Preview:

We shall begin by describing equilibrium state as the stage when there is no change in concentration of the reactants or products and the reaction appears to have stopped. Consequently, we shall learn about the equilibrium that exists between various phase transitions such as: solid-liquid, liquid-gas and solid-gas.

Next, we shall demonstrate dynamic equilibrium with an experiment. A dynamic equilibrium is a chemical equilibrium between a forward reaction and the reverse reaction where the rate of the reactions becomes equal.

Then, we shall state the Law of mass action as, “at constant temperature, the rate of a chemical reaction is directly proportional to the product of active masses of the reactants”. We will then study about reversible and irreversible reactions in some detail.

In the next section, we will state law of chemical equilibrium as, “At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.”

We shall then study about Homogeneous and Heterogenous Equilibria in some detail. The equilibrium that exists in a reaction with homogeneous reactants and products, that is, with reactants and products being in the same physical state is called homogeneous equilibrium and when the reactants and products are in different physical states, we call it heterogenous equilibrium.

In the next section, we shall learn an important principle called Le Chatelier’s principle which states that, change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change. This principle governs the effects due to change in concentration, temperature, pressure and so on, on the equilibrium of a reaction.

Subsequently, we shall talk about ionic equilibrium as the equilibrium that exists between ionised and unionised ions in a reaction. Then we shall discuss in brief, various theories explaining acids and bases like, Arrhenius Concept, Brönsted-Lowry’s theory of Acids and Bases and; Lewis theory of acids and bases.

Then, we shall discuss about the pH scale used to measure the acidity or basicity of substances. pH of a solution is defined as the negative logarithm to the base 10 of hydronium ion (or hydrogen ion) concentration in mol $L^{-1}$.

Then we shall discuss about polybasic acids as those acids which have more than one ionizable proton per molecule of the acid.

Next, we define Common Ion Effect as the shift in equilibrium on adding a substance that provides more of an ionic species already present in the dissociation equilibrium.

We will discuss another important phenomenon called, Salt Hydrolysis as the process of interaction between water and cations or anions or both of salts.

In the last section, we will learn about Buffer Solutions as the solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali. We shall end the chapter with the study of solubility of ionic solids in water.

7.0 Introduction:

Most of the chemical reactions when carried out in closed containers do not go to completion under a given set of conditions of temperature and pressure. In such cases, as the reaction proceeds, the concentration of the reactants decreases and that of products increases.

A stage is reached when there is no change in concentration of the reactants or products and the reaction appears to have stopped. This state of system is called, the Equilibrium state. At equilibrium, the rate of evaporation is equal to the rate of condensation. It may be represented by:

$$ \ce { H_2O (l) <=> H_2O(vap)}$$

The double half arrows indicate that the processes in both the directions are going on simultaneously. The mixture of reactants and products in the equilibrium state is called an equilibrium mixture.

Equilibrium is not static but is dynamic. A dynamic equilibrium is a chemical equilibrium between a forward reaction and the reverse reaction where the rate of the reactions are equal.

Based on the extent to which the reactions proceed to reach the state of chemical equilibrium, these may be classified into three groups:

(i) The reactions that proceed nearly to completion and only negligible concentrations of the reactants are left.

(ii) The reactions in which only small amounts of products are formed and most of the reactants remain unchanged at equilibrium stage.

(iii) The reactions in which the concentrations of the reactants and products are comparable, when the system is in equilibrium.

Note box: Some important aspects of equilibrium involving physical and chemical processes along with the equilibrium involving ions in aqueous solutions is called as Ionic Equilibrium.

Questions for the section:

1. What do we mean by: “Equilibrium state”?

2. What is an equilibrium mixture?

3. What is dynamic equilibrium?

7.1 Equilibrium in Physical Processes

The most familiar examples of physical processes are the phase transformations:

$$ \ce { solid <=> liquid }$$

$$ \ce { liquid <=> gas }$$

$$ \ce { solid <=> gas } $$

7.1.1 Solid-Liquid Equilibrium

When a mixture of ice and water mixture is kept at 0⁰C and 1 atmosphere pressure, in a perfectly insulated Thermosflask, though some molecules of ice may melt into water, equal number of water molecules freeze into ice so that the mass of ice and water remains constant. There exists an equilibrium between ice and water at 0⁰C and 1 atmosphere pressure.

$$ \ce { H_2O_{(solid)} <=> H_2O_{(liquid)} }$$

At equilibrium, rate of melting = rate of freezing

The temperature at which the solid and liquid are in equilibrium, at 1 atmosphere is known as normal freezing point of that substance.

