrated solution, while solution B is a dilute solution, because the amount of H2SO4 present in 1 dm^3 of water is greater in A than in B.

Physical Properties of Acids

1. They turn blue litmus paper red.
2. The dilute acids possess a sour taste. The presence of acids accounts for the sour taste of unripe fruits, vinegar and rancid (stale) milk.
3. The strong acids are good electrolytes, while the weak acids are weak electrolytes.
4. The concentrated forms of strong acids are corrosive.

Precaution: You are NEVER expected to add water to a concentrated acid, as it may cause severe acid burns. If you must dilute a concentrated acid, ALWAYS ADD THE ACID TO A LARGER VOLUME OF WATER.

Remember, when cooking you DO NOT add water to hot oil, rather the reverse is done. The same principle is applicable when dealing with concentrated acids and water.

Chemical Properties of Acids

There are three major properties that all acids exhibit, irrespective of their nature, and these include:

(a) Action on active metals: Active metals displace the hydrogen atoms in acids to form salts with the liberation of hydrogen gas.

acid + active metal ----> salt + hydrogen

Examples:
2HCl(aq) + Zn(s) ----> ZnCl2(aq) + H2(g)

H2SO4(aq) + 2Na(s) ----> Na2SO4(aq) + H2(g)

H2SO4(aq) + Mg(s) ----> MgSO4(aq) + H2(g)

It is important to note that unlike other acids, trioxonitrate (V) acid does not undergo this reaction with the metals, except in the action of very dilute trioxonitrate (V) acid (about 1%) on calcium, magnesium or manganese.

2HNO3(aq) + Ca(s) ----> Ca(NO3)2(aq) + H2(g)

This is due to its highly oxidizing nature, which causes the hydrogen to be oxidized to water, while the acid, itself, is reduced to nitrogen (IV) oxide or nitrogen (II) oxide gas, depending on the reaction conditions, as shown in its reaction with copper.

4HNO3(aq) + Cu(s) ----> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) (hot concentrated HNO3)

8HNO3(aq) + 3Cu(s) ----> 3Cu(NO3)2(aq) + 2NO(g) +4H2O(l) (cold concentrated HNO3)

(b) Action on bases: All acids react with a base to form salt and water. This is known as a neutralization reaction.

acid + base ----> salt + water

Examples:
HCl(aq) + NaOH(aq) ----> NaCl(aq) + H2O(l)

H2SO4(aq) + CaO(s) ----> CaSO4(aq) + H2O(l)

HNO3(aq) + KOH(aq) ----> KNO3(aq) + H2O(l)

CH3COOH(aq) + NaOH(aq) ----> CH3COONa(aq) + H2O(l)

The products are formed by the exchange of negative ions and radicals between the acids and the bases.

(c) Action on trioxocarbonate (IV), (CO3--): All acids react with trioxocarbonate (IV) salts to produce salt and water, and liberate carbon (IV) oxide.

acid + trioxocarbonate (IV) ----> salt + water + CO2

Examples:
2HCl(aq) + Na2CO3(aq) ----> 2NaCl(aq) + H2O(l) + CO2(g)

H2SO4(aq) + CaCO3(s) ----> CaSO4(aq) + H2O(l) + CO2(g)

2HNO3(aq) + K2CO3(aq) ----> 2KNO3(aq) + H2O(l) + CO2(g)

CH3COOH(aq) + PbCO3(s) ----> (CH3COO)2Pb(s) + H2O(l) + CO2(g)

Methods of Acids Preparation

The following are some of the methods employed in preparing acids in the laboratory:

1. Dissolving an acid anhydride in water. Acid anhydrides are nonmetallic oxides, mostly gases, which dissolve in water to form acids. Examples include CO2, SO2, P4O10 etc

Examples:
(a) Carbon (IV) oxide dissolves in water to form trioxocarbonate (IV) acid.

CO2(g) + H2O(l) ----> H2CO3(aq)

(b) Sulphur (IV) oxide dissolves in water to form trioxosulphate (IV) acid.

SO2(g) + H2O(l) ----> H2SO3(aq)

SO3(g) + H2O(l) ----> H2SO4(aq)

(d) Phosphorus (V) oxide dissolves in water to produce tetraoxophosphate (V) acid.

P4O10(s) + 6H2O(l) ----> 4H3PO4(aq)

(e) Nitrogen (V) oxide and nitrogen (III) oxide dissolve in water to form trioxonitrate (V) acid and dioxonitrate (III) acids respectively.

N2O5(g) + H2O(l) ----> 2HNO3(aq)

N2O3(g) + H2O(l) ----> 2HNO2(aq)

2. Using a strong acid to displace a volatile or weak acid from its salts.

(a) Concentrated tetraoxosulphate (VI) acid is used to displace a volatile gas like hydrogen chloride from a chloride salt.

H2SO4(aq) + NaCl(s) ----> Na2SO4 + HCl(g)

The hydrogen chloride produced, dissolves readily in water to form hydrochloric acid.

(b) Concentrated tetraoxosulphate (VI) acid can also be used to displace weak trioxocarbonate (IV) acid from a trioxocarbonate (IV) salt.

H2SO4(aq) + CaCO3(s) ----> CaSO4(aq) + H2CO3(aq)

The trioxocarbonate (IV) acid, which is not stable, readily decomposes to water and carbon (IV) oxide.

H2CO3(aq) ----> H2O(l) + CO2(g)

(c) Concentrated tetraoxosulphate (VI) and hydrochloric acids are used to displace trioxonitrate (V) acid from excess trioxonitrate (V) salts.

H2SO4(aq) + 2KNO3(s) ----> K2SO4(aq) + 2HNO3(g)

HCl(aq) + NaNO3(s) ----> NaCl(aq) + HNO3(g)

3. Direct combination of elements. Most acids that are binary compounds, such as the hydrogen halides can be prepared by combining the constituent elements at appropriate conditions.

Examples:
(a) Hydrogen burns rapidly in chlorine in the presence of activated charcoal as a catalyst to form hydrogen chlorine gas, which then dissolves in water to produce hydrochloric acid.

H2(g) + Cl2(g) + activated charcoal + heat ----> 2HCl(g)

HCl(g) + H2O(l) ----> H3O+(aq) + Cl-(aq)

(b) Hydrogen combines with bromine vapour in the presence of platinum and heat to form hydrogen bromide gas, which then dissolves readily in water to form hydrobromic acid.

H2(g) + Br2(g) + platinum + heat ----> 2HBr(g)

HBr(g) + H2O(l) ----> H3O+(aq) + Br-(aq)

The hydrogen bromide gas produced, dissolves readily in water to form hydrobromic acid.

Uses of Acids

Some uses of acids include:

1. Manufacture of textiles, paints, fertilizers, drugs and other chemicals. E.g. H2SO4, HCl, HNO3.

2. As drying and dehydrating agents. E.g. H2SO4

3. In food preservation and dyeing of textiles. E.g. CH3COOH

4. As solvents in qualitative analysis. E.g. Dilute HCl, dilute HNO3

5. As an oxidizing agent. E.g. Conc. HNO3

6. In oil refineries for refining of some petroleum products. E.g. H2SO4

7. In the manufacture of baking soda, soft drinks and health salts. E.g. H2CO3.

Do These:

Question 1
Define an acid based on the three concepts of acid-base reactions, and give one example each.

Question 2
Give the natural sources of the following organic acids:
(a) Ascorbic acid
(b) Acetic acid
(c) Palmitic acid
(d) Citric acid

Question 3
(a) How would you prove that a given colourless liquid is an acid?
(b) State three chemical properties of acids.

Question 4
(a) Differentiate between a strong acid and a concentrated acid.
(b) Carbon (IV) oxide is described as an acid anhydride. Explain.
(c) State four uses of acids.

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Chemical Kinetics (Part III): Half-Life & Mechanism of Reactions

2016-04-12T11:55:00.001+01:00

Half-life of a Reaction
The half-life, t1/2, of a reaction is the amount of time it will take the concentration, [A], of a reactant to reduce to half its initial concentration, [Ao]. That is to say,
              @ t1/2, [A] = [Ao]/2 ...……(i)
The half-life of the various orders of reaction can be calculated by substituting the above equation into the respective integrated rate laws.
(a) Half-life of a Zeroth-Order Reaction
Recall that the zeroth-order integrated rate law is given by:
                 [A] = [Ao] - kt …………(ii)
                 [Ao] - [A] = kt
@ t1/2, [A] = [Ao]/2
                 kt1/2 = [Ao] - ([Ao]/2)
                 kt1/2 = (2[Ao] - [Ao])/2
                 kt1/2 = [Ao]/2
                   t1/2 = [Ao]/2k …………(iii)
Equation (iii) above, gives the half-life for a zeroth-order reaction, which can be obtained by finding the product of half the initial reactant concentration and the reciprocal of the reaction rate constant ([Ao]/2 × 1/k).
(b) Half-life of a First-Order Reaction
Recall that the first-order integrated rate law is given by:
               ln [A] = ln [Ao] - kt …………(iv)
                     kt = ln [Ao] - ln [A]
@ t1/2, [A] = [Ao]/2
               kt1/2 = ln [Ao] - ln ([Ao]/2)
               kt1/2 = ln ([Ao]/([Ao]/2))
               kt1/2 = ln 2
                 t1/2 = ln 2/k …………(v)
Equation (v) gives the half-life for a first-order reaction, which is independent of the initial concentration of the reactant. It can be obtained by dividing the natural log of 2 (ln 2 = 0.6932) by the reaction rate constant.
Equation (v) is applied when calculating the half-life of most radioactive substances. This is because nuclear or radioactive reactions are generally first-ordered.
(c) Half-life of a Second-Order Reaction
Recall that the integrated rate law for a second-order reaction is given by:
                 1/[A] = 1/[Ao] + kt …………(vi)
                       kt = 1/[A] - 1/[Ao]
@ t1/2, [A] = [Ao]/2
                 kt1/2 = 1/([Ao]/2) - 1/[Ao]
                 kt1/2 = 2/[Ao] - 1/[Ao]
                 kt1/2 = 1/[Ao]
                   t1/2 = 1/[Ao]k …………(vii)
Equation (vii) gives the half-life for a second-order reaction, which can be obtained by finding the reciprocal of the product of the initial concentration of the reactant and the reaction rate constant (1/([Ao] × k)).
Do These:
Question 1
Using the information provided in Example 1 of the Chemical Kinetics (Part II), calculate the half-life of the 1-chloroethane (C2H5Cl) for experiments 1 and 4.
Question 2
Determine the half-life of the 1-bromopropane (C3H7Br) in Example 3 of the Chemical Kinetics (Part II)
Question 3
Calculate the half-life of the peroxydisulfate ion (S2O8--) using the information provided in Example 4 of the Chemical Kinetics (Part II).
Mechanism of Reactions
The mechanism of a chemical reaction is the detailed description of the elementary steps involved in the given reaction. Some reactions proceed directly from the reactants to the products stage, without passing through an intermediate stage. Such reactions are known as one-step reactions, whereas those that pass through the intermediate stage are known as multi-step reactions. An intermediate is actually a very reactive or unstable species produced in one stage and consumed in another stage of a multi-step reaction.
Example of a multi-step reaction is the decomposition of ozone molecules to oxygen gas, according to the equations:
               O3(g) ----> O2(g) + O. …………(a)
               O3(g) + O. ----> 2O2(g) ……...(b)
               2O3(g) ----> 3O2(g) …………….(c)
The mechanism of the two-step reaction is as explained below:
The ozone molecule undergoes a photolytic cleavage (the breaking of bond by the action of UV light) in (a) to produce a highly unstable oxygen atom radical, O. as the intermediate species. The radical is consumed in (b) when it attacks an ozone molecule to produce two molecules of oxygen gas. The addition of (a) and (b) gives the overall reaction (c).
In a multi-step reaction, the elementary step with the slowest rate of reaction is known as the rate determining step of the overall reaction. Remember, a chain, they say, is as strong as its weakest link. Which means that no matter how strong a chain looks, its strength is limited by strength of its weakest link.
The same analogy applies here, because no matter how fast the other reaction steps may be, the overall rate of the reaction is limited by the rate of slowest reaction step. This explains why the rate law of the overall reaction is determined by the stoichiometry of the rate determining step and not the stoichiometry of the overall reaction.
It is important to note that we can propose two or more mechanisms for a particular reaction. However, the reasonable one will be the mechanism that is consistent with the experimentally determined kinetic data for the reaction.
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Chemical Kinetics (Part II): Rate Law & Order of Reaction

2016-03-31T11:21:00.001+01:00

In our previous post, we looked at the overview of the rates of chemical reactions, where we studied the concepts and factors that affect the rates of reactions. In this post, we will be studying the rate law and the different orders of reaction, and how to determine them.
Rate Law
Consider the reaction:
            mA + nB ----> Products
The rate law states that the rate of a reaction is directly proportional to the active masses of the reactants. This implies that the concentration of the reacting species will determine how fast and how far a reaction can go. Using the above equation, the rate law can be expressed as:
            rate & [A]^m[B]^n ..........(i)
            rate = k[A]^m[B]^n ..........(ii)
where,
[A] = concentration of reactant A
[B] = concentration of reactant B
   & = sign of proportionality
   k = rate constant
The rate law is also known as the law of mass action.
Orders of Reaction
In chemical kinetics, an order is the index or power of the concentration of a reactant in a rate law. The overall order of a reaction is the sum of the orders of the respective reactants in the rate law. The three common types of orders of reaction are:
a) zeroth-order
b) first-order
c) second-order
From equation (ii) above, m is the order with respect to reactant A, while n is the order with respect to reactant B, and the overall order of the reaction is given as (m+n), and can either be 0, 1, 2 etc. However, it is important to note that the order of reaction can only be determined experimentally and NOT from the stoichiometric coefficients of the reactants in a given reaction.
For instance, according to the stoichiometry for the decomposition of nitrogen (I) oxide [N2O], to form nitrogen and oxygen in the presence of platinum,
             2N2O(g) ----> N2(g) + O2(g)
             rate = k[N2O]^2 .....(iii)
the order of the reaction is 2, i.e, second order with respect to the reactant, N2O. Meanwhile, the reaction is a zero order reaction when determined experimentally, i.e,
            rate = k[N2O]^0 ......(iv)
Orders of Reactions
Zeroth-Order Reaction
A reaction is said to be zeroth-order, if the rate of the reaction is independent of the concentration of the reactants, i.e,
            rate = k[reactants]^0
From our example on the decomposition of N2O, the rate of reaction is not dependent on the concentration of the reactant, but on the surface area of the platinum catalyst. This explains why the reaction will still proceed at the same rate, even when the concentration of N2O is very low.
So, how can we know that a given reaction is zeroth-order?
Let us consider the hypothetical reaction:
              A ----> Products
For a zeroth-order,
              rate = k[A]^0..............(v)
              rate = k (since [A]^0 = 1) ………(vb)
Recall,
             rate = -d[A]/dt ............(vi)
Therefore,
        -d[A]/dt = k
          d[A]/dt = -k
               d[A] = -k dt..................(vii)
Integrating equation (vii), within the limits of A = Ao and A, and t = 0 and t, we will obtain:
             $d[A] = -k $dt
       [A] - [Ao] = -k (t - 0)
       [A] - [Ao] = -kt
                 [A] = [Ao] - kt ..................(viii)
where [Ao] = initial concentration of A at time, t = 0
              [A] = concentration of A at time, t = t
Equation (vii) is the differential rate law, while equation (viii) is the integrated rate law of a zeroth-order reaction.
Therefore, a plot of the reactant concentration, [A] against time, t, will give a straight-line graph with a negative slope, -k, equal to the rate constant of the reaction, and a positive intercept on the y-axis, which is equivalent to the initial concentration of the reactant, [Ao].
Similarly, from equation (v), a plot of the reaction rate against concentration of reactant will give a straight-line graph, of uniform rate of reaction. From equation (vb), the rate constant of a zeroth-order reaction is equivalent to its rate of reaction. Hence, its unit is mol dm^-3 s^-1.
Thus, any reaction, whose graphs possess the abovementioned features can be said to be a zeroth-order reaction.
First-Order Reaction
A first-order reaction is one, whose rate is directly proportional to the concentration of only one reactant.
Let us consider the following reactions:
            A + B ----> Products .................(ix)
                  A ----> Products ........................(x)
From the above equations, if the rate of reaction is dependent on the concentration of A only, then, the rate law of equation (ix) will reduce to that of equation (x), as shown:
               rate = k[A]^1[B]^0
               rate = k[A] ..................(xi)
Substituting equation (vi) into (xi) gives
        -d[A]/dt = k[A]
          d[A]/dt = -k[A] .................(xii)
Rearranging equation (xii) gives us
          d[A]/[A] = -k dt .................(xiii)
Integrating equation (xiii), within the limits of A = Ao and A, and t = 0 and t, we will obtain:
   $(1/[A])d[A] = -k $dt
ln [A] - ln [Ao] = -k(t - 0)
ln [A] - ln [Ao] = -kt ...................(xiv)
    ln ([A]/[Ao]) = -kt
           [A]/[Ao] = e^-kt
                   [A] = [Ao]e^-kt ..........(xv)
From equation (xiv), equation (xv) can be written as:
               ln [A] = ln [Ao] - kt .......(xvi)
where [A] and [Ao] retain their usual meanings.
Equation (xii) is the differential rate law, while equations (xv) and (xvi) are the integrated rate laws of a first-order reaction. While the former is the exponential form, the latter is the logarithmic form.
Therefore, a plot of the natural logarithm of the concentration of the reactant, ln [A] against time, t, will give a logarithmic curve with a negative slope, -k, equal to the rate constant of the reaction, and a positive intercept on the y-axis, which is equivalent to the natural logarithm of the initial concentration of the reactant, ln [Ao].
Similarly, from equation (xii), a plot of the rate of reaction against the concentration of reactant will give a straight-line graph, passing through the origin, with a negative slope equal to the rate constant of the reaction. The unit of the rate constant of a first-order reaction is s^-1 (i.e, mol dm^-3 s^-1/mol dm^-3)
Hence, any reaction, whose graphs show these features can be classified as a first-order reaction.
Second-Order Reaction
A reaction is said to be second-order when its rate is directly proportional to the square of the concentration of only one reactants, or the product of the concentration of two reactants.
Consider the hypothetical reaction below:
            2A ----> Products .................(xvii)
           A + B ----> Products ……………(xviib)
From the above, the reaction rate can depend on the concentration of A only or those of A and B. Hence, the rate law can be stated as:
               rate = k[A]^2 ..................(xviii), or
               rate = k[A]^1[B]^1............(xix)
For our scope, we will focus on equation (xviii), where the rate of reaction is dependent on the concentration of only one reactant.
So, substituting equation (vi) into (xviii) gives
        -d[A]/dt = k[A]^2
          d[A]/dt = -k[A]^2 .................(xx)
Rearranging equation (xx) gives us
          d[A]/[A]^2 = -k dt .................(xxi)
Integrating equation (xxi), within the limits of A = Ao and A, and t = 0 and t, we will obtain:
   $(1/[A]^2)d[A] = -k $dt
-([A]^-1) - (-[Ao]^-1) = -k(t - 0)
-1/[A] - (-1/[Ao]) = -kt ...................(xxii)
    -1/[A] + 1/[Ao] = -kt
Rearranging the above equation gives us
                    1/[A] = 1/[Ao] + kt .............(xxiii)
where [A] and [Ao] retain their meanings.
Equation (xx) gives the differential rate law, while equation (xxiii) is the integrated rate law of a second-order reaction.
Thus, a plot of the reciprocal of the concentration of the reactant, 1/[A] against time, t, will give a straight-line graph with a positive slope, k, equal to the rate constant of the reaction, and a positive intercept on the y-axis, which is equal to the reciprocal of the initial concentration of the reactant, 1/[Ao].
Similarly, from equation (xx), a plot of the rate of reaction against the square of the concentration of reactant will give a parabolic curve, with a tangent equal to the rate constant, k. The unit of a second-order reaction is dm^3 mol^-1 s^-1 (i.e, mol dm^-3 s^-1/mol^2 dm^-6).
Therefore, any reaction, whose graphs possess these features can be said to be a second-order reaction.
Summary
Reaction Order      Differential Law       Integrated Law
Zeroth                     d[A]/dt = -k                   [A] = [Ao] - kt
First                         d[A]/dt = -k[A]           ln [A] = ln [Ao] - kt
Second                   d[A]/dt = -k[A]^2        1/[A] = 1/[Ao] + kt
How to Determine the Rate Law and Rate Constant from a Set of Experimental Data
Given a set of experimental data of a reaction, the following steps can be used to determine the rate law and the rate constant of the reaction.
1. Compare the reactant concentrations and reaction rates of two experiments to see the effect of concentration on the rate of reaction.
(a) If a change in reactant concentration has no effect on the reaction rate, then, it is a zeroth-order reaction.
(b) If a change in the reactant concentration leads to a corresponding change in the rate of reaction, then, it is first order with respect to the reactant.
(c) If the reaction rate is quadrupled when the reactant concentration is doubled, then, the reaction is second order with respect to the reactant.
Alternatively, you can use the concentration and rate data from two experiments to find the reaction order with regards to a given reactant in the rate law.
2. Write the appropriate rate law for the reaction.
3. Use the measured concentration and rate data from any of the experiments to find the rate constant using the rate law.
Example 1
At high temperatures, 1-chloroethane produces hydrogen chloride and ethene by the following reaction:
C2H5Cl(g) + heat ----> HCl(g) + C2H4(g)
Using the data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the 1-chloroethane, and determine the rate constant for the reaction.
Exp    [C2H5Cl]o (M)    Initial Rate (M s^-1)
1             0.0050                     0.8 × 10^-8
2             0.0075                     1.2 × 10^-8
3             0.0150                     2.4 × 10^-8
4             0.0200                     3.2 × 10^-8
Answer
From Exp 1 and 3, as the concentration is tripled, the reaction rate also triples; and from Exp 1 and 4, the rate of reaction quadruples as the concentration quadrupled. Therefore, the reaction is a first-order with respect to 1-chloroethane, because the rate is directly proportional to its concentration.
Hence, the rate law is:
            Rate = k[C2H5Cl]
To find k, substitute the data from Exp 1 into the rate law:
          0.8 × 10^-8 = k (0.005)
                            k = 8 × 10^-9/5 × 10^-3
                               = 1.6 × 10^-6 s^-1
Alternatively, we can use the data from Exp 1 and 3 to find the order of reaction with respect to 1-chloroethane using:
            Rate = k[C2H5Cl]^m
where, m = order of reaction with respect to C2H5Cl.
2.4 × 10^-8/0.8 × 10^-8 = k(0.015)^m/k(0.005)^m
                                       3 = (0.015/0.005)^m
                                   3^1 = 3^m
             Equate the powers of 3
                                      m = 1
Therefore, the reaction is first-order with respect to the reactant, and the rate law is:
            Rate = k[C2H5Cl]
Example 2
If a sample of the 1-chloroethane in Example 1 above, with an initial concentration of 0.0100 mol dm^-3 is heated at 650°C,
(a) what will be its concentration after 5 hours?
(b) How many hours at 650°C must elapse for the concentration to decrease to 0.0025 mol dm^-3.
Answer
Here, we will make use of the integrated law for a first-order reaction:
             ln [A] = ln [Ao] - kt
Recall that we have already calculated the rate constant, k = 1.6 × 10^-6 s^-1. So, substitute the given parameters into the equation and solve for the unknown.
(a) [Ao] = 0.0100 mol dm^-3, [A] =?
            t = 5 × 3600s = 18000s
ln [A] = ln 0.01 - (1.6 × 10^-6)(1.8 × 10^4)
                  ln [A] = -4.6052 - 0.0288
                  ln [A] = -4.6340
                      [A] = e^-4.6340
                            = 0.00972 mol dm^-3
(b) [Ao] = 0.0100 mol dm^-3, [A] = 0.0025 mol dm^-3, t =?
ln 0.0025 = ln 0.01 - (1.6 × 10^-6)t
               t = (ln 0.01 - ln 0.0025)/(1.6 × 10^-6)
                 = -4.6052 - (-5.9915)/(1.6 × 10^-6)
                 = 1.3863/(1.6 × 10^-6)
                 = 866437.5 s/3600 s hr^-1
                 = 240.68 hrs
Example 3
1-Bromopropane is a colourless liquid that reacts with S2O3-- according to the equation:
C3H7Br + S2O3-- ----> C3H7S2O3- + Br-
The reaction is first order in 1-bromopropane and first order in S2O3--, with a rate constant of 8.05 x 10^-4 mol dm^-3 s^-1.
(a) If the reaction began with 0.4 mol dm^-3 of C3H7Br and an equivalent concentration of S2O3--, what will be the initial reaction rate?
(b) If the concentration of each reactant was decreased to 0.2 mol dm^-3, what will be the initial reaction rate?
Answer
Since the reaction is first order to C3H7Br and S2O3--, then, the rate law of the reaction will be given as:
             Rate = k[C3H7Br][S2O3--], and the overall order = 1 + 1 = 2
(a) If [C3H7Br] = [S2O3--] = 0.4 mol dm^-3, k = 8.05 × 10^-4 dm^3 mol^-1 s^-1, then:
Initial rate = 8.05 × 10^-4 × 0.4 × 0.4
                   = 1.288 × 10^-4 mol dm^-3 s^-1
(b) If [C3H7Br] = [S2O3--] = 0.2 mol dm^-3, then:
Initial rate = 8.05 × 10^-4 × 0.2 × 0.2
                   = 3.22 × 10^-5 mol dm^-3 s^-1
From the above calculations, we can see that a reduction of the reactants concentrations by half decreases the rate of reaction by a quarter, which agrees with the behaviour of a second-order reaction.
Example 4
The peroxydisulfate ion (S2O8--) is a potent oxidizing agent that reacts rapidly with iodide ion in water as represented by the equation:
S2O8--(aq) + 3I-(aq) ----> 2SO4--(aq) + I3-(aq)
The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant
Exp    [S2O8--]o (M)   [I-]o (M)     Initial Rate (M s^-1)
1            0.27                   0.38               2.05
2            0.40                   0.38               3.06
3            0.40                   0.22               1.76
Answer
Here, we are going to use the alternative method by using the data from Exp 1 and 2, and Exp 2 and 3 to determine the orders of reaction with respect to the reactants and the rate law. So, using:
            Rate = k[S2O8--]^m[I-]^n
where, m = order of reaction with regards to S2O8--, and n = order of reaction with respect to I-
(a) From Exp 1 and 2;
3.06/2.05 = k(0.40)^m(0.38)^n/k(0.27)^m(0.38)^n
3.06/2.05 = (0.40/0.27)^m
         1.49 = 1.48^m
             m = 1.01, approximately 1
Similarly, from Exp 2 and 3
3.06/1.76 = k(0.40)^m(0.38)^n/k(0.40)^m(0.22)^n
3.06/1.76 = (0.38/0.22)^n
         1.74 = 1.73^n
              n = 1.01, approximately 1
Therefore, the rate law is:
            Rate = k[S2O8--][I-], and the overall order of reaction = 1 + 1 = 2
(b) To find k, substitute the data from Exp 1 into the rate law, i.e,
          Rate = k[S2O8--][I-]
where rate = 2.05; [S2O8--] = 0.27; [I-] = 0.38
                 2.05 = k(0.27)(0.38)
                       k = 2.05/0.1026
                          = 19.98 dm^3 mol^-1 s^-1
Do These:
Question 1
Sulfuryl chloride (SO2Cl2) decomposes to SO2 and Cl2 according to the reaction:
SO2Cl2(g) ----> SO2(g) + Cl2(g)
Data for the reaction at 320°C are given in the table. Calculate the reaction order with respect to SO2Cl2, and determine the rate constant for the reaction.
Exp    [SO2Cl2]o (M)     Initial Rate (M s^-1)
1             0.0100                    2.2 × 10^-7
2             0.0150                    3.3 × 10^-7
3             0.0200                    4.4 × 10^-7
4             0.0250                    5.5 × 10^-7
Question 2
Iodide reduces iron (III) according to the equation:
2Fe3+(aq) + 2I-(aq) ----> 2Fe2+(aq) + I2(s)
Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4.
(a) What is the reaction order with respect to each reactant?
(b) Determine the overall rate law.
(c) Determine the overall order of reaction.
Question 3
The reaction of nitrogen (II) oxide, (NO) and oxygen gas, (O2) is found to be second order with respect to NO and first order with respect to O2.
(a) What is the overall reaction order?
(b) What is the effect of doubling the concentration of each reactant on the reaction rate?
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Nigerian Education: the Ichabod, the Culprits & the Way Forward

2016-03-10T13:20:00.001+01:00

By Charles Offiong

A few days ago, I was discussing with a childhood friend of mine, who also doubled as my classmate. Our subject of discussion was the state and quality of our education vis-a-vis what was obtained in our days. We were lucky to attend one of the foremost secondary schools in Nigeria, the Hope Waddell Training Institution, Calabar, founded in 1895, which prided itself as the oldest in South-Eastern Nigeria and one of the best in West Africa. Apart from her, our first generation universities were ranked among the best in Africa. However, considering the current circumstances, it will be foolhardy of me to presume that my alma mater and those other institutions still maintain these bragging rights...

The importance of education to our national development and its need to occupy the pole position of any government cannot be overemphasized. The talk about the declining standards of the Nigerian education is what the blind had seen the dumb told the deaf, so it is no longer a news. The issue is how we got here.

The focus of this discourse will be on the rot in the foundational level of our educational system, especially the secondary education, as its importance in educational development has been underestimated. For those of us, who are not bible students, Ichabod is an expression of regret at a loss of former glory or higher standards. Its etymology is from 1 Samuel 4:21, where Eli’s daughter-in-law, Phinehas' wife, named her child Ichabod, saying ‘The glory is departed from Israel', when she heard that the ark of the Lord had been taken away by the Philistines...

Every year, the two major bodies saddled with the responsibility of conducting terminal examinations for pre-tertiary institutions in the country, the West African Examination Council (WAEC) and the National Examination Council (NECO) announce the results of their respective examinations. After reading out the somewhat disappointing statistics for the exams, the airwaves start buzzing for the next one week or thereabout with the analysis of the results and the possible culprits. While some blame the students, others put the blame on the teachers. A few point fingers at the parents and others on the government. A very little group tends to make any negative comment about the school owners and the society.

The puzzle is: Who truly deserves the blame? I mean who is responsible for this rot in our educational system. Thank God, of all the groups listed above, none of them are known to be spirits. So, unlike Ola Rotimi's, 'THE GODS ARE NOT TO BLAME', someone or group(s) of people have to be culpable in this case. In an attempt to answer this question, and as a stakeholder in this sector, I had to engage other stakeholders in meaningful discussions; from which I was able to put the pieces together to help you solve the puzzle, by looking at the roles each group plays in the departure of the glory of our educational system.

Students
Most of our students today suffer from what I call the 'Misplaced Priority Syndrome' (MPS). They exhibit a nonchalant attitude towards their studies, and usually want to be spoon-fed. They hardly perform their tasks by themselves, and are always seeking shortcuts. I call them the Xerox machines, because they are always waiting to 'photocopy' anything that seems like the solution to a task without necessarily thinking through its reasonability. Some of them are so daft that they can even copy their benefactors' details alongside the solution.

This group of students prefer to waste their time watching films, soaps and musical videos on TV. No thanks to Telemundo, Soundcity, MTV etc. They can memorise the names of 20 actors and 3 films or series each of these actors has featured in, the year of release of those movies, and also narrate everything that happened. Also, they can mime the contemporary hip-hop and R&B songs from A to Z with the names of the artistes. However, tell them to recite the 7-times table or ask them to read and summarise the 'Village Boy' in just 300 words...verily, I say unto you, it will be easier for a camel to pass through the eye of a needle, than for an average Nigerian student to do any of these.

Besides music and films, they are also into gaming. They spend their days playing different types of games that add no value to their lives. To worsen the issue, they are now easily distracted by the social media. They can spend hours on the internet gossiping, posting pictures and getting themselves indoctrinated on the social media by one radical group or another, at the expense of their studies. So, tell me how they will pass their exams!

Make no mistake about this, all work and no play makes Jack a dull boy, but all play and no work makes Jack a foolish toy!

Teachers
Teachers are the custodians of knowledge and they have the burden of imparting same to the students. A good teacher is expected to possess the following qualities, though not necessarily limited to these:

a) Sound knowledge and understanding of the subject
b) Smart and hardworking
c) Ability to express himself/herself fluently in spoken and written English
d) Good class management, teaching and interpersonal skills
e) Disciplined

Teaching is a vocation, however, many rush into it out of frustration. No thanks to the current economic situation! It goes beyond dishing out notes, giving assignments, and marking them. It requires a follow-up or a follow-through to ensure the students do and know what is expected of them. It involves being interested in the well-being of your students and inspiring them to be the best they can be. Good teachers are like demigods before their students. Their words are like laws to those students. These teachers know their onions, which makes it easy for them to positively influence their students.

Unfortunately, majority of our teachers today suffer from the 'NEMO DAT QUOD NON HABET' syndrome. How can they give what they don't have?...and this has a direct implication on the standard of education. From the feelers I get, these teachers have low self-esteem and a poor understanding of the subject. During classes, if not for the uniforms, you would hardly differentiate them from the students because of their poor class management skills. The teaching atmosphere is like a market, which makes me wonder how the students assimilate under such condition.

Surprisingly, this group of teachers are always the first to complete their schemes of work without considering the students' understanding of the contents of the schemes. They show little or no interest in the students' welfare forgetting that the latter tend to show more interest in a subject, if they know that the teacher cares about them.

If you carry out a random check when classes are going on in most schools, the number of errors you will discover will baffle you. Such errors range from grammatical to lack of understanding of the subject matter. I have overheard students ask themselves, "If the ones, who are entrusted with the responsibility of correcting our mistakes, are the ones making these mistakes, what then will become of us?"

Don't get me wrong, I am not saying a teacher must be perfect. No!, far from it; but what will you say, if four out of ten sentences made by a teacher are wrong! Cases are abound where engineering graduates could not explain or solve simple quadratic equations at interviews for teaching jobs. Or, what do you have to say about an English Master degree holder, who claimed that adverb and verb have the same meaning. According to her, one can be used in place of the other...Huh! From the information I gathered, most of these people go to these interviews as experienced teachers, which leaves one pondering, who their former or current employers are!

Private School Entrepreneurs
The essence of private schools system was to make up for the inadequacies of the public schools system, which include lack of highly qualified and competent teachers, high students-to-teacher ratio, lack of functional facilities etc. Apart from a handful of them that are living the standards expected of them, others are nothing more than glorified public schools.

Their owners are more of businessmen than educationists. Most of them do not understand what it means to run an educational organisation. They know little or nothing about educational quality control and assurance. They pay little or no attention to the quality of materials or documents sent out by the school. These materials are always true reflections of the standard of the schools. My heart bleeds whenever I go through schools' newsletters or report cards bastardized with grammatical errors. I often ask rhetorically, whether such schools have competent English tutors, and what they teach their students. In some cases, those letters are written by the English teachers...which brings us to the same question: Who recruited them?

These businessmen are more interested in the population of students in their schools than the quality of education acquired by the students. That is why they encourage and indulge in every form of examination malpractice just to achieve that goal. Most of them mandate the teachers to deliver a certain class of result in the external examinations, by hook or crook. This promotes laziness among the students and teachers alike, because as a student, if I know that my teachers will be there to provide the answers during the examinations, why will I waste my time studying for it? And as a teacher, if I know that despite my efforts in class, I will still have to teach the students in the examination hall, why should I stress myself in class? I would rather save that energy for the examination hall jooor!

In fact, the level of malpractice in these so-called private schools supersedes that obtained in public schools. The results are used as marketing tools to attract gullible parents. Meanwhile, the students who claim to own these results can hardly defend them.

In most private schools, window dressing is their specialty. They are poorly equipped for any standard and effective teaching to take place. Yet, they are always given a pass mark during supervision by officials of the Ministry of Education. In fact, it is a crime for any student to repeat a class in some private schools. Is it possible for all the students in a school to be above average or of average performance? Obviously not! The fear of losing the students will make them pad their scores and make their report cards look appealing to the eyes of the foolish parents, who are always against their wards repeating a class. Also, they will never write the true comments about the children's characters in school, just to earn continued patronage from the unsuspecting parents. If these students do not repeat and get to understand what is expected of them in the lower classes, how do you expect them to cope in the upper classes?

Through these acts, they indirectly encourage indiscipline in these schools by promoting a student's world. They tend to attach more importance to the students than the teachers forgetting that without good teachers, you cannot have excellent students.

To crown it all, they offer poor staff welfare packages and working conditions. Firstly, they overload the teachers with multiple subjects. Imagine one teacher teaching three to four subjects, how do you expect such a person to be effective? Secondly, the teachers are paid peanuts, despite one person doing three people's jobs. To worsen it, those peanuts are not paid regularly, because staff salary is not of high priority to them. How do you expect to get the best out of an individual, when you are not willing to give your best? At the end of the day, the students suffer! Those of them that try to pay anything meaningful, are slave-drivers, who expect you to live and die in the school. You do not have time for any extracurricular activities apart from school activities. Some even demand that you be in school on Sundays....FOR WHAT NOW?!

Parents/Guardians
Parents are the first teachers of a child. They are the ones, who are expected to teach the students about values, discipline and dignity of labour. It is the duty of the parents to ensure their wards do their assignments. In achieving this, they are expected to guide them and not do it for them. They are meant to closely monitor the activities and progress of their children in school by regularly checking their notes (to ensure they are updated) and results (to know when the children are backsliding or losing focus), and query any discrepancies, if the need arises.

However, most parents today, like their wards, do not know how to set their priorities right. Some of them relegate their children's education to the background. They do not pay their school fees on time, but prefer to spend their money on wears for parties and expensive gadgets. Others chase money at the detriment of their wards' well-being. They continually hide under the deception that money is the answer to all things. To them, their responsibilities in their children's education start and end in the payment of school fees. They do not even have the time to go through their ward's notebooks or engage them at home to test their understanding of the subjects. They rarely make themselves available for the parent-teacher fora. There are cases of parents, who do not bother to go through their children's report cards throughout their stay in a particular institution? How do you expect such students to give their best, when there is nothing or no one to push them? Some of them can hardly control these children at home; yet, they would expect the schools to wield a magic wand because they pay their wards' fees. From what I gathered, these parents, who expect this magic, are the same people who cry wolf, if their children are disciplined.

Studies have shown that the academic performance of students is directly proportional to their mental maturity, of which age is a factor (though, there are a few exceptions). The rate at which children are being rushed through school these days is quite alarming. In our days, the average age of admission into secondary school was 12, but today, it is 9. Many students gain admission into secondary schools at age 8, which means that they will be 11 years old, in SS1. The question now is: how many of them at that age can easily comprehend the concepts of chemistry, physics, biology, government etc? Yet, their parents seek double promotions for them in their primary education, even when they do not deserve a single promotion! What is the sense in making a child graduate at 14, only to make him/her wait at home for one or more years before gaining admission into the university because he/she is underage?...i.e, if and only if, the student got the prerequisite O' level result at his/her first sitting!

Society
The society does not help matters. It offers the largest forms of distraction, which requires a highly disciplined child to stay focused. The airwaves reverberate more with entertainment and fashion, and less with education. Much emphasis is laid on success without stress, when little value is placed on industriousness and dignity of labour. The corporate world dole out millions of naira as prizes in talents hunts and fashion shows, compared to those in educational competitions. I have never heard of any quiz or essay-writing competition in Nigeria, where the winner was awarded one million naira or given a car; or have you? So, how can you convince and encourage the younger generation to focus on their education, when they see dropouts celebrated more than their educated counterparts on a daily basis?

Government
This is the supposed 'big daddy', who has failed in his role to provide standard infrastructures and make policies that will enhance the development of the sector.

Hope Waddell, like other public schools, is bigger than two communities put together in terms of land mass. It had a standard and two semi-standard football fields, a basketball court, a chapel, an assembly hall, blocks of storey buildings used as classrooms, dormitories and refectory, and a large expanse of land as staff quarters. We never had any reason to take our inter-house sports competitions and other extracurriculars outside the school compound. Our labs were as big as three 12x12 bedrooms, combined together and also well-equipped.

Our teachers were topnotch. In fact, the man, who taught us mathematics in JS2, was an Indian - Mr. Rajah Panchalingam; while his father, Mr. V. C. Panchalingam, who was the head of the science department, taught us physics in SS3. This is without prejudice to the fantastic indigenous teachers like Mrs Idy Akubuiro, our chemistry teacher and Mrs Mfon, our SS1-2 physics teacher, to mention but a few. No teacher was told to handle more than two subjects, if at all. The teachers were enthusiastic about their jobs, because the conditions of service were quite friendly. We had quite a number of foreign students from the neighbouring West African countries like Cameroon, Ghana, Liberia, Sierra Leone etc as schoolmates. The governors', commissioners', directors' and other public officials' children were also our school and classmates. In fact, then, it seemed like a 'taboo' to attend a private school.

As students, we were exposed to all disciplines in our lower senior secondary. As science students, we offered Literature-in-English, Igbo language and Economics; while our arts and commercial students offered chemistry and biology in SS1. The curriculum was very comprehensive and the policies were consistent.

All these things are now history because the government has failed in its responsibilities. The educational sector is completely bastardized. The system is now suffering from infrastructural decay and inconsistent policies. Most public schools are now eyesores with dilapidated roofs and buildings, and abandoned projects. Instead of completing or rehabilitating these structures, the government officials will rather wish they collapse completely, as this will favour them and their cohorts, who are busy setting up private schools with looted funds.

Ministers of Education now want to satisfy their egoistic tendencies by introducing policies, without considering their pros and cons. Students are now confused about their subject combinations. Biology and agricultural science are no longer general subjects. Even the science students cannot offer them with further mathematics or geography at the same time, because they are no longer considered as regular science subjects. New subjects are being introduced and tested without the availability of trained professionals in those disciplines. How then can the students excel?

Way Forward
From the above analysis, you will agree with me that the task of restoring the lost glory of our educational system is a collective responsibility. We are all culpable for its deplorable state, because all of us are stakeholders in the business of education. If you are not a student, then, you are a teacher; otherwise, you are a parent or guardian. If you do not belong to any of these categories, then, you must be a government official or a school owner. If you are none of these, then you must be working in the corporate world.

It is time to make a pragmatic shift from merely paying lip service to a conscious attempt of taking the Nigerian education to its once enviable position in the league of nations. The government should increase the annual budgetary allocation for education to between 16-21%, as recommended by UNESCO, the arm of United Nations that oversees education, science and culture. They should make the public schools to compete favourably with the top private schools, in terms of facilities and staff welfare. This can be achieved by passing a law mandating all children of public office holders to attend public schools.

In addition to the government increasing their supervision on private schools, I would recommend that the sector be sanitized through the introduction of mergers & acquisitions (M&A) to allow real professionals run the business of education the way it should be. Alternatively, all schools should be categorized as Grade A, B or C (renewable biennially) based on the available facilities and quality of teachers; with the high-end institutions regarded as Grade A, the middle-class as Grade B and the low-end schools classified as Grade C. These grades will be boldly printed on every signage and document bearing the name of the school. By so doing, unsuspecting parents will be able to make informed decisions before patronizing any school.

Our teachers should be made to undergo regular and effective developmental trainings. They should also carry out more personal developments on their own to ensure they have adequate knowledge needed to excel in their respective disciplines. Less emphasis should be laid on paper qualifications, and more on competencies. The idea of the rewards of teachers being in heaven should be abolished. They should be made to enjoy the fruits of their labour right here on earth as obtained in other climes. We should learn to appreciate those good ones for their selflessness in laying the right foundations for us to build on.

Parents should make their children's education of utmost priority by paying their school fees regularly. They should endeavour to follow up on their progress in school, by ensuring that the children do their homework and partake in other school projects regularly. They should create time to attend any forum organised by the schools, which will enhance the children's academic performance. These fora give them the opportunity to receive firsthand information about the growth of their wards academically.

Parents should constantly ask for their wards' report cards and query any observed drop in performance. They should closely monitor the kinds of friends they keep, to ensure that such companies are value-adding. Through this, the children will be kept on their toes, when they know the expectations of their parents. Lastly, they should ensure that the homes are peaceful and conducive for studying, as a troubled or divided home causes the child to be psychologically imbalanced.

The corporate world should support and make education enticing by building and equipping more facilities across public schools and increasing the monetary rewards in educational competitions to match those in entertainment. They should try and encourage creativity in science and arts (not singing and dancing only) by funding innovative ideas. I know a few of them, which are trying, but they can do more!

Our media houses should produce more creative and educative programmes, and give them more airtime than what they currently do. More celebrities should be made out of individuals in the educational field, because the youth want to hear more inspiring stories or news about educated minds. They should play down the get rich quick syndrome and let the young minds understand that hard work generally pays.

Finally, the students should start doing what is right by showing more interest in their studies. They should spend their times on matters that will increase their intellectual capabilities, than on trivial issues. They should learn to be independent when carrying out any tasks and stop waiting to be spoon-fed. By so doing, they will be taking responsibilities for their actions...and if all these are achieved; verily, verily, I say unto you, we shall once again witness the glory of the latter days.