Thus, the normal freezing point of water (and melting of ice) is 0⁰C.

7.1.2 Liquid-Vapour Equilibrium

• Let us consider the example of a transparent box carrying a U-tube with mercury (manometer).
• Drying agent like anhydrous calcium chloride (or phosphorus penta-oxide) is placed for a few hours in the box.
• After removing the drying agent by tilting the box on one side, a watch glass or petri dish containing water is quickly placed inside the box.
• It will be observed that, the mercury level in the right limb of the manometer slowly increases and finally attains a constant value. This indicates that, pressure inside the box increases and reaches a constant value. This variation in pressure can be explained as follows: Initially, there was no water vapour (or very less) inside the box. As water evaporated, the pressure in the box increased due to addition of water molecules in the form of gaseous phase inside the box. Consequently, the volume of water in the watch glass decreases (Figure 7.1).

• The rate of evaporation is constant. However, the rate of increase in pressure decreases with time because the vapours get condensed into water which finally leads to an equilibrium condition when there is no net evaporation.
• This implies that, the number of water molecules from the gaseous state into the liquid state also increases till the equilibrium is attained that is,

Rate of evaporation = Rate of condensation

$$ \ce { H_2O (l) <=> H_2O(vap)}$$

• At equilibrium, the pressure exerted by the water molecules at a given temperature remains constant and is called, the equilibrium vapour pressure of water or just, vapour pressure of water.
• Vapour pressure of water increases with temperature.

Note box: If the above experiment is repeated with methyl alcohol, acetone and ether, it is observed that, different liquids have different equilibrium vapour pressures at the same temperature. The liquid which has a higher vapour pressure is more volatile and has a lower boiling point. In an open system where the watch glass is open to the atmosphere, the rate of evaporation is much higher than the rate of condensation. Thus, equilibrium is not possible to reach in an open system. Water and water vapour are in equilibrium at atmospheric pressure (that is, 1.013 bar) and at 100°C in a closed vessel.

Time taken by a liquid for complete evaporation depends on:

• the nature of the liquid,
• the amount of the liquid;
• the temperature and;
• the speed of wind.

Normal Boiling point

Definition box: Normal Boiling point: For any pure liquid at one atmospheric pressure (1.013 bar), the temperature at which the liquid and vapours are at equilibrium is called normal boiling point of the liquid.

Boiling point of the liquid depends on:

1. the atmospheric pressure; 
2. altitude of the place (at high altitude the boiling point decreases).

7.1.3 Solid – Vapour Equilibrium

Definition box: Sublimation: Conversion of a solid directly into its vapour is called sublimation.

• For example, when solid iodine is placed in a closed vessel, after sometime, the vessel gets filled up with violet vapour and the intensity of colour increases with time.
• After certain time, the intensity of colour becomes constant and at this stage equilibrium is attained.
• Hence, solid iodine sublimes to give iodine vapour and the iodine vapour condenses to give solid iodine.
• The equilibrium between the solid phase and vapour phase of iodine is represented as:

$$ \ce { I_2(solid) <=> I_2(vapour) }$$

Other examples showing this kind of equilibrium are,

$$ \ce { Camphor(solid) <=> Camphor(vapour) }$$

$$ \ce { NH_4Cl (solid) <=> NH_4Cl(vapour)}$$

7.1.4 Equilibrium Involving Dissolution of Solid or Gases in Liquids

i) Solids in liquids

• Take about 100$cm^3$ of water in a beaker.
• Add small quantities of sugar and stir the contents. Sugar dissolves.
• Continue to add sugar and stir.
• If we continue to add more sugar, a stage is reached when no more sugar appears to dissolve.
• The solution which has dissolved the maximum amount of solute is known as saturated solution of that solute in the given solvent at a particular temperature.
• A saturated solution is in equilibrium with the solid solute, at a given temperature.
  $$ \ce {Sugar(solution) <=> sugar(solid) }$$
• At equilibrium, the rate of dissolution of the solute is equal to the rate of crystallisation of the solute from the solution.

Definition box: Solubility: The solubility of a solute in a given solvent is defined as the mass in grams of the solute dissolved in 100 grams of solvent to form a saturated solution at a given temperature.

Solubility is also expressed in terms of number moles of the solute dissolved in a litre of the saturated solution of the solute at a given temperature.

The solubility depends on:

1. The nature of solute.
2. The nature of solvent.
3. The temperature.

ii. Gases in Liquids

• Soda water is a solution of Carbon dioxide in water under pressure.
• In a soda can, some of the carbon dioxide is present dissolved in water and remaining gas is present above the solution.
• There is equilibrium between the molecules in the gaseous state above the solution and the molecules dissolved in the liquid under pressure.

$$ \ce {CO_2 (gas) <=> CO_2(in \space solution )}$$

• This equilibrium gets altered if either the pressure or the temperature is changed. If the soda can is opened, the pressure is released and carbon dioxide fizzes out rapidly and makes the soda ‘flat’ in sometime.
• On the other hand, if the temperature increases, soda bottles often explode. This is due to the decrease in solubility of the gas in water which results in increase in pressure inside the bottle.
  The effect of pressure on the solubility of a gas in a liquid is given by Henry’s law which states that, “the mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the pressure of the gas above the solvent”.
• At a given temperature, the mass of a gas dissolved in a given mass of solvent is directly proportional to the pressure of the gas above the solvent as expressed below:

m ∝ P,, at constant temperature

where, “m” is mass of gas and “P” is pressure of the gas above the solvent.

Note box: It can be generalized that, (i) For solid ⇌ liquid equilibrium, there is only one temperature (called, melting point) at 1 atm (1.013 bar) at which the two phases can coexist. If there is no exchange of heat with the surroundings, the mass of the two phases remains constant. (ii) For liquid⇌ vapour equilibrium, the vapour pressure is constant at a given temperature. (iii) For dissolution of solids in liquids, the solubility is constant at a given temperature. For dissolution of gases in liquids, the concentration of a gas in liquid is proportional to the pressure (concentration) of the gas over the liquid.

These observations are summarised in Table 7.1 below:

7.1.5 General Characteristics of Equilibria Involving Physical Processes

(i) Equilibrium is possible only in a closed system at a given temperature.

(ii) Equilibrium is dynamic.

(iii) All measurable properties of the system remain constant.

(iv) When equilibrium is attained for a physical process, it is characterised by constant value of one of its parameters at a given temperature. Table 7.1 lists such quantities.

(v) The magnitude of such quantities at any stage indicates the extent to which the physical process has proceeded before reaching equilibrium.

Problem: The solubility of iodine in water is 1.1 x $10^{-3}$ mol $L^{-1}$ at 288 K. When 0.2 g of iodine is stirred in 100 mL of water till equilibrium is reached, what will be the mass of iodine that is left undissolved? After equilibrium is reached with 0.2 g of iodine and 100 mL of water,we add 150 mL of water to the system. How much iodine will be left undissolved and what will be the concentration of iodine in solution ?

Solution:

Solubility of $I_2$ in water = $ 1.1 \times 10^{-3} \space mol \space L^{-1}$

$$ =1.1 \times 10^{-3} \times 254 \space gL^{-1}$$

$$ = 0.2794 \space gL^{-1}$$

(Mol. mass of $I_2$ = 254 g $mol^{-1}$)

The mass of $I_2$ dissolved in 100 mL of water = $ \frac {0.2794}{1000} \times 100 $

= 0.02794 g

The mass of $I_2$ left undissolved = 0.2 – 0.02794

= 0.17206 g

When 150 mL of water is added, the total volume of water becomes 100 + 150 = 250 mL

$ I_2$ is dissolved in 250 mL of water = $ \frac { 0.2794}{1000} \times 250 $

= 0.06985 g

The mass of $I_2$ undissolved = 0.2- 0.06985

=0.13015 g

Molar concentraion of $I_2$ in solution = $ \frac {0.06985}{254} \times \frac {1000}{25)}$

= 0.0011 mol $L^{-1}$

$ = 1.1 \times 10^{-3} \space mol \space L^{-1}$

Questions from section 7.1:

1. Explain with an example, Solid-Liquid Equilibrium.

2. Explain the experiment demonstrating, Liquid-Vapour Equilibrium.

3. Define Normal boiling point. What are the factors upon which boiling point depends?

4. Define sublimation.

5. Define solubility. What are the factors upon which solubility depends?

6. State Henry’s law.

7. State the general characteristics of equilibria in physical processes.

7.2 Equilibrium in Chemical Processes – Dynamic Equilibrium

Experiment to demonstrate dynamic Equilibrium:

• Equilibrium can be demonstrated by the use of radioactive isotopes.
• Two 100mL measuring cylinders (marked as 1 and 2) and two glass tubes each of 30 cm length is taken. Diameter of the tubes may be same or different in the range of 3-5mm.
• Nearly half of the measuring cylinder-1 is filled with coloured water (for this purpose add a crystal of potassium permanganate to water). The second cylinder is not filled by anything.
• One tube is put in cylinder 1 and another in cylinder 2.
• Then, the upper tip of tube in cylinder 1 is closed with a finger and the coloured water contained in its lower portion is transferred to cylinder 2.
• The transferring of coloured water is continued till the level of coloured water in both the cylinders becomes constant. The transferring is done from both cylinders during the process because, intertransferring coloured solution between the cylinders ensures that, there will not be any further change in the levels of coloured water in two cylinders.
• If we take analogy of ‘level’ of coloured water with ‘concentration’ of reactants and products in the two cylinders, we can say the process of transfer, which continues even after the constancy of level, is indicative of dynamic nature of the process.

Rate or Velocity of a reaction:

Definition box: Rate of a reaction is defined as the change in molar concentration of one of the reactants or the products in unit time.

Rate of a reaction is denoted by $ \frac {dx}{dt}$ where “dx” is the change in concentration of one of the reactants or products in a small interval of time.

For a reaction of the type,

row \bigg( \frac {d[R]}{dt} \bigg)$$

Where,   is the change in concentration of the reactant, “R”. The negative sign shows that, the concentration of the reactant decreases as the reaction progresses.

Since the concentration of product, P increases with time,

$$ Rate = + \frac {d[P]}{dt}$$

The unit of rate of reaction is mol $L^{-1}s^{-1}$.

The Law of mass action

The dependence of rate of reaction on concentration of the reactants at constant temperature is given by the law of mass action. This law was stated by Guldberg and Waage.

The law of mass action states that, “at constant temperature, the rate of a chemical reaction is directly proportional to the product of active masses of the reactants”.

For a reaction of the type,

$$rate, \frac {dx}{dt} \propto [A]$$

$$or \frac {dx}{dt}= k [A]$$

For a reaction of the type,

$$ aA + bB \rightarrow products,$$

$$rate, \frac {dx}{dt} \propto [A]^a[B]^b$$

$$or,\frac {dx}{dt}=k[A]^a[B]^b$$

Where, [A] and [B] are molar concentrations of reactants: A and B; “k” is the rate constant (also known as the velocity constant or specific reaction rate) of the reaction.

If [A] = [B] = 1 mol $L^{-1}$

$$ \frac {dx}{dt} = k $$

Therefore, the rate constant of a reaction is equal to the rate of the reaction when the molar concentration of each reactant is unity.

Reversible and Irreversible Reaction

a) Reversible Reaction: A reaction in which the products of the reaction react to give back the reactants under the same conditions is called a reversible reaction.

Characteristics of a reversible reaction:

• The reactants of the reaction is indicated by a pair of arrows drawn in opposite directions between the reactants and the products as:
  $$ \ce { A + B <=> C + D }$$
• As indicated by the equation, there is accumulation of the products, C and D and depletion of the reactants, A and B (Figure 7.3).

• This leads to a decrease in the rate of forward reaction and an increase in the rate of the reverse reaction.
• The two reactions (forward and reverse) occur simultaneously and at the same rate until the system reaches a state of equilibrium.
• In a reversible reaction, the state of equilibrium can be reached even if we start with only C and D; that is, no A and B being present initially, since equilibrium can be reached from either direction.

$$ \ce {H_2 (g) + I_2 (g) <=> 2HI (g)}$$

$$ \ce {N_2 (g) + 3H_2 (g) <=> 2NH_3 (g)}$$

$$ \ce {2SO_2 (g) + O_2 (g) <=> 2SO_3 (g)}$$

$$ \ce {N_2 (g) + O_2 (g) <=> 2NO (g)}$$

b) Irreversible Reaction

A reaction in which the products do not combine to give back reactants are called

Examples:

1. Thermal decomposition of potassium chlorate,
$$ \ce { 2KClO_{3(s)} -> 2KCl_{(s)} + 3O_{2(g)}}$$

2. Reaction of metals like zinc with dilute sulphuric acid or hydrochloric acid,
$$ \ce {Zn_{(s)} + H_2SO_{4(aq)} -> ZnSO_{4(aq)} + H_{2(g)}}$$
$$ \ce {Mg_{(s)} + 2HCl{(aq)} -> MgCl{_2(aq)} + H_{2(g)}}$$

3. Precipitation reactions between aqueous solutions of electrolytes,
$$ \ce {AgNO_{3(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO_{3(aq)}}$$
$$ \ce {BaCl_{2(aq)} + K_2CrO_{4(aq)} -> BaCrO_{4(s)} + 2KCl_{(aq)}}$$

Questions from section 7.2:

1. Explain the experiment that demonstrates dynamic equilibrium.

2. Define rate of a reaction.

3. State the law of mass action.

4. What is a reversible reaction? State its characteristics.

5. What is an irreversible reaction?

7.3 Law of Chemical Equilibrium and Equilibrium Constant

Let us consider a reversible reaction,

$$ \ce { aA + bB <=> cC + dD }$$

taking place in a closed vessel at constant temperature.