Electrode Potentials: Calculations

2016-02-28T22:31:00.001+01:00

In our last post, we looked at the overview of electrode potentials, where we discussed metal ions/metal systems or half-cells, standard electrode potential and electrochemical cells in depth. Here, we will focus on the calculations involving electrode potentials, which include calculations of the electromotive force (e.m.f) of electrochemical cells, the relationship between e.m.f & free energy and the relationship between e.m.f & equilibrium constant.
Half-Cell Reactions                      Std Reduction Potential, E° (V)
K+(aq) + e- <----> K(s)                              -2.92
Ca2+(aq) + 2e- <----> Ca(s)                     -2.87
Na+(aq) + e- <----> Na(s)                         -2.71
Mg2+(aq) + 2e- <----> Mg(s)                   -2.37
Al3+(aq) + 3e- <----> Al(s)                       -1.66
Zn2+(aq) + 2e- <----> Zn(s)                     -0.76
Fe2+(aq) + 2e- <----> Fe(s)                      -0.44
Sn2+(aq) + 2e- <----> Sn(s)                     -0.14
Pb2+(aq) + 2e- <----> Pb(s)                     -0.13
2H+(aq) + 2e- <----> H2(g)                         0.00
Cu2+(aq) + 2e- <----> Cu(s)                     +0.34
Fe3+(aq) + e- <----> Fe2+(aq)                 +0.77
Hg2+(aq) + 2e- <----> Hg(s)                    +0.79
Ag+(aq) + e- <----> Ag(s)                         +0.80
Au3+(aq) + 3e- <----> Au(s)                    +1.50
Non-metals
I2(s) + 2e- <----> 2I-(aq)                          +0.54
Br2(l) + 2e- <----> 2Br-(aq)                     +1.09
Cl2(g) + 2e- <----> 2Cl-(aq)                     +1.36
F2(g) + 2e- <----> 2F-(aq)                        +2.87
The above table is the electrochemical or electromotive series of some elements showing their standard reduction potentials (SRP), E° in volts. The half-cell reactions represented above are the reduction half-reactions and the reverse reactions will give us the oxidation half-reactions. Similarly, the electrode potential of the reversed reaction will become the standard oxidation potential with an opposite sign.
If we take a closer look at the metallic series, it will be observed that the oxidizing ability or tendency of ions to gain electrons increases downward and decreases upward; while the reducing ability or the tendency of atoms to lose electrons decreases downward and increases upward. This implies that those elements lower in the series (especially those with positive SRP values) will make good cathodes, while those above (especially the ones with negative SRP values) can act as good anodes in electrochemical calls.
For the non-metallic series, the oxidizing ability or the tendency of atoms to gain electrons increases downward, while the reducing ability or the tendency of ions to lose electrons increases upward.
Electromotive Force (e.m.f)
The e.m.f of a cell, E°cell, is a measure of the driving force with which current flows out of the cell. It is the potential difference between the two electrodes when no current is flowing.
Recall the electrochemical cell made up of a Zn2+(aq)/Zn(s) system and a Cu2+(aq)/Cu(s) system, where the zinc electrode is the anode and the copper electrode, the cathode. The cell can be represented symbolically by:
Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
An electrochemical cell can be generally represented by:
Anode(s)/Anode(aq)//Cathode(aq)/Cathode(s)
where the single slash (/) represents the interface between the electrode and its electrolyte, and the double slash (//) represents the salt-bridge. The left-hand side of the salt-bridge represents the oxidation half-reaction, while the right-hand side represents the reduction half-reaction.
Recall that the e.m.f of the cell is given as:
        E°cell = E°cathode - E°anode ...................(i), or
        E°cell = E°reduction - E°oxidation ...............(ii)
Equations (i) and (ii) are used if the standard electrode potentials of the electrode are provided as standard reduction potentials (as in the table above).
However, if the standard electrode potentials are given as standard oxidation potentials (SOPs), i.e, based on the reverse reactions of the half-cell reactions in the table, then equations (iii) and (iv) below are applied.
        E°cell = E°anode - E°cathode ...................(iii), or
        E°cell = E°oxidation - E°reduction ...............(iv)
Example 1
If the electrode potential of copper is +0.34V and that of zinc is -0.76V, calculate the e.m.f of the cell.
Answer
From the above question, zinc will always act as an anode in the presence of copper, and the values of the standard electrode potentials provided are the SRPs, so we apply equation (i) or (ii)
      E°cell = E°cathode - E°anode
                 = +0.34 - (-0.76)
                 = +0.34 + 0.76
                 = +1.1V
It is important to note that an electrochemical cell with a large e.m.f can be produced by using metals that are far apart on the electrochemical series as the electrodes.
Example 2
Calculate the standard cell potential of each of the following cells:
a) Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
b) Mg(s)/Mg2+(aq)//Ca2+(aq)/Ca(s)
Answer
a) Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)
From the above representation, we can infer that the zinc electrode is the anode (since its half-cell is on the left-hand side of the salt-bridge), while the copper electrode is the cathode.
Therefore, using the values of their respective SRPs (Zn = -0.76V, Cu = +0.34V), the solution will be the same as that for Example 1 above.
b) Mg(s)/Mg2+(aq)//Ca2+(aq)/Ca(s)
From the above representation, the magnesium electrode is the anode, while the calcium electrode is the cathode. So, using their SRPs and applying equation (i) or (ii), we will obtain:
      E°cell = E°cathode - E°anode
                 = -2.87 - (-2.37)
                 = -2.87 + 2.37
                 = -0.5V
Example 3
Given the following reactions:
Fe(s) ----> Fe2+(aq) + 2e-; E° = +0.44V
Ag(s) ----> Ag+(aq) + e-; E° = -0.80V
Calculate the e.m.f of a cell that can be produced from the combination of both reactions.
Answer
The above reactions are oxidation reactions, hence, the standard electrode potentials provided are the SOPs. Also, based on their positions in the electromotive series, the iron electrode will act as the anode, while the silver electrode will serve as the cathode.
Hence, applying equation (iii) or (iv) gives:
      E°cell = E°anode - E°cathode
                 = +0.44 - (-0.80)
                 = +0.44 + 0.80
                 = +1.24V
It is important to note that in problems involving two half-reactions of any kind, it is always necessary to determine the number of moles electrons involved in the redox reaction by balancing the electron loss and gain, though, it is does not affect the calculation of the e.m.f
So, considering the reactions above, we will have:
Fe(s) ----> Fe2+(aq) + 2e-; E° = +0.44V (oxidation half-equation)
2Ag+(aq) + 2e- ----> 2Ag(s); E° = +0.80V (reduction half-equation)
Fe(s) + 2Ag+(aq) ----> Fe2+(aq) + 2Ag(s) (overall/net reaction)
Given the above, the e.m.f of the cell can also be calculated based on the net reaction as:
        E°cell = E°reduction + E°oxidation ..........(v)
                   = E°cathode + E°anode ..........(vi)
                   = +0.80 + 0.44
                   = +1.24V
From the above, it can be observed that the standard electrode potential for the anodic/oxidation half-reaction is the standard oxidation potential, while that of the cathodic/reduction half-reaction is the standard reduction potential.
Therefore, when the electrode potentials provided are for the respective redox processes, (i.e, SOP for the oxidation half-reaction, and SRP for the reduction half-reaction), we add both values to obtain the e.m.f of the cell.
Relationship between e.m.f of a Cell, E°cell, and Change in Gibb's Free Energy, DG
Gibb's Free Energy (G), is a measure of the spontaneity of a process. It is an indication that a process will proceed on its own without the interference of an external factor. Examples of spontaneous processes include:
a) the melting of ice
b) the crying of a new born baby
c) the flow of water from a highland to a lowland
d) the flow of electrons from the anode to the cathode
e) the flow of current from a region of higher potential to that of lower potential
To show that a process is spontaneous and feasible, the value of DG (read as delta G or change in free energy), must be negative (DG < 0).
The relationship between free energy and the e.m.f of a cell is given as:
                  DG° = -nFE°cell .................(vii)
           where n = number of electrons transferred in the redox reaction
                      F = Faraday's constant
                         = 96500Cmol^-1
              E°cell = standard e.m.f of the cell
From equation (vii) above, for the process to be spontaneous, the e.m.f of the cell must be positive.
Therefore, using the e.m.f values of the electrochemical cells in our examples above, let us see how to determine the spontaneity of a redox process.
Example 1.1
If the electrode potential of copper is +0.34V and that of zinc is -0.76V, calculate the e.m.f of the cell. Hence or otherwise, determine the spontaneity of the cell.
Answer
From our previous solution, the e.m.f, E°cell = 1.1V
Zn(s) ----> Zn2+(aq) + 2e- (oxidation half-equation)
Cu2+(aq) + 2e- ----> Cu(s) (reduction half-equation)
Zn(s) + Cu2+(aq) ----> Zn2+(aq) + Cu(s) (overall/net reaction)
From the above reactions, the number of moles of electrons transferred = 2 moles
Therefore, substituting these values into equation (vii), we will obtain:
        DG° = -nFE°cell
        DG° = -2 x 96500 x 1.1
                = -212300 J
                = -212.3 kJ
Since DG° < 0, it means the redox reaction is very spontaneous (due to the high negative value) and the cell will work as arranged. That is to say, electrons will flow from the zinc electrode (anode) to the copper electrode (cathode).
Example 2.1
Calculate the standard cell potential of the cell below and comment on the cell reaction:
Mg(s)/Mg2+(aq)//Ca2+(aq)/Ca(s)
Answer
From the earlier solution, the e.m.f, E°cell = -0.5V
Mg(s) ----> Mg2+(aq) + 2e- (oxidation half-equation)
Ca2+(aq) + 2e- ----> Ca(s) (reduction half-equation)
Mg(s) + Ca2+(aq) ----> Mg2+(aq) + Ca(s) (overall/net reaction)
Number of moles of electrons transferred = 2 moles
Therefore, substituting these values into equation (vii), we will obtain:
        DG° = -nFE°cell
        DG° = -2 x 96500 x -0.5
                = +96500 J
                = +96.5 kJ
From the positive value of DG (DG > 0), it means the redox reaction is not spontaneous, and the cell will not work. In other words, magnesium cannot act as an anode in the presence of calcium, because electrons will never flow from the former to the latter by virtue of their positions in the electromotive series. Therefore, to make the cell work, the electrodes must be rearranged.
Ca(s)/Ca2+(aq)//Mg2+(aq)/Mg(s)
      E°cell = E°cathode - E°anode
                 = -2.37 - (-2.87)
                 = -2.37 + 2.87
                 = +0.5V
Hence,
         DG° = -96.5kJ
Example 3.1
Given the following reactions:
Fe(s) ----> Fe2+(aq) + 2e-; E° = +0.44V
Ag+(aq) + e- ----> Ag(s); E° = -0.80V
Determine whether the electrochemical cell will work based on the above arrangement.
Answer
From the earlier solution,
the e.m.f of the cell, E°cell = +1.24V
the number of moles of electrons transferred = 2 moles
But,
        DG° = -nFE°cell
Hence,
        DG° = -2 x 96500 x 1.24
                = -239320 J
                = -239.3 kJ
Since DG < 0, it means the redox reaction is spontaneous, and the electrochemical cell will work based on the stated arrangement.
Relationship between the e.m.f of a Cell, E°cell & Equilibrium Constant, K
Consider a redox reaction that is at equilibrium and represented by the equation:
aA(s) + bB+(aq) <----> aA+(aq) + bB(s)
The equilibrium constant, Keq, of the reaction will be given as:
Keq = [Products]^n/[Reactants]^m
Keq = [A+(aq)]^a.[B(s)]^b/[A(s)]^a.[B+(aq)]^b
but,
The concentration of a solid substance at equilibrium is always taken as a unitary value(i.e, 1) in the presence of aqueous substances.
So, the above equation reduces to:
            Keq = [A+(aq)]^a/[B+(aq)]^b ............(viii)
            Keq = [Anode(aq)]^a/[Cathode(aq)]^b ............(ix)
The change in Gibb's free energy, DG, is related to the equilibrium constant, K, according to the equation:
            DG° = -RT ln Keq .............................(x)
(but, ln (natural log) = 2.303 log)
            DG° = - 2.303 RT log Keq ...............(xi)
Substitute equation (ix) into (xi)
DG° = - 2.303 RT log [Anode(aq)]^a/[Cathode(aq)]^b ............(xii)
Equate (vii) and (xii)
-nFE°cell = - 2.303 RT log [Anode(aq)]^a/[Cathode(aq)]^b
Divide both sides by -nF
E°cell = (2.303RT/nF) log [Anode(aq)]^a/[Cathode(aq)]^b ..........(xiii)
where R = gas constant (8.314 J mol^-1 K^-1)
            T = standard temperature in Kelvin (25°C = 298K)
            n = number of moles of electrons transferred
            F = Faraday's constant
            a = number of moles of the anodic electrolyte
            b = number of moles of the cathodic electrolyte
[Anode(aq)] = concentration of the anodic electrolyte
[Cathode(aq)] = concentration of the cathodic electrolyte
Equation (xiii) gives the relationship between the e.m.f of a cell and the concentrations of the electrolytes at standard conditions.
Recall that in the course of using an electrochemical cell, its e.m.f decreases. This is due to a decrease in the concentrations of the electrolytes. Therefore, we can calculate the e.m.f of a cell (or the electrode potential of an electrode) at any concentration other than the standard concentration of 1.0 mol dm^-3, by using the equation:
Ecell = E°cell - (2.303RT/nF) log [Anode(aq)]^a/[Cathode(aq)]^b ...
.....(xiv)
The above equation is known as the Nernst equation, and it shows the relationship between the e.m.f of a cell (at non-standard concentration), Ecell, the standard e.m.f of the cell, E°cell and the concentrations of the electrolytes.
Example 4
Calculate the cell e.m.f of Al(s)/Al3+(aq)(0.010 mol dm^-3)//Cu2+(aq)/Cu(s)(0.100 mol dm^-3) at 25°C [E°al = +1.66V, E°cu = +0.34V]
Answer
From the arrangement, aluminum is oxidized (anode) and copper is reduced (cathode). So, the given standard electrode potential for Al is the SOP, while that for Cu is the SRP.
2Al(s) ----> 2Al3+(aq) + 6e- (oxidation half-reaction)
3Cu2+(aq) + 6e- ----> 3Cu(s) (reduction half-reaction)
2Al(s) + 3Cu2+(aq) ----> 2Al3+(aq) + 3Cu(s) (overall redox reaction)
Also, the concentrations of the electrolytes indicate that the required cell e.m.f is Ecell, and not the standard cell e.m.f, E°cell. Therefore, we apply the Nernst equation:
Ecell = E°cell - (2.303RT/nF) log [Al3+(aq)]^2/[Cu2+(aq)]^3
but,
E°cell = E°cathode + E°anode
           = + 0.34 + 1.66
           = + 2.00V
Number of moles of electrons transferred, n = 6 mol
Hence,
Ecell = 2.00 - (2.303x8.314x298/ 6x96500) log (0.010)^2/(0.100)^3
         = 2.00 - (5705.8483/579000) log (1x10^-4/1x10^-3)
         = 2.00 - 0.009855 log 10^-1
         = 2.00 - 0.009855(-1)
         = 2.00 + 0.009855
         = 2.009855
         ~ 2.01V
Do These:
Question 1
Calculate the e.m.f of the cell:
Zn(s)/Zn2+(aq)(0.002 mol dm^-3)//Cu2+(aq)/Cu(s)(0.02 mol dm^-3) at 25°C [E°zn = +0.76V, E°cu = +0.34V]
Question 2
What will be the concentration of the electrolyte in the copper half-cell in Example 4 above, if the e.m.f of the cell, Ecell, is 2.06V and the concentration of the electrolyte in the aluminium half-cell is 0.001mol dm^-3 at 25°C?
Question 3
If the overall reaction in an electrochemical cell is given by:
Mg(s) + Fe2+(aq) ----> Mg2+(aq) + Fe(s), calculate the standard e.m.f of the cell.
Question 4
An electrochemical cell was constructed by combining the two half-cells below:
Cu2+(aq) + 2e- ----> Cu(s); E° = +0.34V
Fe2+(aq) + 2e- ----> Fe(s); E° = -0.44V
What is the standard e.m.f of the cell?
Question 5
Ca(s) ----> Ca2+(aq) + 2e-; E° = +2.87V
2Ag(s) ----> 2Ag+(aq) + 2e-; E° = -0.80V. Calculate the standard electrode potential of the calcium-silver cell represented by the half-cell reactions above.
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Electrode Potentials: Overview

2016-02-25T08:34:00.001+01:00

Metal Ions/Metal Systems
If a metal plate or rod is placed in a solution of its own salt, two processes - oxidation and reduction, take place simultaneously. The atoms from the metal's surface, lose electrons and go into solution as ions, as the electrons lost are left on the surface of the metal plate, thereby making it electrons surplus and negatively charged.
                   M(s) ----> M+(aq) + e-
Conversely, the metallic ions in the salt solution gain electrons from the surface of the metal plate and get deposited as metallic atoms. The electrons gained from the surface of the electrode renders it electron deficient and positively charged.
                    M+(aq) + e- ----> M(s)
Depending on the nature of the metal, a particular process will predominate. Hence, a potential difference, known as the electrode potential for the metal ions/metal system [M+(aq)/M(s)] is established between the electrode and the electrolyte.
For instance, if a zinc electrode is placed in a solution of zinc tetraoxosulphate (VI) solution, ZnSO4, the zinc atoms from the electrode, spontaneously lose electrons and go into solution as zinc ions, Zn2+. Hence, the zinc electrode becomes negatively charged on the surface due to excess electrons, while the solution becomes positively charged due to the presence of excess Zn2+ in it. A potential difference, known as the electrode potential of zinc, is therefore set up between the zinc electrode and the solution.
As this continues, the zinc rod keeps diminishing in size until all the metallic atoms are converted to ions.
Zn(s)          ---->              Zn2+(aq)   +   2e-
electrode                          solution
(decreases in size)
Similarly, if a copper electrode is dipped into a solution of copper (II) tetraoxosulphate (VI), CuSO4. The copper (II) ions, Cu2+, from the solution pick up electrons from the copper rod and get deposited as metallic copper atoms on the electrode. Consequently, the copper rod becomes positively charged on the surface due to a deficit of electrons, while the solution becomes negatively charged because of the excess tetraoxosulphate (VI) ions in solution, SO4--.
Also, as this continues, the electrode keeps increasing in size until all the metallic ions in the solution have been converted to metallic atoms.
Cu2+(aq)   +   2e-   ---->        Cu(s)
solution                                   electrode
                                           (increases in size)
The two systems discussed above are known as half cells. If the two electrodes are connected to a galvanometer, by means of a wire, it would be observed that current flows from the zinc electrode to the copper electrode. The zinc electrode serves as the negative terminal or the anode, while the copper electrode serves the positive terminal or the cathode. This is because zinc, which is more electropositive than copper, will always get oxidized by losing electrons in the presence of copper, while the latter will always gain electrons and get reduced in the presence of zinc.
             Zn(s) ----> Zn2+(aq) + 2e- (oxidation half-equation)
             Cu2+(aq) + 2e- ----> Cu(s) (reduction half-equation)
Zn(s) + Cu2+(aq) ----> Zn2+(aq) + Cu(s) (overall equation)
The set-up consisting of two half-cells with different electrode potentials is discussed further under electrochemical cell.
Standard Electrode Potential
The electrode potential of a given metal ions/metal system depends on three factors, namely - the overall energy change, the concentration of ions in solution and the temperature. Due to these, the standard electrode potential is always used to compare the electrode potential of two or more metal ions/metal systems.
The standard electrode potential of a metal, E°, is the value (based on the hydrogen scale) recorded by a voltmeter, when its electrode potential is measured at standard conditions by connecting the metal electrode placed in 1M concentration of its salt solution, to a hydrogen electrode at 25°C (298K) and 1atm (760mmHg/101325Nm^-2), with the hydrogen electrode having an arbitrary value of 0.00V at all temperatures.
Depending on the nature and position of the metal in the electrochemical series, electrons will either flow from it to the hydrogen electrode or vice-versa. If the electrode potential is negative, it indicates that electrons flow from the metal electrode (anode) to the hydrogen electrode (cathode); and if positive, electrons flow from the hydrogen electrode (anode) to the metal electrode (cathode).
Therefore, the electrode potential of the metal ions/metal system can be calculated as:
E°cell = E°cathode - E°anode .......(i), or
E°cell = E°reduction - E°oxidation .......(ii)
where E°cell is the voltmeter reading or the electromotive force (e.m.f.) of the cell.
For instance, if a Zn2+(aq)/Zn(s) half-cell is used, electrons will flow from the zinc electrode to the hydrogen electrode, because zinc is higher than hydrogen in the electrochemical series, and as such, will be oxidized by losing electrons in the presence of hydrogen. The voltmeter records a value of 0.76V, which therefore gives the electrode potential of the zinc as -0.76V. As stated earlier, the negative sign indicates that the zinc electrode served as the anode and was oxidized, while the hydrogen electrode served as the cathode and was reduced. The two processes can be observed by the decrease in size of the anode and the liberation of hydrogen gas at the cathode according to the reactions:
Zn(s) ----> Zn2+(aq) + 2e- (anodic half-reaction/oxidation half-equation)
2H+(aq) + 2e- ----> H2(g) (cathodic half-reaction/reduction half-equation)
Zn(s) + 2H+(aq) ----> Zn2+(aq) + H2(g) (overall reaction)
This can also be confirmed from the calculation of the emf of the cell, by using equation (i),
E°cell = E°cathode - E°anode
0.76 = 0.00 - E°Zn
E°Zn = -0.76V
Or, by using equation (ii),
E° = E°reduction - E°oxidation
0.76 = 0.00 - E°Zn
E°Zn  = -0.76V
On the other hand, when a Cu2+(aq)/Cu(s) half-cell is used, electrons flow from the hydrogen electrode to the copper electrode because copper is lower than hydrogen in the electrochemical series. Therefore, the copper ions in the solution are reduced to metallic copper atoms in the presence of hydrogen. The voltmeter's reading would be seen to be 0.34V, from which the electrode potential of copper is obtained as +0.34V, which shows that the copper electrode served as the cathode, while the hydrogen electrode served as the anode.
H2(g) ----> 2H+(aq) + 2e- (anodic half-reaction/oxidation half-equation)
Cu2+(aq) + 2e- ----> Cu(s) (cathodic half-reaction/reduction half-equation)
H2(g) + Cu2+(aq) ----> 2H+(aq) + Cu(s) (overall reaction )
This can be confirmed by calculating the emf of the cell, by using equation (i),
E° = E°cathode - E°anode
0.34 = E°Cu - 0.00
E°Cu = +0.34V
Or, by using equation (ii),
E° = E°reduction - E°oxidation
0.34 = E°Cu - 0.00
E°Cu  = +0.34V
Electrochemical Cell
The study of electrode potential is a branch of electrochemistry that deals with the generation of electricity using chemical reactions. It is applied in the electrochemical cell, which works based on the difference in the standard electrode potentials of the electrodes in two half-cells.
An electrochemical cell is a device used to convert chemical energy into electrical energy. It generates electricity from chemical reactions. A group of electrochemical cells is known as a battery.
Salt-Bridge & Electrochemical Cells
Let us consider an electrochemical cell made up of the two half-cells - the zinc ions/zinc metal system and the copper ions/copper metal system.
In the Zn(s)/Zn2+(aq) half-cell, where oxidation takes place, the net electrical charge of the electrolyte is positive (Zn2+ > SO4-- ), while the net charge of the electrolyte is negative (SO4-- > Cu2+) in the Cu2+(aq)/Cu(s) half-cell, where reduction takes place.
Therefore, to take care of the imbalance in electrical neutrality between the two half-cells, a salt-bridge is usually used to connect both electrolytes. (Please refer to your textbooks for the diagram).
It is made up of a porous material, e.g. filter paper, saturated with a salt solution, say sodium chloride [NaCl(aq)]. As the electrolyte in the oxidation half-cell becomes increasingly positively charged, the presence of the excess Zn2+ ions are counterbalanced by the Cl- ions, which go into solution from the salt-bridge, when one end of the bridge is dipped into the electrolyte.
Also, the net negative charge of the electrolyte in the reduction half-cell, due to the excess SO4--, is neutralized by the introduction of Na+ ions from the salt-bridge, when its other end is dipped into the electrolyte. As some of the cations and anions from the salt-bridge enter the respective electrolytes, some of the excess cations and anions from the electrolytes move out of their respective solutions via the salt-bridge, thereby, establishing the electrical neutrality, and completing the electrical circuit.
Types of Electrochemical Cells
There are two types of electrochemical cells, namely primary cells and secondary cells.
Primary cells are non-rechargeable. When once the chemicals in the cells are used up, they cannot be regenerated and the electrochemical cells will stop discharging electric current. They will, therefore, need to be changed. Most primary cells are dry cells. Examples of primary cells are Daniell Cell, Leclanché Cell etc.
Leclanché Cell
The Leclanché cell is the commonly used battery in our torches, transistor radios, remote controllers and toys. An example is the Duracell battery.
A typical Leclanché cell is cylindrical in shape and is made up of a solution or paste of ammonium chloride as the electrolyte and carbon rod (graphite) as the cathode. Both are enclosed within a thin zinc sheet, which serves as the anode. In between the cathode and the electrolyte is a paste of manganese (IV) oxide, placed inside a thin film of muslin bag. (Please refer to your textbooks for the diagram).
At the anode, the zinc atoms lose two electrons each to dissolve in the electrolyte as zinc ions
Zn(s) ----> Zn2+(aq) + 2e-(oxidation half-reaction)
The electrons lost by the zinc atoms pass through an external circuit to do some work such as lighting-up a torch bulb, making an electronic bell or device work before reaching the cathode.
At the cathode, the ammonium ions accept the electrons to become ammonia and hydrogen gas. The hydrogen gas is removed by the manganese (IV) oxide, which prevents it from adhering to the cathode.
2NH4+(aq) + 2e- ----> 2NH3(g) + H2(g) (reduction half-reaction)
Zn(s) + 2NH4+(aq) ----> Zn2+(aq) + 2NH3(g) + H2(g) (overall reaction)
The cell stops working immediately the zinc anode or ammonium chloride electrolyte is used up; hence, the need for its replacement.
Secondary Cells are those that store electrical energy, when connected to an alternating current (a.c.) power supply, and discharge the stored energy as a direct current (d.c.), when made to work. That is why they are also known as storage cells. Unlike primary cells, they are rechargeable, because the chemicals in the cells can be regenerated whenever they are used up, through the process of electrolysis. They are made up of wet and dry cells. The wet cells have liquid electrolytes, while the dry cells have electrolytes in molten or pasty form. Examples of secondary cells include the lead accumulator, nickel or lithium-ion accumulator etc.
Lead Accumulator
In the lead accumulator, its anode is made of lead, while the cathode is made of lead (IV) oxide, PbO2; the electrolyte is made up of concentrated tetraoxosulphate (VI) acid, H2SO4. (Please refer to your textbooks for the diagram).
Being a secondary cell, the lead accumulator can be discharged and recharged. An example is the car battery.
Discharging
At the anode, the lead atoms undergo oxidation to form lead (II) ions, Pb2+, by losing two electrons each.
Pb(s) ----> Pb2+(aq) + 2e- (oxidation half-reaction)
Just like the primary cell, the electrons move through an external circuit and perform some work, such as powering the engine of a vehicle, or a mobile phone before arriving at the cathode.
At the cathode, the PbO2 undergo reduction to form Pb2+, by accepting two electrons each in the presence of the hydrogen ions, H+, from the electrolyte.
PbO2(s) + 4H+(aq) + 2e- ----> Pb2+(aq) + 2H2O(l) (reduction half-reaction)
The Pb2+ ions produced at both electrodes combine with the tetraoxosulphate (VI) ions, SO4--, from the electrolyte to deposit lead (II) tetraoxosulphate (VI), PbSO4, which eventually covers the electrodes.
Pb2+(aq) + SO4--(aq) ----> PbSO4(s)
The PbSO4 and the water molecules produced in the half-reactions decrease the density of the acid until it falls below 1.16gdm^-3, due to the removal of the H+ and SO4-- ions from the electrolyte. Similarly, the e.m.f. of the cell drops to 1.8V.
Consequently, the cell stops discharging current because the chemicals have been used up, and need to be regenerated. (At this point, we say the battery has 'run' down and needs to recharged.)
Recharging
Here, the lead accumulator functions as an electrolytic cell. When connected to a power supply, the anode in the discharged cell becomes the cathode, while the cathode becomes the anode. This is because the anode and cathode in the discharged cell are connected to the cathode and anode of the power source respectively. (Refer to your textbooks for diagrams of the lead accumulator during discharging and recharging).
At the anode, the PbSO4 dissociates to form Pb2+ and SO4-- ions. These Pb2+ ions then combine with the water molecules to form PbO2 and H+ ions
                     PbSO4(s) ----> Pb2+(aq) + SO4--(aq)
Pb2+(aq) + 2H2O(l) ----> PbO2(s) + 4H+(aq) + 2e- (oxidation half-reaction)
At the cathode, the PbSO4 dissociates in the same way as it does at the anode. However, the Pb2+ ions produced undergo reduction by accepting two electrons each and are deposited on the cathode as lead metal.
                 PbSO4(s) ----> Pb2+(aq) + SO4--(aq)
              Pb2+(aq) + 2e- ----> Pb(s) (reduction half-reaction)
The 4 moles H+ ions (produced in the oxidation half-reaction) combine with the 2 moles of SO4-- ions (produced from the dissociation of PbSO4 at both electrodes) to regenerate the tetraoxosulphate (VI) acid, H2SO4.
                 4H+(aq) + 2SO4--(aq) ----> 2H2SO4(aq)
                 2H+(aq) + SO4--(aq) ----> H2SO4(aq)
Consequently, the density of the acid in the electrochemical cell returns to the optimal level of 1.25gcm^-3 and its e.m.f. increases back to 2.2V, after recharging.
Similarities between Electrolytic and Electrochemical Cells
1. Both cells are made up of the anode, the cathode and the electrolyte.
2. In both cells, oxidation occurs at the anode, while reduction takes place at the cathode.
3. In both cells, electrons flow from the anode to the cathode.
Differences between Electrolytic and Electrochemical Cells
1. In an electrolytic cell, the anode is the positive terminal, while the cathode is the negative terminal; whereas, in an electrochemical cell, the anode is the negative terminal, and the cathode is the positive terminal.
2. An electrolytic cell is used to convert electrical energy into chemical energy, while an electrochemical cell uses chemical energy to generate electrical energy.
3. In an electrolytic cell, no salt-bridge is required, whereas, a salt-bridge is needed in an electrochemical cell.
4. In an electrolytic cell, there is only one electrolyte, while an electrochemical cell can contain more than one electrolyte.
5. In an electrolytic cell, the two electrodes are in the same compartment, while in an electrochemical cell, the electrodes may be in the same or different compartments.
6. An electrolytic cell is known as a voltameter, while an electrochemical cell is called a galvanic cell.
Do These:
Question 1
a) What do you understand as the electrode potential of a metal?
b) Describe the structure of the Leclanché cell and give the chemical reactions that occur in it.
Question 2
The reactions that occur inside a car battery are said to be redox. Explain this statement using suitable equations to support your points.
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Sight & Vision: Illusion and Mission