The rate of the forward reaction is given by

$$ v_1 = k_1 [A]^a[B]^b$$

and that of the reverse reaction

$$ v_2 = k_2 [C]^c[D]^d$$

At equilibrium,rate of the forward reaction = rate of the reverse reaction,

$$ v_1 =v_2$$

$$ \therefore k_1 [A]^a [B]^b = k_2 [C]^c[D]^d….(1)$$

where [A],[B],[C] and [D] are respectively the concentrations of reactants A,B and products C and D in mol $L^{-1}$.$k_1$ and $k_2$ are the rate constants of the forward and the reverse reactions.

 The equilibrium constant $K_c$ is the ratio of the rate constant of the forward to the rate constant of the reverse reaction.

 $$ K_c = \frac {k_1}{k_2}….(2)$$

 From equations (1) and (2),

 $$ K_c = \frac {k_1}{k_2} = \frac {[C]^c[D]^d}{[Al^a[B]^b}$$

$K_c$ is the equilibrium constant of the reaction when concentration of the reactants and products are expressed in terms of mol $L^{-1}$.

$K_c$ also represents the law of Chemical Equilibrium

$$ K_c = \frac {[C]^c[D]^d}{[A]^a[B]^b}$$

The law of chemical equilibrium can be stated as follows:

“At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.”

Note 1: Equilibrium constant for the reaction $ \ce { H_{2(g)} + I_{2(g)} <=> 2HI_{(g)}}$ is given by

$$ K_c = \frac {[HI]^2}{[H_2][I_2]}$$

The equilibrium constant for the reverse reaction,

$ \ce { 2HI_{(g)} <=> H_{2(g)} + I_{2(g)} }$ at the same temperature is,

$$ K_c = \frac {[H_2][I_2]}{[HI]^2}$$

$$ = \frac {1}{K_c}$$

Thus, equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction.

Note 2:

Equilibrium constant for the reaction, $ \ce { 1/2 H_2(g) + 1/2 I_2(g) <=> HI(g)}$ is given by $$K_c^{11} = \frac {[HI]}{[H_2]^{1/2}[I_2]^{1/2}}$$

$$ = \Biggl\{ \frac {[HI]^2}{[H_2][I_2] } \Biggr\}^{1/2}$$

$$ = K_c^{1/2}$$

Equilibrium constant for the reaction, $ \ce { nH_2 (g) + nI_2 (g) <=> n 2HI(g)}$

Therefore, equilibrium constant for the reaction is equal to $K_c^n$.

These findings are summarised in Table 7.4.

Example: The following concentrations were obtained for the formation of NH$_3$ from N$_2$ and H$_2$ at equilibrium at 500K. [N$_2$] = 1.5 × 10$^{-2}$ M, [H$_2$] = 3.0 × 10$^{-2}$ M and [NH$_3$] = 1.2 × 10$^{-2}$ M. Calculate equilibrium constant.

Solution:

The equilibrium constant for the reaction, 

N$_2$(g) + 3H$_2$(g) $\rightleftharpoons$ 2NH$_3$(g) can be written as, 

$$K_c = \frac{[NH_3 (g)]^2}{[N_2 (g)][H_2 (g)]^3}$$ 

$$= \frac{(1.2 \times 10^{-2})^2}{(1.5 \times 10^{-2})(3.0 \times 10^{-2})^3}$$ 

$$= 0.106 \times 10^4 = 1.06 \times 10^3$$

Example:  At equilibrium, the concentrations of $N_2 = 3.0 \times 10^{–3}M, \space O_2 = 4.2 \times 10^{–3}M$ and $NO = 2.8 \times 10^{–3} M in a sealed vessel at 800K. What will be $K_c$ for the reaction

$$N_2(g) + O_$(g) \rightleftharpoons 2NO(g)$$

Solution:

For the reaction equilibrium constant, 

$K_c$ can be written as, 

$$K_c = \frac{[NO]^2}{[N_2][O_2]}$$

$$= \frac{(2.8 \times 10^{-3} M)^2}{(3.0 \times 10^{-3} M)(4.2 \times 10^{-3} M)}$$ 

$$= 0.622$$

Significance of Equilibrium constant

1. Large value of equilibrium constant (K>>1) implies that, the forward reaction goes to near completion.
2. Small value of equilibrium constant (K<<1) indicates that the reverse reaction is favoured.
3. The value of equilibrium constant does not depend on initial concentration or pressure or volume of the container.
4. The value of equilibrium constant depends on temperature.
5. The value of equilibrium constant is not affected by the use of catalyst. A catalyst alters the rates of both forward and reverse reactions to the same extent so that, $ K = \frac {k_1}{k_2}$ remains constant.

 

Questions from section 7.3:

1. State the law of chemical equilibrium.

2. Explain the significance of equilibrium constant.

7.4 Homogeneous Equilibria

In a homogeneous system, all the reactants and products are in the same phase. For example, in the reaction, $ \ce { N_2(g) + 3H_2(g) <=> 2NH_3(g) }$reactants and products are in the homogeneous gaseous phase. Similarly, for the reactions,

$ \ce {CH_3COOC_2H_5(aq) + H_2O(l) <=> CH_3COOH (aq) + C_2H_5OH(aq)}$

$ \ce { \space and \space Fe^{3+}(aq) + SCN^- <=>Fe(SCN)^{2+}(aq)}$

all the reactants and products are in homogeneous solution phase. We shall now consider equilibrium constant for such homogeneous reactions.

7.4.1 Equilibrium Constant (in terms of partial pressure, $K_p$) in Gaseous Systems

Let us consider a reversible reaction of the type:

$$ \ce { a_A (g) + b_B (g) \rightleftharpoons c_C (g) + d_D (g)}$$

Let $p_A,p_B,p_C$ and $p_D$ be the partial pressures of A,B,C and D repectviely at equilibrium.

Rate of forward reaction ,$v_1 = k_1 \space p_A^a \times p_B^b $

Rate of reverse reaction ,$v_2 = k_2 \space p_C^c \times p_D^d $

At equilibrium , rate of forward reaction = rate of reverse reaction,

$$v_1 = v_2$$

$$ \therefore k_1 \space p_A^a \times p_B^b = k_2 \space p_C^c \times p_D^d $$

But,$K_p = \frac {k_1}{k_2}$

$$ K_p = \frac {p_C^c \times p_D^d}{p_A^a \times p_B^b}$$

7.4.2 Relationship between $K_p$ and $K_c$

Let us consider an equilibrium reaction,

$$ \ce { aA (g) + bB (g) \rightleftharpoons cC(g) + dD(g)}$$

$$ K_p = \frac {p_C^c \times p_D^d }{p_A^a \times p_B^b} ….(1)$$

Where $K_p$ is the equilibrium constant, $p_A,p_B,p_C$ and $p_D$ are the partial perssures of A, B, C and D respectively.

Let [A],[B],[C] and [D] be the molar concentrations of A,B,C and D at equilibrium

$$ K_c = \frac { [C]^c \times [D]^d}{[A]^a \times [B]^b}….(2)$$

For an ideal gas,PV = nRT

$$ \therefore P = \frac {n}{V} RT$$

$$P = CRT$$

where C is the molar concentration (C =$ \frac {n}{V}$ mol $L^{-1}$).Assuming ideal behaviour.

$$ p_A = C_ART = [A]RT$$

$$ p_B = C_BRT = [B]RT$$

$$ p_C = C_BRT = [C]RT$$

and $$ p_D = C_DRT = [D]RT$$

Substituting these values in equation (1),we get

$$ K_p = \frac { \{ [C]RT\}^c \times \{ [D]RT\}^d}{\{ [A]RT\}^a \times \{[B]RT\}^b}$$

$$ = \frac {[C]^c[D]^d}{[A]^a[B]^b} \times (RT)^{(c+d)-(a+b)}$$

$$ K_p = K_c (RT)^{\Delta n}$$

Where ∆n=(Total number of moles of gaseous products)- (Total number of moles of gaseous reactants)

Example: $PCl_5$, $PCl_3$ and $Cl_2$ are at equilibrium at 500 K and having concentration 1.59M. $PCl_3$, $1.59M$ $Cl_2$ and $1.41M$ $PCl_5$.