2016-02-04T07:36:00.001+01:00

By Charles Offiong
These are two words, whose meanings are usually literally misconstrued. In as much as they sound similar, there is a need to put them in the right perspective, so as to erase any form of ambiguity and avoid the trap of mediocrity.
While listening to a programme on a primetime radio station, an analogy was made between two professions, which perfectly describe these words. Using this analogy and other examples, I will endeavour to shed some light on the similarities, differences, and importance of sight and vision. What you will be reading are as seen from my own point of view, and you may not agree with them. However, your contention does not change the fact that there is a need to clearly define or separate these two terms.
According to the dictionary, sight is the ability to see or the perception or view of objects by the eyes. It can be said to be something seen or worth seeing, i.e., a spectacle. It also means having a mental view, opinion or judgment about something. My emphasis shall be on the physical definition of sight, which is driven by how good a person's eyes are.
Sightedness is limited by how far the eyes can see, subject to any obstacles within the line of vision. By line of vision, I mean the distance between the eyes and the object being looked at. In other words, if the eyes can ordinarily see a tree of about 5m tall standing at about 1km (1000m) away from them, and if there is an obstacle, say a tower of 25m in height located 5m away from the eyes in the direction of the tree; then, the eyes will not be able to see the tree because their sight has been restricted to a mere 5m by the tower.
In as much as it helps in reducing our 'dependence' on other people for our locomotion and other activities in life, sight is deceptive in some cases. This is because the things we see and fall in love with at a glance, most times, do not always turn out to be what they seem. This has led many into committing avoidable errors and has destroyed destinies.
Do you remember Lot, Abraham's nephew? He was a victim of not-all-that-glitters-is-gold. When it was time for him to be separated from his uncle, he allowed his sight to influence his decision. Looking at the lushness of Sodom and Gomorrah (S&G), he hastily chose to make it a dwelling place for himself and his family. The rest is history, as we know how he lost everything, including his wife, who, unfortunately, was also a victim of sightedness. Yes, she was the one, who became that rock of sodium chloride after looking back. The only things that Lot safely left S&G with, were his morally-bankrupt daughters, who eventually got him drunk and made him commit incest with them.
Many people, like Lot, jump boats in their carriers, relationships and businesses because of the lofty promises the enticers offer, only to find out that it is the same sh*t everywhere, and that the grass always looks greener on the other side. So, in cases like this, relying on what you see or perceive only is not enough to make such life-changing decisions, we also need the GOD factor; which is always present as that faint little voice within us.
A person, who is moved by what he sees, suddenly becomes handicapped, when there is a damage to his instruments of sight, the eyes. This is because sight, which is ephemeral, can only last as long as the eyes are in good condition. Any tragic occurrence to them can throw the individual into a void of darkness and 'helplessness'.
A typical example is Samson, who being born with so much grace and valour, failed to realise why he was created; simply because he was driven by sight. The things he was instructed not to do, were the very things he did and when the grace finally departed from him and his eyes plucked out, he became a helpless puppet in the hands of his enemies; until he realized his mistakes and cried out to God for a second chance. Though, his request was granted, we know how he ended because of Delilah.
Vision, on the other hand, is something imaginary that one thinks he sees. It is an ideal or a goal toward which one aspires. It is a creation of fancy.
From the above, it is obvious that we do not need our eyes to have a vision, but our minds because we can only imagine with our subconscious and not with our eyes. To be visionary, you must be 'abnormal', because most of the things you will do while trying to achieve that vision will be viewed as madness by those around you, especially those moved by sight. It is only in a state of 'insanity' that you will be able to wade through the flood of discouragements coming from the 'sane' members of your family, community, colleagues, friends and fellow students. To be a visionary means breaking away from the norm and living your dreams!
Sight and vision have a few things in common. Firstly, both involve the act of seeing. However, while one makes use of the eyes for seeing, the other sees with the mind. While the eyes can only see clearly during the day or at night in the presence of light, the mind keeps on imagining (seeing) things irrespective of the time or condition. Also, both can be lost by a loss in focus of the object or dream. The eagle is one bird blessed with the power of focus. When it sights a prey on the ground, from the top of a mountain or tree, it does not lose focus of it until it has made the kill. The eagle knows that a slight shift in focus can make it lose the prey and cannot predict when it will find the next one.
One interesting case that we can use to buttress this point is the story of Joseph. At a very young age, he had a vision of becoming a famous and successful leader. He never allowed the challenges he faced to distract him from realizing his dreams, even when he was offered the 'pleasure' of Madam Portiphar on a platter of gold. The circumstance Joseph overcame and got exalted to the palace (from the prison) was the same Samson fell for and got his destiny truncated.
Sight is impeded by obstacles, whereas, vision can only be stopped by the refusal of the individual to dream. In as much as a person will love to see so much with his eyes, he can only achieve so little in the presence of obstacles. In fact, most times, the eyes leave the object they are meant to see and magnify the obstacle. With the mind, the opportunities are boundless and the abilities are limitless; instead of obstacles, it sees stepping stones; instead of mountains, it sees plains because it works by faith! The mind calls unto those things which are not as if they were!
This was the difference between #Team-Joshua-and-Caleb and #Team-Other-Israelites when they were sent by Moses to go and spy the land of Canaan. While Joshua and Caleb saw the inhabitants as bread before the children of Israel, the others saw themselves as grasshoppers before the inhabitants of Canaan. In fact, they even magnified their status by calling them giants. How do you view your challenges in life? Are they as bread before you, or are you as grasshopper before them?
A man driven by sight is like a hunter, who lives for the now, and believes that 'tomorrow will take care of itself'. He sets the trap, and spends his time monitoring to see when it makes a catch. When a game is eventually caught, he is ready to consume everything with his 'friends and family'. He hardly thinks about preserving for the future, he believes that the games will always be available whenever he needs them. He knows nothing about the rainy day. On the other hand, a man of vision is like a farmer, who lives for the future. He is ready to spend his today to secure the future. He cultivates the land, sows the seeds, nurtures the plants and harvests them when mature. He sows a seed, and reaps a bunch of hundreds of seeds. Yet, he does not consume everything, as he preserves some for the next season and sells some in the market, if necessary. The hunter is shortsighted, while the farmer is foresighted.
This reminds me of the weird twins. Yes, you guessed right - Esau and Jacob! I call them weird because they are the only set of twins recorded to have fought for supremacy right from their mother's womb till their adulthood. I have never heard nor read of any other. Bros Esau, who walked by sight, was asked by his brother, Jacob - a man of vision, to sell his birthright to him, for a plate of porridge. He lived for the now, and what mattered to him was how to satisfy his hunger. Due to his shortsightedness, he could not understand the importance of a birthright, and as such, did not hesitate to sell his future for his stomach. The rest is history. At this point, I will like to ask a rhetorical question: Would that hunger have killed Esau, if he had chosen not to eat the porridge? Your answer is as good as mine.
My dear readers, it is important to note that true blindness is not the lack of sight, but the lack of vision. No wonder the Bible says, "My people perish for lack of vision" (paraphrased). Though, one is physically blind, as long as he can dream or allow his imaginations to run wild within his subconscious, then, he is illimitable. Being handicapped is not a crime, but it suddenly becomes one when it is used as an excuse for failure. There are individuals abound, both dead and alive, who were/are blind, but accomplished in their various endeavours. They include: Ray Charles, Helen Keller, Steve Wonders and our very own Cobhams Asuquo, to mention but a few. So, if the physically blind do not have any excuse to fail, what about you - the able bodied?
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Faraday's Laws of Electrolysis

2016-01-31T14:59:00.001+01:00

During electrolysis, substances like oxygen gas, chlorine gas, bromine etc. are liberated at the anode depending on the electrolytes used, while substances like hydrogen gas, copper, silver etc. are liberated or deposited at the cathode. The volume of gases liberated or mass of metals deposited depend on the amount of electricity that is passed, be it carried out on one or more electrolytes. These relationships were summarized by Michael Faraday into what are now known as the Laws of Electrolysis.
Faraday's 1st Law of Electrolysis
This law states that the mass of a substance deposited or liberated at the electrodes during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. Mathematically, this can be expressed as:
                          m α Q ..................................(i)
where, m = mass in grams (g); and Q = quantity of electricity in Coulombs (C)
but,
                         Q = I x t ................................(ii)
where, I = current in Amperes (A); and t = time in seconds (s)
From equation (ii), it implies that 1 Coulomb = 1 Ampere x 1 sec
                                                                             = 1 Ampere sec (As)
Substituting equation (ii) into (i) will give us:
                          m α It ...................................(iii)
Removing the sign of proportionality, by introducing the equal sign and a constant of proportionality, equation (iii) changes to:
                         m = EIt ...........................................(iv)
where, E = electrochemical equivalence, and most times has a value of 1. Its unit is grams per Coulomb (g C^-1)
Equation (i) can also be written as:
                          m = EQ ............................................(v)
From equation (v), a plot of mass (m) against quantity of electricity (Q) gives a straight line graph passing through the origin, with a slope equal to the electrochemical equivalence of the element involved.
(Please refer to your textbooks for the verification of Faraday's 1st Law of Electrolysis)
The amount of electricity required to discharge one electron during electrolysis is known as the Faraday and is equivalent to 96500Coulombs, i.e., 1F = 96500C
The quantity of electricity required to deposit 1 mole of various classes of ions is summarized below:
For a univalent monoatomic ion, (e.g., Na+, Ag+, Au+) = 1mol of e- = 1F is required
      a divalent monoatomic ion, (e.g., Zn2+, Cu2+, S2-) = 2mol of e- = 2F are required
      a trivalent monoatomic ion, (e.g., Al3+, P3-) = 3mol of e- = 3F are required
    a univalent diatomic ion, (e.g., Cl-, F-, Br-) = 2mol of e- = 2F are required
      a divalent diatomic ion, (e.g., O2-) = 4mol of e- = 4F are required
     a trivalent diatomic ion, (e.g., N3-) = 6mol of e- = 6F are required
Examples
Question 1
What is the mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s? [Ag = 108, 1F = 96500 C]
Answer
Quantity of electricity passed = Current x Time
                                                   = 10 A x 4830s
                                                   = 48300 C
Silver ions undergo reduction by gaining one electron at the cathode, according to the equation:
                          Ag+(aq) + e- ----> Ag(s)
                        108g          1mol       108g
From the above equation, 1 mole of electron is required to deposit 1 mole of Ag,
i.e., 1F = 1 mole of Ag
but, 1F = 96500C and 1mole = 108 g of Ag
Therefore,
If, 96500 C of electricity are required to deposit 108 g of Ag, then
    48300 C will be required to deposit xg of Ag
                            x = 48300 x 108/96500
                               = 54.05 g of Ag
Question 2
If a given quantity of electricity deposits 0.65g of Zn2+, what amount of Hg2+ would be deposited by the same quantity of current? [Zn = 65, Hg = 201]
Answer
                      Zn2+(aq) + 2e- ----> Zn(s)
                     1mol            2F             1mol
                      65g             2F              65g
From the above equation, 2F of electricity are needed to deposit 1 mole of zinc metal, and 1 mole of zinc = 65g.
If, 65g of zinc are deposited by 2F of electricity, then
0.65g of zinc will be deposited by x F of electricity
                                x = 0.65 x 2/65
                                   = 0.02F
For mercury, Hg;
                      Hg2+(aq) + 2e- ----> Hg(l)
                      1mol            2F            1mol
                      201g            2F            201g
If, 2F of electricity deposit 201g of Hg
0.02F of electricity will deposit xg of Hg
                             x = 0.02 x 201/2
                                = 2.01g of Hg
Question 3
What current will deposit 3.25g of zinc in 2hrs? [Zn = 65, F = 96500 C mol^-1]
Answer
From the equation of the reduction of zinc ion to zinc metal in question 2 above, we can see that:
65g of zinc = 2F = 2 x 96500C (recall 1F = 96500C)
If, 65g of Zn are deposited by 2 x 96500C of electricity
3.25g of Zn will be deposited by xC of electricity
                              x = 3.25 x 193000/65
                                 = 9650C
but,
Quantity of electricity (C) = Current (A) x Time (s)
                                9650C = I x 2 x 3600s               (1hr = 3600s)
                                          I = 9650/7200
                                            = 1.34A
Note that we chose to work with the quantity of electricity directly in Coulombs because it will be faster to calculate the value of the current from it than using Faraday.
Question 4
What quantity of aluminium is deposited when a current of 10A is passed through a solution of an aluminium salt for 1930s? [Al = 27, F = 96500 C mol^-1]
Answer
Current passed = 10A; Time taken = 1930s
Quantity of electricity passed = current x time
                                                   = 10 x 1930
                                                   = 19300C
Aluminium ions are reduced by gaining 3 electrons each at the cathode,
                       Al3+(aq) + 3e- ----> Al(s)
                       1mol           3F            1mol
                       27g             3F             27g
If, 3 x 96500C of electricity deposit 27g of Al, then
          19300C of electricity will deposit xg of Al
                                x = 19300 x 27/289500
                                   = 1.8g of Al
Question 5
0.444g of a divalent metal is deposited when a current of 0.45A is passed through a solution of its salt for 50mins. Calculate the relative atomic mass of the metal.
Answer
Quantity of electricity passed = Current x Time
                                                    = 0.45A x 50 x 60
                                                    = 1350C
Being a divalent metal, 2F of electricity will be required to deposit 1 mole of the metal according to the equation:
                       M2+(aq) + 2e- ----> M(s)
                       1mol          2F             1mol
If, 1350C of electricity deposit 0.444g of the metal, then
2 x 96500C of electricity will deposit x g of the metal
                                 x = 193000 x 0.444/1350
                                    = 63.48g
Faraday's 2nd Law of Electrolysis
This law states that when the same quantity of electricity is passed through different electrolytes, the relative number of moles of the element discharged are inversely proportional to the charges on the ions of the element.
That is to say, if the same quantity of electricity is passed through two electrolytic cells containing silver trioxonitrate (V), AgNO3, solution and copper (II) tetraoxosulphate (VI), CuSO4, solution respectively, 2 moles of Ag will be deposited in the cell containing AgNO3, while 1 mole of Cu will be deposited in the one containing CuSO4 solution, as illustrated below:
                    Ag+(aq) +   e- ----> Ag(s)
                    1mol            1F           1mol
                  Cu2+(aq) + 2e- ----> Cu(s)
                   1mol           2F             1mol
From the above, 1F of electricity is required to deposit 1 mole of Ag because Ag+ is univalent. Similarly, 2F of electricity are required to deposit 1 mole of Cu because Cu2+ is divalent. Since the same quantity of electricity is passed, the same number of moles of electrons must be transferred. So, balancing the number of electrons involved, the above equation becomes:
                2Ag+(aq) +   2e- ----> 2Ag(s)
                2mol               2F             2mol
                 Cu2+(aq) +   2e- ----> Cu(s)
                 1mol              2F             1mol
At the end of the electrolysis:
Ag - Charge of ion : No of Moles deposited = +1: 2
Cu - Charge of ion : No of Moles deposited = +2:1
(Please check your textbooks for the verification of Faraday's 2nd Law of Electrolysis)
Examples
Question 6
A copper and silver voltameter are connected in series. If a current of 4A flow for 1hr 30mins, calculate the masses of copper and silver deposited [Cu = 63.5, Ag = 108, F = 96500C mol^-1]
Answer
Quantity of electricity passed = Current x Time
                                                   = 4 x [(1 x 3600) + (30 x 60)]
                                                   = 4 x (3600 + 1800)
                                                   = 4 x 5400
                                                   = 21600C
                     Cu2+(aq) + 2e- ----> Cu(s)
                     1mol             2F             1mol
                     63.5g            2F             63.5g
From the above,
2F of electricity are required to deposit 1 mole of copper, i.e.,
2 x 96500C of electricity are required to deposit 63.5g of copper
      21600C of electricity will deposit x g of copper
                                 x = 21600 x 63.5/193000
                                    = 7.11g of copper
                      2Ag+(aq) + 2e- ----> 2Ag(s)
                      2mol            2F            2mol
                      2x108g        2F            2x108g
Conversely,
2F of electricity are required to deposit 2 moles of silver
2 x 96500C of electricity are required to deposit 2 x 108g of silver
      21600C of electricity will deposit x g of silver
                                  x = 21600 x 216/193000
                                     = 24.17g of silver
Note that if we convert the masses of copper and silver deposited into moles, we will obtain 0.11 and 0.22 moles respectively. This gives a ratio of 1:2, compared to the ratio of their charges, which is 2:1.
Question 7
A voltameter containing silver trioxonitrate (V) solution was connected in series to another containing copper (II) tetraoxosulphate (VI) solution. When a current of 0.2amp was passed through the solutions, 0.780g of silver was deposited. Calculate (i) the mass of copper deposited in the copper voltameter. (ii) the quantity of electricity used. (iii) the time of current flow [Cu = 63.5, Ag = 108, 1F = 96500C]
Answer
                      2Ag+(aq) + 2e- ----> 2Ag(s)
                      Cu2+(aq) + 2e- ----> Cu(s)
From the above equations, 2F of electricity are required to deposit 1 mole of Cu and 2 moles of Ag.
(i) For every 63.5g of Cu deposited, 2 x 108g of Ag are deposited, so
                           xg of Cu will be deposited when 0.780g of Ag are deposited
                                x = 0.780 x 63.5/216
                                   = 0.229g of Cu
(ii) If 2 x 108g of Ag are deposited by 2 x 96500C of electricity, then
            0.780g of Ag will be deposited by xC of electricity
                                x = 0.780 x 193000/216
                                   = 696.94C
(iii) Quantity of electricity = Current x Time
                              696.94C = 0.2A x t
                                         t = 696.94/0.2
                                             = 3484.7s
but, 60s = 1min
                                             = 58.07mins
Question 8
A current is passed through three electrolytic cells connected in series containing solutions of silver trioxonitrate (V), copper (II) tetraoxosulphate (VI) and brine respectively. If 38.1g of copper are deposited in the second electrolytic cell, calculate
(a) the mass of silver deposited in the first cell,
(b) the volume of chlorine liberated in the third cell at 27°C and 900mmHg pressure. [Ag = 108, Cu = 63.5, 1F = 96500C, Molar Gas Volume at s.t.p. = 22.4dm^-3]
Answer
                      2Ag+(aq) + 2e- ----> 2Ag(s)
                      Cu2+(aq) + 2e- ----> Cu(s)
                       2Cl-(aq) ----> Cl2(g) + 2e-
                       2mol              22.4dm^3   2F
From the above equations, 2F of electricity are required to deposit 1 mole of copper, 2 moles of silver and liberate 1 mole (22.4dm^3) of chlorine gas.
(a) For every 63.5g of Cu deposited, 2 x 108g of Ag are deposited, so
                for 38.1g of Cu deposited, x of Ag will be deposited
                                 x = 38.1 x 216/63.5
                                    = 129.6g of Ag
(b) For every 63.5g of Cu deposited, 22.4dm^3 of Cl2 are liberated at s.t.p., so
                for 38.1g of Cu deposited, x dm^3 of Cl2 will be liberated
                               x = 38.1 x 22.4/63.5
                                  = 13.44dm^3 of chlorine gas at s.t.p.
We will now convert this volume to that at 27°C and 900mmHg, using the general gas equations:
                             P1V1/T1 = P2V2/T2
where P1 = 760mmHg, T1 = 273K, V1 = 13.44dm^3, P2 = 900mmHg, T2 = 27 + 273 = 300K, V2 = ?
                               V2 = P1V1T2/P2T1
                                     = 760 x 13.44 x 300/900 x 273
                                     = 3064320/245700
                                     = 12.47dm^3
Therefore, the volume of chlorine gas liberated in the third cell at 27°C and 900mmHg is 12.47dm^3.
Do These
Question 1
Calculate the volume of oxygen liberated at 570K and 1.82 x 10^5 Nm^-2 when a current of 2.5A is passed through dilute sodium chloride solution for 3mins. [Molar Volume of a Gas = 22.4dm^3; Standard Pressure = 1.01 x 10^5 Nm^-2; Standard Temperature = 273K; 1F = 96500C]
Question 2
A solution of copper (II) tetraoxosulphate (VI) was electrolysed between pure copper electrodes and the following results were obtained:
Mass of anode before experiment      = 14.40g
Mass of anode after experiment         =    8.00g
Mass of cathode before experiment   = 11.50g
From the information provided,
(i) calculate the mass of the cathode after experiment
(ii) write an equation for the reaction at the I. anode II. cathode
(iii) state whether the colour of the solution would change during the electrolysis. Give a reason for your answer.
(iv) if the electrolysis was carried out for 2 hours 40 minutes with a current of 2.0 Amperes, determine the value of the Faraday.
Question 3
State Faraday's Laws of electrolysis and show how you would verify the Laws
Question 4
If three electrolytic cells with each containing a solution of gold trioxonitrate (V), zinc tetraoxosulphate (VI) and brine respectively are connected together and a given quantity of electricity passed through them. If 26g of zinc are deposited in the second electrolytic cell, calculate
(a) the mass of gold deposited in the first cell,
(b) the volume of chlorine liberated in the third cell at 17°C and 800mmHg pressure. [Au = 197, Zn = 65, 1F = 96500C, Molar Gas Volume at s.t.p. = 22.4dm^-3]
(c) the quantity of electricity passed
(d) the current flow if the electrolysis took 3 hours 50 minutes.
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Electrolysis of Some Typical Electrolytes (Part II)

2016-01-24T11:48:00.001+01:00

In our last post: Electrolysis of Some Typical Electrolytes (Part I), we studied the electrolysis of acidified water, dilute sodium chloride and brine under different conditions. Here, we will be looking at the electrolysis of copper (II) tetraoxosulphate (VI), CuSO4, solution.