Calculate $K_c$ for the reaction, $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ 

Solution:

The equilibrium constant $K_c$ for the above reaction can be written as, 

$$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(1.59)^2}{(1.41)} = 1.79$$

Example:  The value of $K_c = 4.24$ at 800K for the reaction,  $CO (g) + H_2O (g) \rightleftharpoons CO_2 (g) + H_2(g)$ 

Calculate equilibrium concentrations of $CO_2$, $H_2$, $CO$ and $H_2O$ at 800 K, if only $CO$ and $H_2O$ are present initially at concentrations of 0.10M each.

Solution:

For the reaction, 

CO (g) + H$_2$O (g) $\rightleftharpoons$ CO$_2$ (g) + H$_2$ (g) 

Initial concentration: 

0.1M \hspace{1cm} 0.1M \hspace{1cm} 0 \hspace{1cm} 0 

Let x mole per litre of each of the product be formed. 

At equilibrium: 

(0.1-x) M \hspace{0.5cm} (0.1-x) M \hspace{0.5cm} x M \hspace{0.5cm} x M 

where x is the amount of CO$_2$ and H$_2$ at equilibrium. 

Hence, equilibrium constant can be written as, 

$$K_c = \frac{x^2}{(0.1-x)^2} = 4.24$$ 

$$x^2 = 4.24(0.01 + x^2 – 0.2x)$$ 

$$x^2 = 0.0424 + 4.24x^2 – 0.848x$$ 

$$3.24x^2 – 0.848x + 0.0424 = 0$$ 

a = 3.24, b = -0.848, c = 0.0424 

(for quadratic equation ax$^2$ + bx + c = 0, 

$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$ 

x = (0.848 ± √(0.848)$^2$ – 4(3.24)(0.0424)) / (3.24 × 2) 

x$_1$ = (0.848 – 0.4118) / 6.48 = 0.067 

x$_2$ = (0.848 + 0.4118) / 6.48 = 0.194 

the value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration.

Hence the equilibrium concentrations are,

[CO$_2$] = [H$_2$] = x = 0.067 M 

[CO] = [H$_2$O] = 0.1 – 0.067 = 0.033 M

Example: For the equilibrium, 2NOCl(g) $\rightleftharpoons$ 2NO(g) + Cl$_2$(g) the value of the equilibrium constant, $K_c$ is 3.75 × 10$^{-6}$ at 1069 K. Calculate the $K_p$ for the reaction at this temperature? 

Solution:

We know that, 

$$K_p = K_c (RT)^{\Delta n}$$ 

For the above reaction, 

$\Delta n = (2 + 1) – 2 = 1$ 

$$K_p = 3.75 \times 10^{-6} (0.0831 \times 1069)$$ 

$$K_p = 0.033$$

Questions from section 7.4:

1. Arrive at the expression for equilibrium constant in terms of partial pressure for a gaseous system.

7.5 Heterogeneous Equilibrium

The equilibrium reactions in which reactants and products are present in different phases are called heterogeneous equilibrium reactions.

For example,

$$ \ce { H_2O (l) <=> H_2O (g)}$$

$$ \ce { CaCO_3 (s) <=> CaO (s) + CO_2 (g)}$$

In equilibrium expression ,the concentrations of pure solids and liquids are taken as unity.

$$ K_c = [CO_k2(g)] mol \space L^{-1}$$

$$ or \space K_p = p_{CO_2}bar$$

Problem:  Write the equilibrium expressions for each of the following equilibria:

(i) $Ag_2O_(s) + 2HNO_3 (aq) \rightleftharpoons 2AgNO_3 (aq) + H_2O_{(l)}$

Solution:

$K_c = \frac{[AgNO_3]^2}{[HNO_3]^2 }$ (units : No units)

Note: $H_2O_{(l)}$ is present in large excess and therefore $H_2O_{(l)} =1$

(ii) $CrO_4^{2-} (aq) + Pb^{2+} (aq) \rightleftharpoons PbCrO_4 (s)$

Solution:

$$ K_c = \frac {1}{[CrO_4^{2-} (aq)][Pb^{2+}(aq)]}mol^{-2}L^2 $$

(iii) $3Fe (s) + 4H_2O (g) \rightleftharpoons Fe_3O_4 (s) + 4H_2 (g)$

Solution:

$$ K_p = \frac {p^4_{H_2}}{p^4_{H_2O}}(units : No \space units)$$

(iv) $CH_3COOC_2H_5 (aq) + H_2 O (l) \rightleftharpoons CH_3COOH(aq) + C_2H_5OH (aq)$

Solution:

$$ K_c = \frac {[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5]}$$

(v) $CH_3COOH_{(l)} + C_2H_5OH_{(l)} \rightleftharpoons CH_3COOC_2H_{5(l)} + H_2O_{(l)}$

Solution:

$$ K_c = \frac {[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$$

Example: The value of $K_p$ for the reaction, CO$_2$ (g) + C (s) $\rightleftharpoons$ 2CO (g) is 3.0 at 1000 K. If initially $p_{CO_2} = 0.48$ bar and $p_{CO} = 0$ bar and pure graphite is present, calculate the equilibrium partial pressures of $CO$ and $CO_2$. 