In solution, copper (II) tetraoxosulphate (VI) undergoes complete ionization to form copper (II) ions, Cu2+, and tetraoxosulphate (VI) ions, SO4--, according to the equation:

CuSO4(aq) ----> Cu2+(aq) + SO4--(aq) .....................(i)

Note that the two minus signs attached to the SO4 stand for 2-

Electrolysis of Dilute Copper (II) tetraoxosulphate (VI) Using Inert (Platinum or Carbon) Electrodes

The ions present in copper (II) tetraoxosulphate (VI) solution are Cu2+, SO4-- and H+, OH-; with the latter pair coming from the dissociation of water. Expectedly, the OH- and SO4-- ions migrate to the anode, while the H+ and Cu2+ ions migrate to the cathode.

Electrolysis of copper(II) tetraoxosulphate(VI) in the laboratory

The electrolysis of dilute copper (II) tetraoxosulphate (VI) solution using inert electrodes gives the same products as the electrolysis of acidified water and dilute sodium chloride, which are 2 volumes of hydrogen gas and 1 volume of oxygen gas.

(See the electrolysis of acidified water in the Electrolysis of Some Typical Electrolytes (Part I) for the full anodic and cathodic half-reactions and equations).

However, in dilute copper (II) tetraoxosulphate (VI) solution, even though the Cu2+ is lower in the electrochemical series than the H+, the H+ still gets discharged in preference to the former, because of its higher concentration, which overrides their positions in the electrochemical series, especially when they are so close to each other. (See Electrolysis: Overview)

Summary:
Electrolyte - Dilute copper (II) tetraoxosulphate (VI), CuSO4
Ions Present - Cu2+, SO4-- and H+, OH-
Concentrations - High (H+, OH-), Low (Cu2+, SO4--)
Nature of Electrodes - Inert
Product at the Cathode - Hydrogen gas
Product at the Anode - Oxygen gas
Gas Volume Ratio - 2:1
Effect on Electrolyte - The electrolyte becomes concentrated, due to the removal of water molecules in the form of hydrogen and oxygen. This is indicated by the increased intensity of the electrolyte's colour (light blue ----> deep blue).

Electrolysis of Concentrated Copper (II) tetraoxosulphate (VI) Using Inert (Platinum or Carbon) Electrodes

The electrolytic cell used here must be designed in a way that will enable the collection of gas at the anode and the removal of deposited metal at the cathode. (Please refer to your textbooks for a standard labelled diagram)

At the anode, despite the higher concentration of the SO4-- ions, the OH- ions are still preferentially discharged because of the wide gap separating them in the electrochemical series. So, the positions of ions in the series overrides their concentration. Hence, expectedly, they undergo oxidation to form 2H2O and O2 gas.

                                 4OH-(aq) ----> 2H2O(l) + O2(g) + 4e- .................(ii) (Anodic half-reaction)

At the cathode, the Cu2+ is preferentially discharged over H+, because it is lower in the electrochemical series and also, of higher concentration. So, each Cu2+ undergoes reduction by gaining two electrons to become deposited as metallic copper atoms:

                                  Cu2+(aq) + 2e- ----> Cu(s) ...........................(iii)

We will multiply equation (iii) by 2 to balance the number of moles of electrons transferred

                                 2Cu2+(aq) +  4e- ----> 2Cu(s) .......................(iv) (Cathodic half-equation)

Combining equations (ii) and (iv) will give the overall electrolytic reaction:

                               2Cu2+(aq) + 4OH-(aq) + 4e- ----> 2Cu(s) + 2H2O(l) + O2(g) + 4e-

                 2Cu2+(aq) + 4OH-(aq) ----> 2Cu(s) + O2(g) + 2H2O(l) .................(vi) (Overall equation)

From equation (vi) above, it can be seen that for every 2 moles of copper metal deposited at the cathode, 1 volume of oxygen is liberated at the anode. The water molecule produced in the equation is a by-product.

Note: As the Cu2+ and OH- are being removed from the solution in form of Cu metal and O2 gas, the H+ and SO4-- still remain in solution. These combine to form tetraoxosulphate (VI) acid, which makes the resulting solution after the electrolysis more acidic. The blue colour of the electrolyte also fades because of the removal of the Cu2+ from solution.

Summary:
Electrolyte - Concentrated copper (II) tetraoxosulphate (VI), CuSO4
Ions Present - Cu2+, SO4-- and H+, OH-
Concentrations - High (Cu2+, SO4--), Low (H+, OH-)
Nature of Electrodes - Inert
Product at the Cathode - Copper metal
Product at the Anode - Oxygen gas
Mole Ratio - 2:1
Effect on Electrolyte - The electrolyte becomes more acidic and faint in colour.

Electrolysis of Copper (II) tetraoxosulphate (VI) Using Copper Anode and Carbon/Platinum or Copper Cathode

Here, the concentration of the electrolyte is irrelevant as the copper anode interferes with the electrolytic process and influences the ion that is discharged, because the electrolyte is a salt of copper.

So, at the anode; in as much as, the OH- and SO4-- ions migrate there, none of them is discharged. Rather, the metallic copper atoms on the surface of the electrode go into solution by losing two electrons each to form Cu2+ ions, as shown:

                                           Cu(s) ----> Cu2+(aq) + 2e- .....................(vii)

The Cu2+ ions then migrate to the cathode, through the electrolyte to pick up the two electrons that were lost at the anode and get deposited as copper metal on the cathode, neglecting the H+ and Cu2+ from the electrolyte:

                                           Cu2+(aq) + 2e- ----> Cu(s) .....................(iii)

The above reactions indicate that nothing special actually takes place than the anode being transferred to the cathode, because combining equations (vii) and (iii) to obtain the overall equation will give zero. However, the reaction can be summarized as shown:

                                      Anode                Electrolyte               Cathode
                                                     -2e-                          +2e-
                                      Cu(s)      ---->    Cu2+(aq)      ---->    Cu(s)

Over time, as the copper atoms on the anode keep dissolving in the electrolyte and get deposited on the cathode, the anode becomes smaller in size, while the cathode increases in size. The original ions from the salt and water still remain in solution, and as such, the colour, concentration and pH of the electrolyte remain unchanged until the whole anode is used up or replaced with another material.

Electrolysis of copper(II) tetraoxosulphate(VI) using copper anode

This process is employed industrially in the purification of metals (like copper, silver, gold etc.)and electroplating. Electroplating is the process of coating a substance with a metal using electricity either for beautification or protection. If the substance is coated with gold, it is known as gold-plating, if it is coated with silver, the process is silver-plating and if coated with zinc, it is said be a galvanizing process.

In purification of metals, say copper, the impure copper ore is made the anode while a strip of very pure copper is used as the cathode. The electrolyte is a solution of any soluble salt of copper (II), like copper (II) tetraoxosulphate (VI), copper (II) trioxonitrate (V), Cu(NO3)2, copper (II) chloride, CuCl2 etc. The electrodes are connected to a power supply that supplies AC current. As the process proceeds, the copper atoms from the anode are transferred to the cathode. The anode decrease in size while the cathode, which is made up of copper only increases in mass. The impurities from the ore settle at the bottom of the electrolytic tank. At the end of the process, the pure copper cathode is removed and dried.

In electroplating, say silver-plating, the substance to be silver-plated e.g. a spoon, is made the cathode, while a strip of pure silver is used as the anode. The electrolyte is a solution of a soluble salt of silver, e.g. silver trioxonitrate (V), AgNO3. The electrodes are connected to an AC power supply. As the process proceeds, the silver atoms from the anode go into solution by losing one electron each to form silver ions, Ag+ (since silver is univalent). The silver ions, then migrate to the cathode where they gain one electron each to become silver atoms, which are then deposited on the surface of the spoon, according to the equation:

                                      Anode                Electrolyte               Cathode
                                                     -e-                           +e-
                                      Ag(s)      ---->    Ag+(aq)      ---->    Ag(s)

Depending on the thickness of the electroplating, the power supply is switched off when the desired level of electroplating has been attained, and the silver-plated spoon is removed and allowed to dry.

Summary:
In the purification of metals from their ores, the ore to be purified is the anode, while the cathode is a very pure form of the metal, whose ore is to be purified and the electrolyte is a solution of any soluble salt of the metal.

In electroplating, the substance to be electroplated is the cathode, while the metal that is to be used is the anode, and the electrolyte is a solution of any soluble salt of the anode.

Do These

Question 1
a) Draw the diagram of the electrolytic cell used in the purification of copper.
b) (i) Which of the electrodes in your diagram increases in mass during the electrolysis? Give reasons for your answers. (ii) State with reason the site of oxidation.
c) What will the electrolyte be? Explain why its colour does not change in intensity during the electrolysis.

Question 2
a) Explain why a crystal of sodium chloride is a poor conductor of electricity but its solution in water readily conducts electricity.
b) Write the equation which represents the dissociation of a salt formed from a trivalent cation A, and a divalent anion B, when dissolved in water.

Question 3
During a practical class, a student was asked to electroplate a stainless steel spoon using a silver rod as the electrode and silver trioxonitrate (V) solution as the electrolyte. The following were the results she obtained:

Mass of spoon                                                                 =       103.75g
Mass of silver-plated spoon                                            =                  g
Mass of the silver rod before electrolysis                       =       100.35g
Current                                                                            =              30A
Time taken                                                                      =         30mins

a) Draw a diagram of the set-up the student had used to electroplate the spoon. Which of the electrodes should the spoon be?
b) The student had forgotten to fill up one of the values in her results sheet. Find that value, and fill it in the empty space.
c) What was the mass of the silver rod after the electrolysis?

(To completely answer Q3, you might need to see our next post: Faraday's Laws of Electrolysis)

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Electrolysis of Some Typical Electrolytes (Part I)

2016-01-19T17:05:00.000+01:00

Recall that electrolytes are substances that conduct electricity in their aqueous or molten state, and get decomposed in the process. We are going to study the electrolysis of some electrolytes such as dilute tetraoxosulphate (VI) acid, dilute sodium chloride, concentrated sodium chloride, copper (II) tetraoxosulphate (VI) solution etc.

Electrolysis of Acidified Water Using Platinum Electrodes

The electrolysis of acidified water is carried out using the Hofmann voltameter. (Please click on the link or refer to your textbooks for a standard labelled diagram).

Hofmann Voltameter

Acidified water, which is simply dilute tetraoxosulphate (VI) acid, H2SO4, ionizes according to the equation:

                                     H2SO4(aq) ----> 2H+(aq) + SO4--(aq)

Meanwhile, the water molecules undergo partial dissociation to form H+ and OH-,

                                     H2O(l) <----> H+(aq) + OH-(aq)

Therefore, the acidified water contains three ions, namely: SO4--(from the acid), OH-(from the water) and H+(from the acid and water). Since the acid is a dilute solution, its ions (H+, SO4--) are of lower concentration than those from the water (H+, OH-). Also, the platinum electrodes used are considered as inert with respect to this electrolytic process. So, being negatively-charged, the SO4-- and OH- ions migrate to the anode, while the H+ ions migrate to the cathode.

At the anode, the OH- is discharged in preference to the SO4-- because it is higher in the electrochemical series and also of higher concentration. Therefore, each OH- ion donates one electron to the anode and becomes oxidized according to the equation:

                                      OH-(aq) ----> OH + e- ………(i)

Since a neutral OH cannot exist on its own, it combines with another OH to form a molecule of H2O and an atom of O; and for that to happen, equation (i) must occur twice as illustrated in the equations below:

                                      OH-(aq) ----> OH + e- ....................................(x 2)

                                      2OH-(aq) ----> 2OH + 2e- ………(ii)

but, 2OH = H2O + O because that is the only stable set of products that it can produce in electrolysis. Thus, substituting the value of 2OH in equation (ii) gives:

                                    2OH-(aq) ----> H2O(l) + O + 2e-………(iii)

Remember that oxygen is a diatomic molecule, so equation (iii) must occur twice to enable the O atoms to combine and produce O2.

                                   2OH-(aq) ----> H2O(l) + O + 2e- ............................ (x 2)

                                   4OH-(aq) ----> 2H2O(l) + O2(g) + 4e-………(iv)

Equation (iv), therefore, gives the oxidation half-equation or the anodic half-reaction.

At the cathode, the H+ ions, being the only cations are readily reduced by gaining one electron each according to the equation:

                                   H+(aq) + e- ----> H .........................(v)

being diatomic, equation (v) must occur twice, for a hydrogen molecule, H2 to be formed.

                                   H+(aq) + e- ----> H ................................................(x 2)

                                  2H+(aq) + 2e- ----> H2(g) ........................(vi)

Equation (vi), then, gives the reduction half-equation or the cathodic half-reaction.

If we take a closer look at the oxidation and reduction half-equations, we can observe that the 4e- were lost, while only 2e-were gained. So, to account for the electron deficit, we will multiply equation (vi) by 2 to obtain:

                                  4H+(aq) + 4e- ----> 2H2(g) ........................(vii)

Since the number of moles of electrons transferred has been balanced, we can then combine equations (iv) and (vii) to obtain the overall equation:

                                 4H+(aq) + 4OH-(aq) + 4e- ----> 2H2(g) + 2H2O(l) + O2(g) + 4e-................(viii)

                                 4H2O(l) ----> 2H2(g) + O2(g) + 2H2O(l)

By netting-off the terms that appear on both sides of the equation, we obtain equation (ix) as the final overall equation.

                                 2H2O(l) ----> 2H2(g) + O2(g) ................(ix)

From the above equation, it shows that at the end of the electrolysis of acidified water, 2 volumes of hydrogen gas are liberated at the cathode, while 1 volume of oxygen is liberated at the anode. It means that water is broken down into hydrogen and oxygen in the ratio of 2:1 respectively.

Summary:
Electrolyte - Dilute tetraoxosulphate (VI) acid, H2SO4
Ions Present - H+, SO4-- and OH-
Concentrations - High (OH-), Low (SO4--)
Nature of Electrodes - Inert
Product at the Cathode - Hydrogen gas
Product at the Anode - Oxygen gas
Gas Volume Ratio - 2:1

Electrolysis of Dilute Sodium Chloride Using Platinum Electrodes

Sodium chloride, NaCl, ionizes according to the equation:

                                       NaCl(aq) ----> Na+(aq) + Cl-(aq)

being dilute, the solution will also contain the H+ and OH- ions from the water molecules

Therefore, dilute sodium chloride solution contains four ions, which are: Na+ and Cl-(from the salt), H+ and OH-(from water). Being a dilute solution, it implies that the Na+ and Cl- ions from the salt are of lower concentration than the ions from the water. The platinum electrodes used are inert because they do not partake in the electrolytic process. As expected, the Cl- and OH- ions migrate to the anode, while the Na+ and H+ ions migrate to the cathode.

At the anode, the OH- is preferentially discharged over the Cl- because it is higher in the electrochemical series and also of higher concentration, and the oxidation process and the products will be similar to the one discussed in the electrolysis of acidified water above.

At the cathode, the H+ is discharged in preference to the Na+ because its position is lower in the electrochemical series than the Na+. So, it undergoes reduction as described in the electrolysis of acidified water to produce hydrogen gas.

Just like the electrolysis of acidified water, the end-products of the electrolysis of dilute sodium chloride are 2 volumes of hydrogen gas at the cathode and 1 volume of oxygen gas at the anode.

Summary:
Electrolyte - Dilute sodium chloride, NaCl
Ions Present - Na+, Cl- and H+, OH-
Concentrations - High (H+, OH-), Low (Na+, Cl-)
Nature of Electrodes - Inert
Product at the Cathode - Hydrogen gas
Product at the Anode - Oxygen gas
Gas Volume Ratio - 2:1

Electrolysis of Brine Using Graphite (Carbon) Electrodes

Brine is simply concentrated sodium chloride solution, and contains the same types of ions as the dilute solution, except that the Na+ and Cl- ions will be of higher concentrations than the H+ and OH- ions.

At the anode, the Cl- is preferentially discharged over the OH- because of its higher concentration. Here, the concentration of ions override their positions in the electrochemical series because their distance apart is not wide. So, the chloride ions get oxidized by donating their electrons to the anode to become chlorine atoms, which then combine in pairs to form molecules of chlorine gas, according to the equations:

                                      Cl-(aq) ----> Cl + e- .............................(x)

                                      Cl-(aq) ----> Cl + e-

                                     2Cl-(aq) ----> Cl2(g) + 2e- ..................(xi) (Oxidation half-equation)

At the cathode, the H+ is still discharged in preference to the Na+, despite the higher concentration of the Na+ in the solution. This is because their positions in the electrochemical series are very wide apart, and this overrides the concentration of ions in solution. So, it undergoes reduction  to produce hydrogen gas.

                                    2H+(aq) + 2e- ----> H2(g) ................(vi) (Reduction half-equation)

Since the number of moles of electrons transferred in the electrolytic process is balanced, we can combine the anodic and cathodic half-reactions to obtain the overall reaction.

                                  2H+(aq) + 2Cl-(aq) + 2e- ----> H2(g) + Cl2(g) + 2e- ............................(xii)

                                  2H+(aq) + 2Cl-(aq) ----> H2(g) + Cl2(g) ............................(xiii)

                                  2HCl(aq) ----> H2(g) + Cl2(g) ............................(xiv)

From equation (xiv), we can see that the electrolysis of brine produces equal volumes of hydrogen and chlorine gases liberated at the cathode and anode respectively.

Summary:
Electrolyte - Concentrated sodium chloride, NaCl (Brine)
Ions Present - Na+, Cl- and H+, OH-
Concentrations - High (Na+, Cl-), Low (H+, OH-)
Nature of Electrodes - Inert
Product at the Cathode - Hydrogen gas
Product at the Anode - Chlorine gas
Gas Volume Ratio - 1:1

Electrolysis of Brine Using Mercury Cathode and Graphite (Carbon) Anode

At the anode, just like the electrolysis of brine using carbon electrodes, Cl- will be discharged preferentially, and chlorine gas will be liberated at the anode.

                              2Cl-(aq) ----> Cl2(g) + 2e- ..................(x) (Oxidation half-equation)

At the cathode, the mercury electrode will influence the discharge of ions. Because of its strong affinity to sodium, the mercury cathode combines with the sodium ion, Na+, to form sodium amalgam (Na/Hg), thereby making the energy required for the discharge of Na+ to be lower than that required for the H+, according to the equation:

                                        Na+(aq) + Hg(l) + e- ----> Na/Hg(l) ................(xv)

To balance the number of moles of electrons transferred in the process, we multiply equation (xv) by 2, to obtain:

                              2Na+(aq) + 2Hg(l) + 2e- ----> 2Na/Hg(l) ................(xvi) (Reduction half-equation)

Combining the balanced oxidation and reduction half-equations, the overall reaction becomes:

                                     2Na+(aq) + 2Hg(l) + 2Cl-(aq) ----> 2Na/Hg(l) + Cl2(g) ...........(xvii)

but, the sodium amalgam is not quite useful, and to obtain a more useful product from it, it is treated with water, to form sodium hydroxide, hydrogen gas and mercury according to the equation:

                                   2Na/Hg(l) + 2H2O(l) ----> 2NaOH(aq) + H2(g) + 2Hg(l) .........(xviii)

At the end of the process, the mercury regenerated in equation (xviii) can be reused if the need arises. The sodium hydroxide is produced as a by-product, while the hydrogen gas is liberated at the cathode.

This process is applied industrially in the electrolysis of brine to produce chlorine gas, hydrogen gas and sodium hydroxide.

Summary:
Electrolyte - Concentrated sodium chloride, NaCl (Brine)
Ions Present - Na+, Cl- and H+, OH-
Concentrations - High (Na+, Cl-), Low (H+, OH-)
Nature of Electrodes - Anode: Graphite (Inert); Cathode: Mercury (Reactive)
Product at the Cathode - Hydrogen gas and Sodium hydroxide
Product at the Anode - Chlorine gas
Gas Volume Ratio - 1:1

From the above, does it mean it is not possible to achieve the deposition of sodium metal only at the cathode?

The answer is: Yes, it is possible!

The deposition of sodium metal only at the cathode can be achieved by using molten sodium chloride (melted sodium chloride without water) and graphite (carbon) electrodes. The anodic half-reaction will remain the same as equation (x) above, while the cathodic half-reaction will be:

                                          2Na+(aq) + 2e- ----> 2Na(s) ...............(xix)

Combining both half-reactions will give us the overall equation:

                                         2Na+(aq) + 2Cl-(aq) + 2e- ----> 2Na(s) + Cl2(g) + 2e-

                                         2Na+(aq) + 2Cl-(aq) ----> 2Na(s) + Cl2(g) ...............(xx)

                                         2NaCl(l) ----> 2Na(s) + Cl2(g) ...........(xxi)

This principle is employed industrially in the extraction of sodium metal using molten sodium chloride.

We will study the electrolysis of copper (II) tetraoxosulphate (VI), CuSO4, solution in our next post: Electrolysis of Some Typical Electrolytes (Part II).

Do These:

Question 1
a) Define the term electrolysis.
b) State three factors that affect the discharge of ions during electrolysis, and discuss any two.
c) Briefly describe what happens when a concentrated solution of sodium chlorine is electrolyzed using carbon electrodes. What will the outcome be if mercury is used as the cathode?

Question 2
a) Explain briefly each of the following terms: i) anode ii) inert electrode iii) voltameter
b) Sodium and aluminium are extracted by the electrolysis of molten sodium chloride and alumina, Al2O3, respectively. Write balanced equations for the reactions at the anode and cathode during the extraction of: i) sodium ii) aluminium

(Hint: Alumina ionizes according to the equation - Al203(aq) ----> 2Al3+(aq) + 3O2-(aq). There are no water molecules and each of the aluminium ions will gain three electrons to become deposited as aluminium metal atoms, while the oxides will donate two electrons each to become oxygen atoms. Remember, oxygen is a diatomic molecule)

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Electrolysis: Overview

2016-01-19T16:30:00.000+01:00

Electrolysis is a word formed from two Latin words - 'electrum', which means electricity and 'lysis', which means to breakdown. Therefore, electrolysis is defined as the use electrical energy to breakdown a compound into its constituent elements. For instance, using electricity to decompose sodium chloride into sodium metal and chlorine gas, as depicted by their state symbols (s) and (g) in the equation below:

                      2NaCl(aq) ----> 2Na(s) + Cl2(g)

It is a redox process, and one needs a good understanding of redox reactions to understand it. (Kindly refer to our posts on Redox Reactions: Overview, Balancing Redox Equations and Oxidation Numbers)

Before we continue, it is imperative to understand the meanings of some terminologies in electrolysis.

Definition of Terms

Electrolyte
This is a substance which conducts electricity in the aqueous or molten state and gets decomposed in the process. An aqueous solution is formed when a substance dissolves in water as a solvent to form a solution, e.g, aqueous sodium chloride, NaCl(aq), whereas a molten state is when a solid substance is made to melt, the melted state is the molten state. The difference between aqueous and molten states is the former involves the use of water, while the latter involves the use of heat only to breakdown the force of cohesion between the solid particles.

Strong Electrolytes
These are substances which dissociate completely into ions when dissolved in water. Examples include sodium chloride, potassium iodide and strong acids like hydrochloric acid, tetraoxosulphate (VI) acid etc.

Weak Electrolytes
These are substances that undergo partial dissociation into ions when dissolved in water. Examples are weak organic acids like ethanoic acid or weak alkalis like calcium hydroxide etc.

Electrodes
These are metallic plates, rods or wires through which electric current enters and leaves the electrolyte.

Cathode
This is the negative electrode or terminal by which current leaves the electrolyte. It is the terminal where the reduction process occurs and is always connected to the negative terminal of the power supply.

Anode
This is the positive electrode or terminal by which current enters the electrolyte. It is the terminal where the oxidation process occurs and is always connected to the positive terminal of the power supply.

Electrolytic Cell
This is an assembly of the electrodes, electrolyte and power supply where the electrolysis is carried out. It is also known as voltameter. (Please refer to your textbooks for the diagram of an electrolytic cell)

Ionic Theory
According to the Arrhenius Theory, some substances undergo dissociation when dissolved in water or melted to yield electrically charged atoms or molecules, called ions. This dissociation is also known as ionization. The substances can be electrovalent or ionic in nature, e.g, sodium chloride, copper (II) tetraoxosulphate (VI), iron (II) chloride etc. or polar in nature like acids, alkalis and even water.
                    NaCl(aq) ---> Na+(aq) + Cl-(aq)

                   CuSO4(aq) ---> Cu2+(aq) + SO4--(aq)

                   FeCl2(aq) ---> Fe2+(aq) + 2Cl-(aq)

                   HCl(aq) ---> H+(aq) + Cl-(aq)

                   2H2O(l) <---> H3O+(aq) + OH-(aq)

The Arrhenius Theory is also known as the ionic theory and is the basis for electrolysis.

Mechanism of Electrolysis & Electrolytic Reactions

When these substances undergo the ionization in water, they form positive and negative ions which move about randomly in the solution. However, immediately a potential difference is set up by passing electric current through the solution with the help of electrodes attached to a power supply, the migration of the ions occurs. Based on the law of electrostatic attraction, i.e, unlike charges attract while like charges repel; the positively-charged ions are attracted to the negative electrode or cathode, while the negatively-charged ions move to positive electrode or anode. The positive ions that migrate to the cathode are known as the cations (cathode ions), while negative ions that migrate to the anode are called anions (anode ions).

It is important to note that it is the mobile ions in the solution and not electrons that are responsible for the conduction of electricity by electrolytes.

At the anode, the anions lose their electrons to the electron-deficient terminal and get oxidized to the non-metallic atoms. This process is known as oxidation and is illustrated below:

                  A-(aq) - e- ---> A …………(i)

                  A-(aq) ---> A + e- …………(ii) (oxidation half-reaction)

If A is a diatomic molecule, it means two of equation (ii) can be combined to give equation (iii), which is the oxidation half-equation for a diatomic gas molecule.

                 2A-(aq) ---> A2 + 2e- ………(iii) (oxidation half-reaction)

From equation (i), the anion, A- loses one electron since it is univalent to form an atom A. The loss of the electron, e- is depicted by a subtraction sign before the electron on the left hand side of the equation. However, when writing chemical equations we do not use minus signs, so in applying the principle of balancing equations in mathematics, we move the electron to the right hand side and the sign changes to a plus sign. Equation (ii), therefore, is the conventional way of writing an oxidation half-equation.

On the other hand, at the cathode, the cations gain electrons from the electron-surplus terminal and get reduced to metallic atoms. This is a reduction process and is illustrated below:

                M2+(aq) + 2e- ---> M(s) …………(iv) (reduction half-equation)

In equation (iv), the cation, M2+ gains two electrons because it is divalent to form a metallic atom M. The gain of the electron is indicated by an addition sign before the electron on the left hand side of the equation. Thus, equation (iv) is the conventional way of writing a reduction half-equation.

If M is a univalent metal ion, then equation (iv) will be:

               M+(aq) + e- ---> M(s) …………(v) (reduction half-equation)

What do you think the reduction half-equation for a trivalent metal will be?

Note that the electrons gained by the cations are the ones that were lost by the anions, and passed through the external circuit to the cathode; and as such, the number of electrons lost must equal the number gained. So, if a univalent anion of a diatomic non-metal and a divalent cation of a monoatomic metal are involved in the reaction, then, we will make use of equations (iii) and (iv) in forming the overall redox equation:

               2A-(aq) ---> A2(g) + 2e- ………(iii) (oxidation half-reaction)

               M2+(aq) + 2e- ---> M(s) …………(iv) (reduction half-equation)

               2A-(aq) + M2+(aq) ---> A2(g) + M(s)

or better rearranged as:

               M2+(aq) + 2A-(aq) ---> M(s) + A2(g) (overall equation)

Now, let's use calcium chloride as typical example of an electrolyte to explain these jargons.
Calcium chloride undergoes ionization to give calcium ions as the cations and chloride ions as the anions as shown:

               CaCl2(aq) ---> Ca2+(aq) + 2Cl-(aq)

Therefore, it can undergo electrolysis to produce calcium metal and chlorine gas as explained further. In the presence of an electric current, the chloride ions migrate to the anode, where they undergo oxidation by losing their electrons to the positive electrode to form chlorine atoms. Since chlorine atoms cannot exist on their own, they combine together to form diatomic molecules of chlorine gas, as shown:

              Cl-(aq) ---> Cl + e-

              Cl-(aq) ---> Cl + e-

              2Cl-(aq) ---> Cl2(g) + 2e (oxidation half-equation)

On the other hand, the calcium ions migrate to the cathode, where they pick up two electrons each and get reduced to calcium metals, as represented by the equation below:

              Ca2+(aq) + 2e- ---> Ca(s) (reduction half-equation)

Since the number of moles of electrons transferred in the process is balanced, then, both equations can be combined to give:

              Ca2+(aq) + 2Cl-(aq) ---> Ca(s) + Cl2(g) (overall equation)

Therefore, at the end of the process, calcium is deposited at the cathode, while chlorine gas is liberated at the anode.

Nevertheless, in practice, the process is not as simple and straightforward as described above, because in a solution of any substance, there are more than two types of ions. Remember, water also undergoes dissociation, so the water molecules in the solution will also partake in the electrolytic reactions.

This implies that for the aqueous calcium chloride solution, there are four ions that will partake in the electrolytic process, namely: Ca2+, Cl- (from the CaCl2) and H+, OH- (from the H2O). So, at any of the electrodes, there will be two types of ions competing for deposition or discharge. At the anode, the Cl- and the OH- will compete for discharge, while the H+ and Ca2+ will compete for deposition or discharge at the cathode. Only one type of ions eventually gets discharged at each electrode and for that to occur, there are certain conditions that must be met. These conditions are the factors that affect the preferential discharge of ions during electrolysis.

Factors Affecting the Preferential Discharge of Ions

These include:

a) Position of ions in the electrochemical series
b) Concentration of ions in the electrolyte
c) Nature of electrodes

Position of Ions in the Electrochemical Series (ES)
The electrochemical is a table showing the reduction potentials of various metals or non-metals  arranged in a particular order as shown:

Cations
K+       -   Potassium
Na+     -   Sodium
Ca2+   - Calcium
Mg2+ -   Magnesium
Al3+    -   Aluminum
Zn2+   -   Zinc
Fe2+   -    Iron
Sn2+   -   Tin
Pb2+   -   Lead
H+       -  Hydrogen
Cu2+   - Copper
Hg2+   - Mercury
Ag+     - Silver
Au+     - Gold
Pt2+    -   Platinum

Anions
OH-         - Hydroxide
I-             - Iodide
Br-          - Bromide
Cl-          - Chloride
(NO3)-   - Trioxonitrate (V)
SO4--     - Tetraoxosulphate (VI)
F-            - Fluoride

For cations, ions lower down in the ES usually get discharged in preference to the ones higher up in the series. This is because the ions of the lower metals have the tendency of being reduced by receiving electrons more readily than the ones above, as electropositivity decreases down and increases up the series. Electropositivity is the ability to lose electrons and it is a typical metallic property.

So, in our example, H+ ions will be reduced at the cathode in preference to the Ca2+ ions because the position of the hydrogen ion is lower than that of calcium ion in the series.
Hence, at the cathode, the following reaction will take place:

                          H+(aq) + e- ---> H

Since hydrogen is a diatomic molecule, the final equation will be:

                          2H+(aq) + 2e- ---> H2(g)

The above equation illustrates that hydrogen gas will be liberated at the cathode.

Similarly, for anions, ions higher up are always preferentially discharged over the ones lower in the series. This is because the ions of the higher non-metals have the ability of being oxidized by losing electrons much readily than the ones below. This is because electronegativity, which is the ability of a substance to gain electrons decreases up and increases down the series.

Going by our illustrations, at the anode, the OH- will get discharged in preference to the Cl- because the former is higher than the latter in the ES.

So, the anodic half-reaction will be:

                          OH-(aq) + e- ---> OH

OH cannot exist on its own, it will have to combine with other OH radicals to form water and O atom as shown below:

                          OH-(aq) + e- ---> OH………(a)

                          OH-(aq) + e- ---> OH………(b)

Adding (a) and (b) gives:

                          2OH-(aq) + 2e- ---> H2O + O………(c)

However, O atom cannot exist on its own because oxygen is a diatomic molecule. Therefore, equation (c) must occur twice to make oxygen a molecule.

                         2OH-(aq) + 2e- ---> H2O + O………(c) x 2

                        4OH-(aq) + 4e- ---> 2H2O(l) + O2(g)………(d)

So, from equation (d) above, oxygen gas will be liberated at the anode.

Concentration of Ions
The ion of higher concentration in the electrolytic solution is always discharged in preference to the one of lower concentration. However, this factor is only applicable when the distance between the ions in the ES is not wide, say 3 - 4 steps apart.

So, in our example, if the concentration of the calcium chloride in the solution is high, it means that only Ca2+ and Cl- will be of high concentration, while the H+ and OH- will be of low concentration. In that case, when Cl- and OH- get to the anode, the Cl- will be preferentially discharged over the OH- because of its higher concentration. However, at the cathode, the H+ will be discharged in preference to Ca2+, despite the higher concentration of the Ca2+ in the solution. This is because the distance between them in the ES is quite wide. Hence, in the electrolysis of concentrated CaCl2 solution, chlorine gas and hydrogen gas will be liberated at the anode and cathode respectively.

Nature of Electrodes
This factor overrides the first two because of the high affinity (attraction) of certain electrodes towards some substances.There are two classes of electrodes usually involved in electrolytic reactions, namely: inert and reactive electrodes. The inert electrodes, by nature, are relatively unreactive and so do not partake in the electrolytic reactions, e.g., platinum and carbon, while the reactive electrodes affect the outcome of the electrolytic reaction because of their interference, e.g., copper electrodes used in the electrolysis of a copper salt. However, in the presence of some substances, the inert electrodes can become reactive because of their affinities for such substances.

For instance, mercury has a high affinity for sodium, and will also combine with sodium to form sodium amalgam (Na/Hg). So, in the presence of mercury cathode, sodium ion will always get discharged in preference to any other cation, irrespective of their positions in the ES. Also, platinum readily combines with chlorine to form platinum (II) chloride, and as such will preferentially discharge the chloride ions in the presence of any anion, irrespective of their positions and concentrations. A similar situation occurs between carbon and oxygen. A carbon anode will always favour the discharge of the OH- over other anions to liberate oxygen.

We will see how to apply all these factors when we look at the electrolysis of some typical electrolytes like sodium chloride solution, dilute tetraoxosulphate (VI) acid, copper (II) tetraoxosulphate (VI) solution etc. in our next post.

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Saved by a Hair's Breadth

2016-01-17T21:53:00.001+01:00

By Doreen Eweama

Born into an affluent family, I did not lack anything I needed or wanted. I am Susan Ken, a tall, slender and fair-skinned girl. I have three siblings. I am a brilliant and beautiful twenty-one-year old, the dream child of every parent. Unlike my siblings, I never caused trouble in within and without the neighbourhood, which made my parents proud of me.

I attended one of the most prestigious secondary schools in the country, and as a final year student, I was under a lot of pressure from my parents, teachers and relations. Being serious-minded, I hardly engaged in any activity that would not add any positive value to my life. After months of studying, the time we were all preparing for eventually came - the Senior School Certificate Examination (SSCE). Many of my mates were scared, but I drew confidence and inspiration from my mother's saying, "Prior Planning Prevents Poor Performance", what I also call, 'The 5P's Escape Route From Failure'. So, we sat for the examination within a month and two weeks after, I wrote my Universities & Tertiary Institutions Matriculation Examination (UTME). I got 290 marks out of 400 and was very happy and prayed for my success in the SSCE. I know most of you will be wondering why I was happy at scoring only 290/400? If you have ever written a 'JAMB' examination, you will understand that 290 is an A* performance.

Four months later, the results of our SSCE were released and I had 'A-parallel ', i.e., 9A's. I was so excited, my parents' joy knew no bounds, and for the next four months, I was treated as the queen of the house. On our valedictory service, in spite of the fact that I was not a prefect, I was given a five-star treatment because I won the prize for the best graduating student, which went with a cash benefit of N50,000 (fifty thousand naira only). Everybody wanted to have photo shots with me - my parents, siblings, teachers, school management and my friends. At the end of the day, I was so exhausted.

Two months after my valedictory service, I was invited to the University of Lagos (UNILAG) for my Post-UME. I passed the test with about 88 points and was granted admission on merit to study medicine, my dream course!. My parents paid for my tuition, accommodation and other requirements even before the new session began. On the day of resumption, I was given the sum of N100,000 (one hundred thousand naira only) as my allowance. On my way to the hostel, I met two girls, Bimbo and Vanessa, who told me that they also gained admission into the university to study theatre arts and physiotherapy respectively. I later discovered that Vanessa and I were bunkmates.

We got to know each other and we seemed to share common interests. She was also studious like me. However, unlike me, she was from a poor family, but this did not deter her from being the best she wanted to be; and that was what endeared her to me. We became best of friends, even closer than some siblings.

In our third year on campus, Vanessa made a new friend, named Sandra. Sandra was corrupt, dumb, and whorish. She was the type you would classify as a 'runz girl' on campus. She was ready to sleep with everything in a pair of trouser, provided they have the money to spare. When Vanessa introduced her to me, I knew at a glance that she was loose. I tried to advise her about her companionship with Sandra, but she paid no heed to my warnings. The only choice I had, which was a difficult one to make, was to stop moving around with her, because I did not want to lose focus.

Before you could say, "Jack Robinson!", Vanessa started skipping classes. She began attending clubs and keeping late night. She also had a guy, whom she said was her boyfriend, called Tunde. Sandra introduced him to Vanessa. Tunde was tall, hunky and charming. He was the type of guy, a girl would easily fall for, but as the saying goes, "Never judge a book from its cover". Unknown to Vanessa, Tunde was the leader of one of the notorious cult groups on campus known as the 'Sons of Darkness'.

As usual, I kept on doing well in my courses, and was the favorite of every lecturer in my department. On the other hand, Vanessa's grades started declining due to her new commitment, you guess right - Tunde!. One day, she told me that her younger sister, Priscilla, would be marking her birthday, which will also be used to celebrate her success in her SSCE, and requested that I go with her. On that fateful day, after much persuasion, I reluctantly followed Vanessa to the venue, which was a clubhouse called 'Dance Till You Drop...'.

On getting inside, we met Tunde with some guys that I presumed were his fellow cult group members. I did not see any shadow of Priscilla's, nor anyone from her family. When I asked Vanessa, she said they were on their way to the venue. She made me believe her as she continually faked some phone calls in my presence, as if she was monitoring them. By this time, it was 11:00pm. The impact of the sound system almost broke my ear drums. We had to shout at the top of my our voices before we could hear ourselves.

Tunde and his cohorts offered us two glasses of drink, which they drank but I politely rejected it, because I started smelling a rat. Being her friends, Vanessa took a glass and drank from it. Unknown to her, the edges of the glasses were laden with some sedatives, which meant that putting the glass in your mouth even without the drink will get you sedated. After a while, she started feeling tipsy. It was then it dawned on me what I had gotten myself into. I suggested to Vanessa that it was high time we left the place, but she told me, she did not feel like going yet.

I secretly left the club hall because the air was reeking of cigarettes and alcohol. I needed to get to the restroom first, before finding my way back to the hostel. I was ready to take that risk than remain in that club with those sons of the devil. After finishing from the restroom, on my way out of the clubhouse, I suddenly heard Vanessa's voice in the nearby bush screaming for help. I peeped to check what was going on, only to find her so-called boyfriend, Tunde and his fellow group members taking turns on her until she passed out. I took out my phone and contacted the police, but before they arrived, the culprits had escaped.

She was rushed to the hospital, but the deed had already been done. The doctor conducted several tests on her and found out that her womb had been severely damaged and her vagina had been torn. I contacted her parents and informed them about the incident. They begged me to help take care of her until they come the next day, as they were not in town, when I called. The matter was reported to the university authority, who ordered the police to carry out a full scale investigation and bring the perpetrators to book. So, Tunde and his 'brothers' were eventually arrested, and interrogated.

In their confessional statement, they stated that they got their powers by raping virgins until they pass out, and since they knew I was a virgin, they convinced Vanessa to trick me into coming to the clubhouse that night. However, to their utmost disappointment, I refused to take the drink they offered me. So, they had to vent their frustrations on Vanessa, whom they claimed, did not do her job well (I wonder how!). They were eventually rusticated. Although, I feel sorry for Vanessa, I am eternally grateful to God, because it could have been me.

My dear readers, this is no fairy tale, t is actually what happens to 1 out of every 10 girls on our campuses. Majority of the cases are unreported because they are never known. I would like this to serve as an eye-opener to my fellow teenagers who get carried away on gaining admissions into higher institutions. We tend to drop the values, which were inculcated in us by our loving parents, all in the name of freedom. We quickly make friends with whosoever comes our way; we swallow everything they tell us hook-line-and-sinker without taking time to carry out background checks about them, in order to know more about their dark sides, which they will never reveal to us. We want to be the most talked about babes on campus, neglecting the major reason of our being there. Even when our minds tell us that something is wrong with our companionships, we still ignore our instincts. Imagine what would have been of Susan, if she had ignored her intuitions?

For how long will we keep falling preys to these sex-drunk and evil-minded baboons? For how long will we keep falling victims of cheap emotional blackmails? The time to stop is come, yea, the set time for a change in attitude is NOW!

You May Be Down, But You Are Not Out!

2016-01-10T17:00:00.001+01:00

By Esther Olorunshola

Growing up was a miserable experience for me. My childhood days were nothing good to write about and most times, I wished I was never part of history. Living without a father nor a mother made things worse as I had to fend for myself without the help of any relatives.

Ours was one small and happy family until July 19, 1995 when everything took a U-turn. I was alone with my parents in the car. We were on our way to the beach to mark my 10th birthday. I sat at the owner's corner while my mother sat close to my father. Out of excitement, probably because it was long we went to the beach, I mistakenly covered my dad's face with a cloth, and before he could uncover his face, he lost control of the steering, swerved to the bush and the rest is history.

My name is Goodwill and I am a paraplegic. At age 7, I was struck by mild poliomyelitis, the condition, however, was worsened by the accident that claimed my parents' lives. My uncles saw me as a huge liability and threw me out of the house claiming they could not live with a 'vegetable', who has no future. The only thing I was left with was my wheelchair. Hence, at age 10, I became a paralysed urchin.

In the school I attended, I had just two friends, Geoffrey and Gideon. We were best of friends and we did most things in common. Geoffrey's story is kind of complicated. He was dumped in a dungeon and picked up by a wealthy woman. He enjoyed life also for the first 10 years until he discovered that his supposed mother was a ritualist and he was next in line. He, therefore, had to escape to save his life. Gideon, on the other hand, is from a polygamous home. His mother was the last of five wives, so, she had the least authority. His step-mothers maltreated him like a slave. When he could no longer endure it, he had to leave the house.

We were homeless and also physically challenged. Geoffrey was one-eyed, while Gideon was dumb. We received care and attention from no one. Our mates made mockery of us and made us feel inferior but we kept on encouraging ourselves that the storm would soon be over. After school each day, we worked for a man that decided to pay our school fees as our wages but did not bother himself about our feeding and accommodation. We had to do extra jobs, and even sell in traffic jams just to survive.

Life was harsh on us. We were not dull students but because nobody listened to us when we spoke, we decided to keep our ideas to ourselves. February 28, 1998 happened to be our memorable day in school. We were asked us to write an essay of what we wished to become. We got back home (under the bridge) and we started writing. We all wrote our minds down, praying it came to pass someday.

The next day, with great anger, our teacher walked into the class. After a moment of silence, he shouted, "Can the three unrealistic musketeers come out here?!" With fear we stepped out, he gave us our scripts to read to the class. Geoffrey and I read ours while he assisted Gideon in reading his. He then looked straight into our eyes and said, "You fools should wake up from your slumber! Goodwill, with this condition of yours do you think you can ever become a doctor?" "So, you have not asked yourself, who wants to be treated by a paraplegic?", he continued. Turning to Geoffrey he said, "You wish to become a civil engineer, right? Well, that can only happen in your dreams! Where will you get the money from or you think people would love to award contracts to a one-eyed man?". "As for you Gideon, were you sleeping when you were writing this essay? How can a dumb idiot like you wish to become a pilot?! You are just deceiving yourselves", he said.

"You better stop daydreaming. Everything you wrote are all fantasies and it is my work to set you on the right path. Well, as you can see on your scripts, your grade is F, l give you till tomorrow to represent a more realistic essay like you becoming a conductor, fisherman or a farmer, if not be prepared to fail this year", he concluded. We had never been discouraged and humiliated in our entire lives. The man sounded so certain that we will not make it, but we summoned courage and repeated what we wrote the previous day and added: " Keep your grades, while we keep our dreams"

We graduated from high school in the year 2002 thanks to the man who kept his promise and saw us through secondary school level. We wished to further our education; so, we decided to work and save up some money. Getting a job for ourselves was quite challenging as people turned us down due to our disabilities. After a long search, we found one whose pay was discouraging but just had to manage. After 3 years of working, the money we had was only enough to get us some JAMB forms. Most times, when our classmates passed by and saw us packing dirts, they made mockery of us saying, "Are these not the doctor, civil engineer and pilot? Is this meant to be your work stations?". The torment was becoming unbearable but our desire for success was far greater than our fear of failure.

May 12, 2005 was the day our story changed. Gideon and Geoffrey were cutting grasses on a man's field, and I was cleaning up the cut areas when we got attracted to a poster on the wall with the inscription: 1000 words of creative writing inscribed with the title: "KEEPING YOUR DREAMS ALIVE! - 1000 words of creative writing". Knowing fully well that the story of our lives was more than a thousand words, we decided to give it a try. We worked as a team and to the Glory of God, we came out tops. We were awarded full scholarship to study anywhere in West Africa.

Presently I am a renowned doctor with so many honorary awards. Just last week, the teacher who almost shattered my dreams rushed to my hospital with his son who is epileptic. After treating his son, I went to the reception to see him. He recognised me at once and asked, "Excuse me sir, are you a twin?" I laughed and replied, "I am the same Goodwill you advised to become a conductor. He went down on his knees crying and apologizing. He then said; "No doubt dreams do come true". Geoffrey and Gideon are doing fine as well. They turned out to be what they wanted to be because they were not discouraged.

Dear friends, that you are now at the most difficult part of your lives does not mean that the game of life is over for you. It can only be over if you so decide. Imagine what would have become of Goodwill, Gideon and Geoffrey if they had thrown in the towel after the humiliation they got from the very person, who was meant to inspire them. The fame and honour they now enjoy would not have been. Nature saw their sincere desire to succeed and helped them. It is said that if your intentions are pure, the forces of nature will always come together to favour you. In other words, if there is a will, there will definitely be a way.

Many of us have dropped in our grades in school, even far below our own expectations, and are doubting our abilities to graduate with a 1st Class, or 2nd Class divisions (no prejudice to other divisions of passes) or even gain admission into tertiary institutions. Some of you have lost your lucrative jobs, while others are struggling with their once enviable marriages, businesses and characters. While some are thinking of quitting businesses, others are about to walk out of their marriages and some are even having suicidal thoughts. Before you make that final decision, stop and think twice, bearing in mind that you are not alone in this battle. Do you remember Job from the Holy Scriptures? Despite all that befell him, he still did not allow himself to be used as a tool in the hands of the devil.

In conclusion, there is no great man without a story, as challenges are said to be the food of champions. It is said that if you do not fall, you can never rise. If you do not suffer lack, you will never appreciate abundance. You need to experience failure to really understand what success means. So, though you may be down, it is not yet time to throw in the towel, because you are not out!

Water of Crystallization

2016-01-07T15:44:00.001+01:00

Water of crystallization (WC) is the number of molecules of water present in one mole of a hydrated salt. Example, in iron (II) tetraoxosulphate (VI)-heptahydrate [FeSO4.7H2O], there are seven molecules of water attached to a molecule of FeSO4, that is its water of crystallization.
Some other examples of salts with water of crystallization include:
MgSO4.7H2O - Magnesium tetraoxosulphate(VI)-heptahydrate, (Epsom Salt)
CaSO4.2H2O - Calcium tetraoxosulphate(VI)-dihydrate (Gypsum)
Na2CO3.10H2O - Sodium trioxocarbonate(IV)-decahydrate (Washing Soda)
CuSO4.5H2O - Copper (II) tetraoxosulphate(VI)-pentahydrate (Blue Vitriol)
ZnSO4.7H2O - Zinc tetraoxosulphate(VI)-heptahydrate (White Vitriol)
Cu(NO3)2.3H2O - Copper (II) trioxonitrate(V)-trihydrate
From the above examples, it can be observed that each salt has a definite number of molecules of water of crystallization attached to it.
Therefore, it can also be defined as the definite amount of water some substances chemically combine with to form salt from their aqueous solution, as shown in the formation of CuSO4.5H2O:
CuSO4(s) + 5H2O(l) ---> CuSO4.5H2O(s)
In the equation, five molecules of water of crystallization combine with a molecule of CuSO4 to form the pentahydrate.
However, the molecules of water of crystallization are loosely attached to the hydrated salt and can easily dissociate on heating, leaving behind the anhydrous form of the salt. The water molecules can also be lost through efflorescence, a process whereby hydrated salts lose their water of crystallization when exposed to the atmosphere over a period of time.
The definite number of molecules of water of crystallization present in the hydrated salt gives it a definite crystalline structure and heating it above a particular temperature leads to the loss of the water of crystallization leaving behind a residue of the anhydrous salt, which is amorphous (formless or noncrystalline) in nature. Most times, this is usually characterized by a change in colour.
Generally,
Hydrated Salt ----> Anhydrous Salt + Water of Crystallization
crystalline                   amorphous
Determination of Amount of Water of Crystallization in a Hydrated Salt
CuSO4.5H2O(s) -----> CuSO4(s)   +   5H2O(l)
blue hydrated            white anhydrous
(crystalline)                (amorphous)
Now, let's apply a little mole concept by calculating their molar masses:
Molar mass of CuSO4.5H2O = (1*Cu) + (1*S) + (4*O) + 5[(2*H) + (1*O)]
= (1*63.5) + (1*32) + (4*16) + 5[(2*1) + (1*16)]
= 63.5 + 32 + 64 + 5(18)
= 159.5 + 90
= 249.5 g/mol
Similarly, molar mass of CuSO4 = 159.5 g/mol
and, molar mass of 5H2O = 90 g/mol
           CuSO4.5H2O(s) -----> CuSO4(s)   +   5H2O(l)
Mole ratio        1                :             1             :        5
Molar mass    249.5                     159.5                90
From the above stoichiometry, it implies that 1 mole of the pentahydrate on decomposition will produce 1 mole of the anhydrous salt and 5 moles of water, and vice-versa.
Also, 249.5g of CuSO4.5H2O will decompose to give 159.5g of CuSO4 and 90g of H2O. This implies that given the masses of the hydrated salt and its anhydrous component or water of crystallization, we can easily calculate the percentage by mass of the water of crystallization present in it as shown below:
Mass of hydrated salt = Mass of anhydrous salt + Mass of WC ………(i)
Therefore,
Mass of WC = Mass of hydrated salt - Mass of anhydrous salt ………(ii)
Then,
Percentage by mass of water of crystallization
= (Mass of WC / Mass of hydrated salt) x 100 ………(iii)
Or
[(Mass of hydrated - Mass of anhydrous) / Mass of hydrated] x 100 ………(iv)
So, for CuSO4.5H2O, the % by mass of the water of crystallization content will be:
= (90/249.5) x 100
= 0.36 x 100
= 36.1 %
While, the % by mass of the anhydrous content will be:
= (159.5/249.5) x 100
= 0.64 x 100
= 63.9 %
The above can also be obtained by subtracting the % by mass of the water of crystallization from 100%, i.e,
100% - 36.1% = 63.9%
From the above analysis, we can deduce that in a given hydrated salt,
(Mass of WC / Mass of Anhydrous Salt) = (% by Mass of WC / % by mass of Anhydrous Salt) ………(v)
Hence, for CuSO4.5H2O, equation (v) will give:
90g/159.5g = 36.1%/63.9% = 9/16
Also,
(Reacting mass of WC/Molecular mass of WC) / (Reacting mass of anhydrous salt/Molecular mass of anhydrous salt) = No. of molecules of water of crystallization ………(vi)
Equation (vi) above can be reduced to:
No. of moles of WC / No. of moles of anhydrous salt = No. of molecules of water of crystallization ………(vii)
Similarly,
(Reacting mass of WC/Reacting mass of anhydrous salt) = (Mass of WC in 1 mole of hydrated salt/Molecular mass of anhydrous salt) ………(viii)
Equations (vi), (vii) or (viii) is used when the complete formula of the hydrated salt is not given. Usually, the molecular mass of the anhydrous salt will be given.
Examples
Question 1
5.00g of hydrated salt of barium when heated to a constant weight gave 4.26g of anhydrous salt with a molecular weight of 208. Calculate:
a) the percentage by mass of water of crystallization
b) the number of molecules of water of crystallization in the hydrated salt.
Answer
Mass of hydrated salt = 5.00 g
Mass of anhydrous salt = 4.26 g
Mass of water of crystallization = 5.00 - 4.26
= 0.74 g
Molecular mass of anhydrous salt = 208
a) % by mass of water of crystallization
= (Mass of WC/Mass of hydrated salt) x 100
= (0.74/5.00) x 100
= 0.148 x 100
= 14.8 %
b) Molecular mass of H2O = (2*H) + (1*O)
                                                = (2 x 1) + (1 x 16)
                                                = 2 + 16
                                                = 18
No. of molecules of water of crystallization = (Reacting mass of WC/Molecular mass of WC) / (Reacting mass of anhydrous salt/Molecular mass of anhydrous salt)
No. of WC = (0.74/18)/(4.26/208)
                   = 0.0411/0.0205
                   = 2.005, approximately 2.00
Therefore, the hydrated barium salt has 2 molecules of water of crystallization.
Note that the formula of the hydrated salt is not provided in this question, hence, there is no need for the equation of reaction. Equation (vi), therefore, provides the best solution to the problem.
Question 2
Some copper (II) tetraoxosulphate (VI) pentahydrate was heated at 120°C with the following results:
Mass of crucible = 10.00g
Mass of crucible + CuSO4.5H2O = 14.98 g
Mass of crucible + residue = 13.54 g
How many molecules of water of crystallization were lost? [H = 1, Cu = 63.5, O = 16, S = 32]
Answer
Mass of crucible = 10.00 g ………(a)
Mass of crucible + hydrated salt = 14.98 g ………(b)
Mass of crucible + anhydrous salt = 13.54 g ………(c)
Mass of hydrated salt = (b) - (a)
                                        = 14.98 - 10.00
                                        = 4.98 g …………(d)
Mass of anhydrous salt = (c) - (a)
                                           = 13.54 - 10.00
                                           = 3.54 g …………(e)
Mass of water of crystallization lost = (d) - (e)
                                                                = 4.98 - 3.54
                                                                = 1.44 g
No. of molecules of WC lost = (Reacting mass of WC/Molecular mass of WC) / (Reacting mass of anhydrous salt/Molecular mass of anhydrous salt)
but,
Molecular mass of H2O = 18
Molecular mass of anhydrous CuSO4 = 159.5 (see calculations above)
Substitute the values into the equation above:
No. of molecules of WC lost = (1.44/18) / (3.54/159.5)
                                                  = 0.08/0.022
                                                  = 3.64
Therefore, approximately 4 molecules of water of crystallization were lost.
Question 3
0.499g of CuSO4.xH2O when heated to constant weight gave a residue of 0.346g. Find the value of x. [Cu = 63.5, S = 32, O = 16, H = 1]
Answer
Hydrated salt = CuSO4.xH2O
                CuSO4.xH2O(s) -----> CuSO4(s)   +   xH2O(l)
Mass of hydrated salt = 0.499g
Mass of anhydrous residue = 0.346g
Mass of water of crystallization lost (x) = 0.499 - 0.346 = 0.153 g
Molar mass of anhydrous CuSO4 = 159.5 g/mol
Molar mass of H2O = 18 g/mol
(Always endeavour to show the full workings)
No. of molecules of WC lost (x) = (Reacting mass of WC/Molecular mass of WC) / (Reacting mass of anhydrous salt/Molecular mass of anhydrous salt)
No. of molecules of WC lost (x) = (0.153/18) / (0.346/159.5)
                                                        = 0.0085/0.0022
                                                        = 3.86
Therefore, x is approximately 4
Question 4
1.34g of hydrated sodium tetraoxosulphate (VI) was heated to give an anhydrous salt weighing 0.71g. What is the formula of the hydrated salt? [Na = 23, S = 32, O = 16, H = 1]
Answer
Proposed formula of hydrated salt = Na2SO4.xH2O
On heating, it decomposes according to the equation
                 Na2SO4.xH2O(s) -----> Na2SO4(s)   +   xH2O(l)
Mass of hydrated salt = 1.34 g
Mass of anhydrous salt = 0.71 g
Mass of water of crystallization = 1.34 - 0.71
                                                         = 0.63 g
No. of molecules of WC lost (x) = (Reacting mass of WC/Molecular mass of WC) / (Reacting mass of anhydrous salt/Molecular mass of anhydrous salt)
but,
Molar mass of Na2SO4 = (2*Na) + (1*S) + (4*O)
                                          = (2*23) + (1*32) + (4*16)
                                          = 46 + 32 + 64
                                          = 142 g/mol
No. of molecules of WC lost (x) = (0.63/18) / (0.71/142)
                                                       = 0.035/0.005
                                                       = 7
Since x = 7, therefore, the formula of the hydrated salt is Na2SO4.7H2O
Question 5
A hydrated salt of formula MSO4.xH2O contains 45.3% by mass of water of crystallization. Calculate the value of x. [M = 56, S = 32, O = 16, H = 1]
Answer
Hydrated salt = MSO4.xH2O
Percentage by mass of WC = 45.3 %
Percentage by mass of MSO4 = 100 - 45.3
                                                     = 54.7 %
Mass of anhydrous salt, MSO4 in the hydrated salt
= (1*M) + (1*S) + (4*O)
= (1 x 56) + (1 x 32) + (4 x 16)
= 56 + 32 + 64
= 152 g
Mass of WC, xH2O, in the hydrated salt = x[(2*H) + (1*O)]
                                                                      = x[(2 × 1) + (1 × 16)]
                                                                      = x(2 + 16)
                                                                      = 18x g
Using equation (v),
(Mass of WC / Mass of anhydrous salt) = (% by mass of WC / % by mass of anhydrous salt)
Hence,
18x/152 = 45.3/54.7
18x/152 = 0.828
        18x = 0.828 x 152
               = 125.86
            x = 125.86/18
               = 6.99, approximately 7
Question 6
Calculate the percentage of water of crystallization in ZnSO4.7H2O [Zn = 65, S = 32, O = 16, H = 1]
Answer
Hydrated salt = ZnSO4.7H2O
We will make use of equation (iii), i.e,
Percentage by mass of water of crystallization
= (Mass of water of crystallization / Mass of hydrated salt)
but,
Mass of ZnSO4.7H2O = (1*Zn) + (1*S) + (4*O) + 7[(2*H) + (1*O)]
                                       = (1 x 65) + (1 x 32) + (4 x 16) + 7[(2 x 1) + (1 x 16)]
                                       = 65 + 32 + 64 + 7(18)
                                       = 161 + 126
                                       = 287 g
Mass of WC in hydrated salt = 7 x 18
                                                   = 126 g
Therefore,
Percentage by mass of water of crystallization
= (126/287) x 100
= 0.439 x 100
= 43.9 %
Do These
Question 1
24.6g of iron (II) tetraoxosulphate (VI) crystals when heated gave 12.0g of the anhydrous salt. If it lost its water of crystallization due to the heating, find the:
a) percentage mass of the water of crystallization
b) molecules of water of crystallization in the hydrated salt. [Fe = 56, S = 32, O = 16, H = 1]
Question 2
If Na2CO3.10H2O contains 62% of water of crystallization, what mass of the hydrated salt are expected to be obtained from 7.38g of anhydrous sodium tetraoxosulphate (VI). [Na = 23, S = 32, O = 16, H = 1]
Question 3
18.50g of white vitriol, ZnSO4.7H2O, were heated in a crucible to a constant mass, first at a temperature of 120°C, and then at 140°C. The residue that was obtained at 120°C weighed 12.70g and that obtained at 140°C weighed 10.38g. Calculate the stages of dehydration reached (molecules of water of crystallization lost) at 120°C and 140°C respectively. [Zn = 65, S = 32, O = 16, H = 1]
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Isotopy

2016-01-07T15:40:00.003+01:00

John Dalton in one of his postulations of Atomic Theory stated that atoms of the same element are alike and different from atoms of other elements. This was, however, faulted with the discovery of nucleon mass (mass number) and neutrons by Rutherford and Chadwick respectively.

Studies showed that there are atoms of the same element that have the same atomic number, but different mass numbers as a result of the difference in their number of neutrons. These atoms are called isotopes and the phenomenon is known as isotopy. Examples of elements that exhibit isotopy include chlorine (Cl-35 and Cl-37), carbon (C-12, C-13 and C-14), oxygen (O-16, O-17 and O-18) etc.

In reality, all elements exhibit isotopy because this explains why the relative atomic masses of elements are not whole numbers.

Relative Abundance of Isotopes
In a particular element, say chlorine, the two isotopes - Cl-35 (chlorine-35) and Cl-37 (chlorine-37) are present in different quantities of 75% and 25% respectively. These are their relative abundances.

Therefore, relative abundance is said to be the amount of an isotope present in a given sample of an element. It can be expressed as a percentage, fraction or decimal. The sum of the relative abundances of all the isotopes in an element is always equal to 100 (if expressed as %) or 1 (if expressed as fractions or decimals)

Still using chlorine as an example, the relative abundance of its isotopes can be expressed as shown:

Isotopes        Relative Abundances         Sum
Cl-35                        75 %
Cl-37                        25%
                                                                     100%

Cl-35                        3/4
Cl-37                        1/4
                                                                     4/4 = 1

Cl-35                        0.75
Cl-37                        0.25
                                                                     1.00

The relative abundances can also be expressed as a ratio as stated below:
                             Cl-35 : Cl-37
                              75% :   25%
Ratio                        3    :     1

Calculation of Relative Atomic Mass Using Relative Abundance

Since the isotopes present in a given element are in different proportions and possess different masses, it will be wrong to assume that the atomic mass of the element will be the simple addition of the masses of all the isotopes in the element.

Thus, it is always calculated as a weighted average of the mass numbers of all the isotopes using the formula:

Relative Atomic Mass (R.A.M.)
= (Relative Abundance of Isotope A x Mass No. of Isotope A) + (Relative Abundance of Isotope B x Mass No. of Isotope B) + …

Examples

Question 1
Given that the relative abundances of the two isotopes of chlorine, Cl-35 and Cl-37 are in the ratio of 3:1 respectively. Calculate the relative atomic mass of chlorine.

Answer
Isotopes           Cl-35           Cl-37
Mass number      35                37
Ratio                     3         :         1
Sum of ratio:            3 + 1= 4
Proportion (Relative Abundance) of Cl-35 = 3/4
Proportion (Relative Abundance) of Cl-37 = 1/4

Using,

Relative Atomic Mass
= (Relative Abundance of Isotope-35 x Mass No. of Isotope-35) + (Relative Abundance of Isotope-37 x Mass No. of Isotope-37)

Therefore,

Relative Atomic Mass of Chlorine = (3/4 x 35) + (1/4 x 37)
= 26.25 + 9.25
= 35.5

Question 2
If the relative atomic mass of copper is 63.54, calculate the proportions of the isotopes, Cu-65 and Cu-63 in the metal.

Answer
Relative atomic mass of Cu = 63.54
Let the relative abundance of Cu-65 = x
Let the relative abundance of Cu-63 = y

but,

x + y = 1……(i) [sum of relative abundances must be equal to 1]

from equation (i),

y = 1 - x……(ii)

Using,

Relative Atomic Mass
= (Relative Abundance of Isotope-65 x Mass No. of Isotope-65) + (Relative Abundance of Isotope-63 x Mass No. of Isotope-63)

Therefore,

63.54 = (x × 65) + (y x 63)
63.54 = 65x + 63y………(iii)

Substitute y = 1 - x into equation (iii)

63.54 = 65x + 63(1 - x)
63.54 = 65x + 63 - 63x
63.54 - 63 = 65x - 63x
0.54 = 2x
x = 0.54/2
= 0.27 or 27%

Substitute the value of x into equation (ii)

        y = 1 - x
           = 1 - 0.27
           = 0.73 or 73%

Therefore, the proportions of Cu-65 and Cu-63 are 27% and 73% respectively.

Question 3
An analysis of a sample of iron using mass spectroscopy showed that the relative abundances of the four isotopes of iron are 5.8%, 91.6%, 2.3% and 0.3% for Fe-54, Fe-56, Fe-57 and Fe-58 respectively. From the result, determine the relative atomic mass of iron.

Answer
Isotopes                                  Fe-54    Fe-56    Fe-57    Fe-58
Mass Numbers                         54          56          57           58
Relative Abundances (%)       5.8         91.6       2.3          0.3

Using,

Relative Atomic Mass
= (Relative Abundance of Isotope-54 x Mass No. of Isotope-54) + (Relative Abundance of Isotope-56 x Mass No. of Isotope-56) + (Relative Abundance of Isotope-57 x Mass No. of Isotope-57) + (Relative Abundance of Isotope-58 x Mass No. of Isotope-58)

Therefore,

R.A.M. of Fe = (5.8% x 54) + (91.6% x 56) + (2.3% x 57) + (0.3% x 58)

(To convert percentage to decimal, divide the percentage value by 100)

R.A.M. of Fe = (0.058 x 54) + (0.916 x 56) + (0.023 x 57) + (0.003 x 58)

R.A.M of Fe = 3.132 + 51.296 + 1.311 + 0.174
= 55.913

Question 4
Given that the mean relative mass of oxygen, which contains three isotopes of mass numbers 16, 17 and 18 is 16.0044. Determine the percentage compositions of isotopes 16 and 17, if the relative abundance of the last isotope is 0.2%.

Answer
Relative atomic mass of O = 16.0044
Let the relative abundance of O-16 = x
Let the relative abundance of O-17 = y
Relative abundance of last isotope (O-18) = 0.2% = 0.002

but,

x + y + 0.002 = 1……(i)
x + y = 1 - 0.002
x + y = 0.998……(ii)

from equation (ii),

y = 0.998 - x……(iii)

Using,

Relative Atomic Mass of Oxygen
= (Relative Abundance of Isotope-16x Mass No. of Isotope-16) + (Relative Abundance of Isotope-17 x Mass No. of Isotope-17) + (Relative Abundance of Isotope-18 x Mass No. of Isotope-18)

Therefore,

16.0044 = (x × 16) + (y x 17) + (0.002 x 18)
16.0044 = 16x + 17y + 0.036………(iv)

Substitute y = 0.998 - x into equation (iv)

16.0044 = 16x + 17(0.998 - x) + 0.036
16.0044 = 16x + 16.966 - 17x + 0.036
16.0044 - 16.966 - 0.036 = 16x - 17x
-0.9976 = -x
x = 0.9976 or 99.76%

Substitute the value of x into equation (iii)

        y = 0.998 - x
           = 0.998 - 0.9976
           = 0.0004 or 0.04%

Therefore, the proportions of O-16 and O-17 are 99.76% and 0.04% respectively.

Do These

Question 1
The relative atomic mass of neon is 21.5. Its two isotopes, Ne-20 and Ne-X, occur in nature in a ratio of 1:3. Find the value of X.

Question 2
What is the relative atomic mass of carbon, if the relative abundances of its three isotopes 12, 13 and 14 are 98.7%, 1.1% and 0.2% respectively?

Question 3
a) What are isotopes? Name any three elements that exhibit isotopy and give their respective isotopes.
b) Explain what you understand as relative abundance.
c) Why is the relative atomic mass of chlorine 35.5

Question 4
The abundance in nature of three isotopes of magnesium of mass numbers 24, 25, and 26 are 78.6%, x and y. Calculate the values of x and y, if the relative atomic mass of magnesium is 24.327

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Solubility

2016-01-07T13:26:00.001+01:00

Definition of Terms
Solute
This is a substance that dissolves in a solvent to form a solution. It can be solid, liquid or gas. Examples include common salt, sugar, copper (II) tetraoxosulphate (VI) etc.
Solvent
This is a substance that dissolves a solute to form a solution. It can be liquid or gas. Examples are water, ethanol, benzene etc.
Solution
This is a mixture of solute and solvent, i.e,
         Solute + Solvent = Solution
It can be homogeneous with a uniform composition, e.g, an unsaturated solution of a sodium chloride, or heterogeneous with non-uniform composition like an unsaturated solution of the same salt.
Saturated Solution
A saturated solution is one that holds as much solute as it can dissolve in the presence of undissolved solute particles at a given temperature. In a saturated solution, the dissolved and undissolved solutes are in equilibrium at a given temperature.
Unsaturated Solution
An unsaturated solution is one that can still dissolve more solutes at a given temperature until it becomes saturated.
For instance, the solubility limit of sodium chloride is 36g dm^-3, which means that as long as the mass of sodium chloride in 1dm^3 of its solution is below 36g, the solution will remain unsaturated. However, immediately the mass is 36g and above, the solution will be considered to be saturated because any mass in excess of 36g will be precipitated out of solution as undissolved particles. This implies that if the mass of sodium chloride in the solution is 37g; 1g which is the excess will be precipitated out as undissolved particles.
Supersaturated Solution
A solution which contains more of the solute than it can normally hold at a given temperature is said to be supersaturated. The supersaturated solution is unstable at a given temperature and the solute particles can crystallize out of the solution at the slightest touch or addition of a dry crystal of the solute. At times, dust particles can cause the crystallization of the solute particles. Also, merely scratching the inside of the glass vessel holding the solution with a glass rod or spatula can cause the solute particles to crystallize out. This is known as seeding, and is employed in crystallization as a separation technique.
Solubility
The solubility of a solute in a given solvent is the amount of a solute in moles or grams required to saturate 1 dm^3 of the solvent at a definite temperature. Its units are moles per dm^3 (mol dm^-3), g per dm^3 (g dm^-3) or g per 100g of solvent.
It is a way of comparing the level to which different solutes dissolve in a given solvent at a particular temperature.
Generally, the solubility of some substances increases with increase in temperature, especially most solids. This is why your hot water dissolves more sugar or beverages than cold water. However, the solubility of gases decreases with increase in temperature and increases with increase in pressure. Gases like hydrogen chloride and ammonia are highly soluble in water and do not follow this rule.
Calculations on Solubility
Mathematically, solubility can be expressed as:
Solubility (mol dm^3) = No. of moles/Volume (dm^3)…………(i)
                             = (No. of moles/Volume (cm^3)) x 1000………(ii)
Equations (i) and (ii) are used when the volume is given in dm^3 and cm^3 respectively.
Also,
Solubility (g dm^-3) = Reacting mass/Volume (dm^3)…………(iii)

                        = (Reacting mass/Volume (cm^3)) x 1000………(iv)
but,
No. of moles = Reacting mass/Molar mass…………(v)
Substituting equation (v) into (i) and (ii), then;
Solubility (mol dm^-3)
= (Reacting mass/Molar mass) x 1/Volume (dm^3)……(vi)
= (Reacting mass/Molar mass) x 1000/Volume (cm^3)……(vii)
Equations (vi) and (vii) are used when the molar mass with the reacting mass is given.
Solubility Curve
The solubility of a given varies with changes in temperature, therefore, a plot of solubility against on the y-axis against temperature on the x-axis is known as a solubility curve. (Please refer to your textbooks for standard solubility curves)
A study of the solubility curves for some solutes in a given solvent over a range of temperatures shows that the solubility of:
a) KNO3 increases very rapidly over a short range of temperature.
b) NaCl is independent of changes in temperature.
c) KClO3 also increases very rapidly like KNO3, but over a wider range of temperature.
d) KCl is linearly proportional to temperature.
d) Ca(OH)2 is relatively independent of temperature.
Applications of Solubility Curves
a) They are used in pharmaceutical companies to determine the amounts of drugs that must be dissolved in a particular quantity of solvent to give a prescribed drug mixture.
b) They are used by chemists and other research workers to determine the most suitable solvents that can be used to extract essential oils from their natural sources at various temperatures. Example of an essential oil is the eucalyptus oil.
c) It is applied in fractional crystallization.
Examples
Question 1
In an experiment to determine the solubility of a given salt Z, the following data were obtained:
Mass of empty dish = 14.32g
Mass of dish + saturated solution of salt Z = 35.70g
Mass of dish + salt Z = 18.60g
Temperature of solution = 30°C
Molar Mass of salt Z = 100
Density of solution Z = 1.00g cm^-3
Determine the solubility of salt Z in (i) g dm^-3 (ii) mol dm^-3
Answer
Mass of saturated solution
= Mass of dish + saturated solution of salt Z - Mass of empty dish
= 35.70g - 14.32g
= 21.38g
Mass of salt Z = Mass of dish + salt Z - Mass of empty dish
                           = 18.60g - 14.32g
                           = 4.28g
Molar Mass of salt Z = 100
Amount in moles of salt Z = Mass of salt Z/Molar mass of salt Z
                                               = 4.28/100
                                               = 0.0428 mole
Density of solution Z = 1.00g cm^-3
Volume of solution = Mass of solution/Density of solution
                                   = 21.38g /1g cm^-3
                                   = 21.38 cm^3
Therefore, solubility of salt Z in:
i) g dm^-3 = (Mass/Volume (cm^3)) x 1000
                   = (4.28/21.38) x 1000
                   = 200.19g dm^-3
ii) mol dm^-3 = (Moles/Volume (cm^3)) x 1000
                         = (0.0428/21.38) x 1000
                         = 2.00mol dm^-3
Question 2
16.55g of lead (II) trioxonitrate (V) was dissolved in 100g of distilled water at 20°C, calculate the solubility of the solute in mol per dm^3. [Pb = 207, N = 14, O = 16]
Answer
Mass of Pb(NO3)2 = 16.55g
Molar mass of Pb(NO3)2 = (1*Pb) + 2[(1*N) + (3*O)]
                                             = (1 x 207) + 2[(1 x 14) + (3 x 16)]
                                             = 207 + 2(14 + 48)
                                             = 207 + 124
                                             = 331g mol^-1
Volume of water = 100cm^3 (density of water = 1g cm^-3)
Solubility (mol dm^-3) = (Reacting mass/Molar mass) x 1000/Volume (cm^3)
                                        = (16.55/331) x (1000/100)
                                        = 0.05mol dm^-3
Question 3
Potassium trioxochlorate (V) has a solubility of 1.8 mol per dm^3 at 35°C. On cooling this solution to a temperature of 20°C, the solubility was found to be 0.3 mol per dm^3. What mass of KClO3 was crystallized out? [K = 39, Cl = 35.5, O = 16]
Answer
Molar mass of KClO3 = (1*K) + (1*Cl) + (3*O)
                                      = (1 x 39) + (1 x 35.5) + (3 x 16)
                                      = 39 + 35.5 + 48
                                      = 122.5g mol^-1
Solubility of salt at 35°C = 1.8mol dm^-3
Solubility of salt at 20°C = 0.3mol dm^-3
Amount in moles of KClO3 crystallized out on cooling from 35°C - 25°C = 1.8 - 0.3
                      = 1.5 moles
Therefore, mass of salt crystallized out on cooling = Amount in moles x Molar mass
                                                                                        = 1.5 X 122.5
                                                                                        = 183.75g
Question 4
The solubility of a salt of molar mass 101g at 35°C is 0.35 mol per dm^3. If 3.50g of the salt is dissolved completely in 250cm^3 of water in beaker, what will the resulting solution be - unsaturated, saturated or supersaturated?
Answer
Solubility of salt at 35°C = 0.35mol dm^-3
Molar mass of salt = 101g
Mass of salt in solution = 3.50g
Volume of water = 250cm^3
Solubility of salt in saturated solution = (Reacting mass/Molar mass) x 1000/Volume (cm^3)
= (3.5/101) x (1000/250)
= 0.138mol dm^-3
Since the calculated solubility of the salt solution in the beaker at 35°C, is less than the standard solubility of the salt at the same temperature, it means that the salt has not reached its solubility limit at that given temperature. Hence, the resulting solution will be unsaturated.
Question 5
The solubility of Na3AsO4.12H2O is 38.9g per 100g H2O. What is the percentage of Na3AsO4 in the saturated solution? [As = 75, Na = 23, O = 16, H = 1]
Answer
Molar mass of Na3AsO4.12H2O = (3*Na) + (1*As) + (4*O) + 12[(2*H) + (1*O)]
= (3 x 23) + (1 x 75) + (4 x 16) + 12[(2 x 1) + (1 x 16)]
= 69 + 75 + 64 + 12(2 + 16)
= 208 + 216
= 424g mol^-1
From the above, molar mass of Na3AsO4 = 208g mol^-1
Mass of Na3AsO4.12H2O in 100g of water = 38.9g
Volume of water = 100cm^3 (density of water - 1g cm^-3)
Therefore, solubility of Na3AsO4.12H2O (g dm^-3)
= (Mass/Volume (cm^3 )) x 1000
= (38.9/100) x 1000
= 389g dm^-3
but,
424g of Na3AsO4.12H2O contains 208g of Na3AsO4
Therefore,
389g of Na3AsO4.12H2O will contain x of Na3AsO4
x = (389 x 208)/424
   = 190.83g of Na3AsO4
Percentage of Na3AsO4 in the saturated solution
= (Mass of Na3AsO4 in solution/Mass of Na3AsO4.12H2O in solution) x 100
= (190.83/389) x 100
= 49.1%
Question 6
80g of a salt, XCl2, was placed in
40cm^3 of water to give a saturated solution at 25°C. If the solubility of the salt is 8 mol dm^-3 at that temperature, what is the mass of the salt that will be left undissolved at the given temperature? [X = 24, Cl = 35.5]
Answer
Molar mass of XCl2 = (1*X) + (2*Cl)
                                    = (1 x 24) + (2 x 35.5)
                                    = 24 + 71
                                    = 95g mol^-1
At 25°C,
Solubility of XCl2      = 8.0mol dm^-3
i.e, 1 dm^3 of solution contains 8 x 95 = 760g
If, 1000cm^3 of water = 760g of XCl2
Then, 40cm^3 of water = x g of XCl2
                                      x = 760 x 40/1000
                                         = 30.4g of XCl2
From the above, based on the solubility of the salt, 40cm^3 will only dissolve 30.4g
Therefore, the mass of the salt that will be left undissolved is 80 - 30.4 = 49.6g
Question 7
The solubility of potassium trioxochlorate (V) is 1500g per 1000g water at 80°C and 600g per 1000g water at 45°C. Calculate the mass of KClO3 that will crystallize out of solution if 200g of the saturated solution at 80°C is cooled to 45°C.
Answer
Masses at 80°C,
1000g of water + 1500g of KClO3 = 2500g of saturated solution
Masses at 45°C,
1000g of water + 600g of KClO3 = 1600g of saturated solution
Mass of salt deposited on cooling from 80°C to 45°C
= 2500 - 1600 = 900g
If 2500g of saturated solution deposit 900g of solute on cooling
Then, 200g of the saturated solution will deposit x g of solute
x = 900 x 200/2500
   = 72g of KClO3.
Do These
Question 1
If 24.4g of lead (II) trioxonitrate (V) were dissolved in 42g of distilled water at 30°C; calculate the solubility of the solute in g per dm^3
Question 2
Calculate the solubility in mol per dm^3 of 50g of CuSO4 dissolved in 100g of water at 120°C. [Cu = 64, S = 32, O = 16]
Question 3
If 10.5g of lead (II) trioxonitrate (V) in 20cm^3 of distilled water at 18°C, what is the solubility of the solute in mol per dm^3?
Question 4
117g of NaCl was dissolved in 1dm^3 of distilled water at 25°C. Determine the solubility in mol per dm^3 of sodium chloride at that temperature. [Na = 23, Cl = 35.5]
Question 5
a) What do you understand by terms saturated solution and solubility?
b) When 50cm^3 of a saturated solution of sugar of molar mass 345g at 50°C was evaporated to dryness, 34.5g of dry solid was obtained. Calculate the solubility of sugar ay 50°C in a) g per dm^3 b) mol per dm^3
Question 6
3.06g of a sample of potassium trioxochlorate (V), KClO3, was required to make a saturated solution with 10cm^3 of water at 25°C. What is the solubility of the salt at that temperature?
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Gas Laws (Part II): Gay-Lussac's Law & Avogadro's Law

2015-12-28T21:42:00.001+01:00

The Gay-Lussac's Law of Combining Volumes states that when gases react, they do so in volumes, which are in simple ratio to one another and to the volume of the product, if any; provided temperature and pressure remain constant. It applies to only gases, which means that solid and liquid reactants and products are always ignored when applying this law.

For instance, hydrogen burns in oxygen at 100°C to form steam according to the equation:

       2H2(g) + O2(g) ---> 2H2O(g)
         2mol      1mol           2mol
         2vol        1vol            2vol
         2cm^3    1cm^3        2cm^3

From the above, it implies that at 100°C, when water is in its gaseous state, 2 volumes of hydrogen gas (dm^3 or cm^3) will combine with 1 volume of oxygen gas to form 2 volumes of steam, to give a simple mole ratio of 2 : 1 : 2. Therefore, 50cm^3 of hydrogen will need 25cm^3 of oxygen to produce 50cm^3 of steam. Similarly, 15cm^3 of oxygen will require 30cm^3 of hydrogen to form 30cm^3 of steam. In other words, if you start with (2 + 1 = 3) volumes of gases (hydrogen + oxygen); after the reaction, you will be left with 2 volumes of residual gas (steam).

Now, what will happen if this same reaction is carried out at a temperature lower than 100°C, say 50°C. The product will not be steam but water. This is because at the reaction temperature of 50°C, water is liquid, and not gaseous. So, the equation of reaction becomes:

        2H2(g) + O2(g) ---> 2H2O(l)
         2mol      1mol           2mol
         2vol        1vol                -
         2cm^3    1cm^3            -

From the analysis, it is only hydrogen and oxygen that contribute to the gaseous volume of the reaction, the water [H2O(l)] has no effect. Therefore, if you start with (2 + 1 = 3) volumes of gases (hydrogen + oxygen); after the reaction, you will be left with zero or no volume of residual gas.
According to Gay-Lussac, the gaseous reactants have no relationship with the liquid product, in terms of gas volumes.

Examples

Question 1
65cm^3 of hydrogen are sparked with 30cm^3 of oxygen at 100°C and 1atm. Calculate the total volume of residual gases at:
a) 100°C
b) room temperature

Answer
a)                            2H2(g) + O2(g) ---> 2H2O(g)
                                2vol           1vol             2vol
Before sparking     65cm^3    30cm^3       -
On sparking            60cm^3   30cm^3       60cm^3
After sparking          5cm^3      0cm^3       60cm^3

Therefore, volume of residual gas left at 100°C will be
= 5cm^3 of hydrogen + 60cm^3 of steam
= 65cm^3 of gases

Analysis
From the stoichiometry, 2 volumes of hydrogen will burn in 1 volume of oxygen to produce 2 volumes of steam, right? Going by the volume of gases available, what volume of oxygen will we need to completely spark 65cm^3 of hydrogen? That will be 1/2 of 65 = 32.5cm^3, isn't it? But the volume of oxygen available is only 30cm^3, which means it is not enough to completely spark the available hydrogen. So, we make use of the available oxygen, which implies that only 30 x 2 = 60cm^3 of hydrogen will react, while 65 - 60 = 5cm^3 of hydrogen will be left unreactive.

Since the oxygen is not enough to completely burn the hydrogen gas, it means it was used up and nothing was left unreactive, right? Also, from the stoichiometry, 30cm^3 of oxygen will produce 60cm^3 of steam, won't it? This implies that since all the oxygen reacted with hydrogen, it means 60cm^3 of steam was produced. Recall that no steam was produced before the sparking. Therefore, 60cm^3 of steam will be left as a residual gas, in addition to the 5cm^3 of unreactive hydrogen to give us a total of 60 + 5 = 65cm^3 of residual (left-over) gas.

b) Equation of reaction:
                               2H2(g) + O2(g) ---> 2H2O(l)
                                2vol      1vol             -
Before sparking     65cm^3    30cm^3       -
On sparking          60cm^3    30cm^3       -
After sparking          5cm^3      0cm^3        -

Therefore, volume of residual gas at room temperature will be
= 5cm^3 of hydrogen

Analysis
The reaction is similar to the one above, except that it occurs at room temperature, and at this temperature, water is a liquid and not gas. [Note that room temperature is the normal temperature of the room (without air-conditioner or heater) at a particular time]. In that case, it is only the reactants that will account for the residual gases if any. Since 5cm^3 of hydrogen is the only volume of gas left after the reaction, it therefore, accounts for the volume of residual gas as 5cm^3.

Question 2
60cm^3 of a mixture of carbon (II) oxide and hydrogen yielded 20cm^3 of carbon (IV) oxide after explosion with excess of oxygen. What is the percentage of carbon (II) oxide in the mixture?

Answer
2CO(g) + O2(g) ---> 2CO2(g)
2vol 1vol 2vol

From the question, no concrete information is provided for oxygen, so will have to ignore it and focus on CO and CO2.

No information is provided on the actual volume of CO that reacted with the excess oxygen, but we can use the volume of CO2 produced to get that as shown below:

If 2vol of CO2 = 2vol of CO
Then, 20cm^3 of CO2 = 20cm^3 of CO.

The percentage of CO in the mixture is given by:
(Volume of CO that reacted/Volume of mixture) x 100
= (20/60) x 100
= 0.3333 x 100
= 33.33%

It is important to note that since the product of the explosion between the mixture and oxygen was CO2, it means that it was only CO that took part in the reaction.

Question 3
25cm^3 of hydrogen was sparked with 20cm^3 of oxygen at atmospheric pressure. What is the total volume of the residual gas at room temperature?

Answer
Temperature of reaction = room temperature (>0°C and <100°C), this means that the product was water and not steam.
                               2H2(g) + O2(g) ---> 2H2O(l)
                                2vol           1vol             2vol
Before sparking     25cm^3    20cm^3        -
On sparking            25cm^3   12.5cm^3     -
After sparking          0cm^3      7.5cm^3      -

Therefore, total volume of residual gas left at room temperature was
= 7.5cm^3 of oxygen

Analysis
From the stoichiometry, 20 x 2 = 40cm^3 of hydrogen will be required to completely react with the 20cm^3 of oxygen; but the volume of hydrogen available is just 25cm^3, and it will require 25/2 = 12.5cm^3 of oxygen to completely react as shown above.

Question 4
20cm^3 of carbon (II) oxide is mixed and sparked with 80cm^3 of air containing 21% oxygen. If all the volumes are measured at s.t.p., calculate the volume of the resulting gases.

Answer
Volume of air = 80cm^3
Percentage of oxygen in air = 21%
Actual volume of oxygen = (21/100) x 80
= 16.80cm^3
Remaining components of air that do not react = Volume of air - Volume of oxygen
= 80 - 16.8
= 63.2cm^3 (made up of nitrogen, carbon (IV) oxide, water vapour and noble gases)

At s.t.p., carbon (II) oxide, oxygen and carbon (IV) oxide are all gases. So, the equation of reaction is:

                                2CO(g) + O2(g) ---> 2CO2(g)
                                 2vol          1vol           2vol
Before sparking     20cm^3    16.8cm^3        -
On sparking            20cm^3   10cm^3       20cm^3
After sparking          0cm^3      6.8cm^3      20cm^3

Total volume of resulting gas = Volume of excess oxygen + Volume of CO2 produced + Volume of unreactive components of air
= 6.8cm^3 + 20cm^3 + 63.2cm^3
= 90cm^3

Question 5
2H2(g) + O2(g) ---> 2H2O(g)
In the reaction above, what volume of hydrogen would be left over when 300cm^3 of oxygen and 1000cm^3 of hydrogen are exploded in a sealed tube?

Answer
                                 2H2(g) + O2(g) ---> 2H2O(g)
                                   2vol          1vol            2vol
Before explosion 1000cm^3   300cm^3        -
On explosion        600cm^3    300cm^3       600cm^3
After explosion     400cm^3       0cm^3        600cm^3

Therefore, the volume of hydrogen that would be left = 400cm^3

Question 6
35cm^3 of hydrogen was sparked with 12cm^3 of oxygen at 110°C and 760mmHg to produce steam. What percentage of the total volume of gas left after the reaction is hydrogen?

Answer
                                   2H2(g) + O2(g) ---> 2H2O(g)
                                   2vol          1vol              2vol
Before sparking        35cm^3   12cm^3          -
On sparking              24cm^3    12cm^3       24cm^3
After sparking          11cm^3       0cm^3        24cm^3

Total volume of residual gas = Volume of unreactive hydrogen + Volume of steam produced
= 11cm^3 + 24cm^3
= 35cm^3

Percentage of hydrogen in residual gas
= (Volume of excess (unreactive) hydrogen/Total volume of residual gas) x 100
= (11/35) x 100
= 0.3143 x 100
= 31.43%

Avogadro's Law
This law states that equal volumes of all gases contain the same number of molecules at the same temperature and pressure. It is related to Gay-Lussac's Law, as its application shows that the volume of gases can be converted into number of molecules at constant temperature and pressure. It implies that if we know the equal volumes of two gases A and B, and the number of molecules of one of the gases, we can easily determine the number of molecules of the other gas.

For instance, when hydrogen gas combines with oxygen gas to form steam according to the equation:

                               2H2(g) + O2(g) ---> 2H2O(g)
Mole ratio               2         :       1         :        2
Gay-Lussac's          2vol   :      1vol     :      2vol
Avogadro's              2        :       1         :        2
                          molecules   molecule     molecules

Similarly, the reaction between nitrogen and hydrogen to form ammonia can be expressed as:

                                N2(g) + 3H2(g) ---> 2NH3(g)
Mole ratio               1         :       3         :        2
Gay-Lussac's          1vol   :      3vol     :      2vol
Avogadro's              1        :       3         :        2
                          molecule   molecules     molecules

Also, the reaction between carbon (II) oxide and oxygen to produce carbon (IV) oxide:

                               2CO(g) + O2(g) ---> 2CO2(g)
Mole ratio               2         :       1         :        2
Gay-Lussac's          2vol   :      1vol     :      2vol
Avogadro's              2        :       1         :        2
                          molecules   molecule     molecules

Hydrogen reacts with fluorine to form hydrogen fluoride according to the equation:

                                 H2(g) + F2(g) ---> 2HF(g)
Mole ratio               1         :       1         :        2
Gay-Lussac's          1vol   :      1vol     :      2vol
Avogadro's              1        :       1         :        2
                          molecule   molecule     molecules

In the above equations, we can see that all the volumes of the gases are converted to molecules, based on the assumption that
1 mol of a gas = 1 vol of the gas = 1 molecule of the gas, at constant temperature and pressure.

Examples

Question 7
8g of methane, CH4 occupies 11.2dm^3 at s.t.p. What volume would 22g of propane, C3H8 occupy under the same condition?

Answer
Convert the given masses to moles
Molar mass of CH4 = (1*C) + (4*H)
                                   = (1*12) + (4*1)
                                   = 12 + 4
                                   = 16g/mol

Molar mass of C3H8 = (3*C) + (8*H)
                                     = (3*12) + (8*1)
                                     = 36 + 8
                                     = 44g/mol

Reacting mass of CH4 = 8g
Number of moles of CH4 = 8/16
= 0.5mol

Reacting mass of C3H8 = 22g
Number of moles of C3H8 = 22/44
= 0.5mol

If 0.5mol of CH4 occupies 11.2dm^3
Then, 0.5mol of C3H8 will also occupy 11.2dm^3

According to Avogadro's Law,since CH4 and C3H8 contain the same number of moles under the same condition, it implies that they will occupy the same volume.

Question 8
Calculate the volume of oxygen required to completely burn 20cm^3 of ethane (C2H6)

Answer
Equation of reaction:
             2C2H6(g) + 7O2(g) ---> 4CO2(g) + 6H2O(g)
Mole ratio    2       :    7             :     4             :   6
Vol ratio    2vol         7vol              4vol            6vol

From the above,
2cm^3 of C2H6 = 7cm^3 of oxygen
20cm^3 of C2H6 = x cm^3 of oxygen
                     2 * x = 20 * 7
                         2x = 140
                           x = 140/2
                              = 70cm^3 of oxygen

Question 9
What is the volume of propyne (C3H4) required to completely burn in 20cm^3 of oxygen?

Answer
Equation of reaction:
              C3H4(g) + 4O2(g) ---> 3CO2(g) + 2H2O(g)
Mole ratio    1       :    4             :     3             :   2
Vol ratio      1vol       4vol              3vol            2vol

From the stoichiometry,
1cm^3 of C3H8 = 4cm^3 of oxygen
x cm^3 of C3H8 = 20cm^3 of oxygen
                     4 * x = 20 * 1
                         4x = 20
                           x = 20/4
                              = 5cm^3 of propyne

Do These

Question 1
20cm^3 of hydrogen gas are sparked with 20cm^3 of oxygen gas at 373K (100°C) and 1 atmosphere. The resulting mixture is cooled to 298K (25°C) and passed over calcium chloride. Calculate the volume of the residual gas.
A. 40cm^3 B. 20cm^3 C. 30cm^3 D. 10cm^3

Question 2
10cm^3 of hydrogen fluoride gas (HF) reacts with 5cm^3 of dinitrogen difluoride gas (N2F2) to form 10cm^3 of a single gas (NHF2).
a) Write the most likely equation of the reaction.
b) What volume of gas will be left if 50cm^3 of HF reacts with 40cm^3 of N2F2?

Question 3
Calculate the minimum volume of oxygen required for the complete combustion of a mixture of 10cm^3 of CO and 15cm^3 of H2.
A. 25.0cm^3 B. 12.5cm^3 C. 10.0cm^3 D. 5.0cm^3

Question 4
20 volumes of a mixture of hydrogen and carbon (II) oxide were mixed with 13 volumes of oxygen and the mixture exploded. On cooling, the resulting volume was found to be 9. Determine the composition of the mixture.

Question 5
One mole of propane is mixed with five moles of oxygen. The mixture is ignited and the propane burns completely. What is the volume of the products at s.t.p? [H = 1, C = 12, O = 16, G.M.V = 22.4dm^3 per mol]

Question 6
50cm^3 of carbon (II) oxide was exploded with 150cm^3 of air containing 20% oxygen by volume. Which of the reactants was in excess?
A. Carbon (II) oxide B. Oxygen C. Carbon (IV) oxide D. Nitrogen

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