Solution:

For the reaction, 

let ‘x’ be the decrease in pressure of $CO_2$, then 

$CO_2 (g) + C(s) \rightleftharpoons 2CO(g) $

Initial pressure: 0.48 bar \hspace{1cm} 0 

At equilibrium:  (0.48 − x) bar $\hspace{2cm}$ 2x bar 

$$K_p = \frac{p_{CO}^2}{p_{CO_2}}$$ 

$$K_p = (2x)^2 / (0.48 – x) = 3$$ 

$$4x^2 = 3(0.48 – x)$$ 

$$4x^2 = 1.44 – x$$ 

$$4x^2 + 3x – 1.44 = 0$$ 

a = 4, b = 3, c = -1.44 

$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$ 

= [−3 ± √(3)$^2$ − 4(4)(−1.44)] / 2 × 4 

= (−3 ± 5.66)/8 

= (−3 + 5.66)/8  (as value of x cannot be negative hence we neglect that value) 

x = 2.66/8 = 0.33 

The equilibrium partial pressures are, 

$p_{CO} = 2x = 2 \times 0.33 = 0.66$ bar 

$p_{CO_2} = 0.48 – x = 0.48 – 0.33 = 0.15$ bar 

Questions from section 7.5:

1. What is aheterogeneous equilibrium reaction?

7.6 Applications of Equilibrium constant

We shall first discuss the important features of equilibrium constant as explained below:

1. Expression for equilibrium constant is applicable only when concentrations of the reactants and products have attained constant value at equilibrium state.

2. The value of equilibrium constant is independent of initial concentrations of the reactants and products.

3. Equilibrium constant is temperature dependent having one unique value for a particular reaction represented by a balanced equation at a given temperature.

4. The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction.

• The equilibrium constant K for a reaction is related to the equilibrium constant of the corresponding reaction, whose equation is obtained by multiplying or dividing the equation for the original reaction by a small integer.

Applications of Equilibrium Constant:

1. It helps to calculate the equilibrium concentrations of reactants or products.
2. It helps in predicting the direction of reaction.
3. It helps to predict the extent of a reaction on the basis of magnitude as explained:

• If $K_c > 10^3$, products predominate over reactants, that is, if $K_c$ is very large, the reaction proceeds nearly to completion.
  Let us consider the following examples:

(a) The reaction of $H_2$ with $O_2$ at 500 K has a very large equilibrium constant, $K_c$ = 2.4 × $10^{47}$.

(b) $ \ce { H_2(g) + Cl_2(g) <=> 2HCl(g) }$, at 300K has $K_c$ = 4.0 × $10^{31}$.

• If $K_c < 10^{–3}$, reactants predominate over products, that is, if $K_c$  is very small, the reaction proceeds rarely.
  Let us consider the following examples:

(a) The decomposition of $H_2O$ into $H_2$ and $O_2$ at 500 K has a very small equilibrium constant, $K_c$ = 4.1 × $10^{–48}$.

(b) $N_2(g) + O_2(g) ⇌ 2NO(g)$, at 298 K has $K_c$ = 4.8 × $10^{–31}$.

• If $K_c$  is in the range of $10^{–3}$ to $10^3$, appreciable concentrations of both reactants and products are present.
  Consider the following examples:

(a) For reaction of $H_2$ with $I_2$ to give HI, $K_c$ = 57.0 at 700K.

(b) Also, gas phase decomposition of $N_2O_4$ to $NO_2$ is another reaction with a value of $K_c$ = 4.64 × $10^{–3}$at 25°C which is neither too small nor too large. Hence, equilibrium mixtures contain appreciable concentrations of both $N_2O_4$  and $NO_2$.

These points are illustrated by Figure 7.6:

7.6.2 Predicting the Direction of the Reaction

The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage as we studied in the previous section. However, for this purpose, we calculate the reaction quotient, Q which is same as the equilibrium constant, $K_c$  except that the concentrations in $Q_c$ are not necessarily equilibrium values.

($Q_c$ – reaction quotient with molar concentrations and; $Q_P$ – reaction quotient with partial pressures)

For a general reaction:

$$ \ce { aA + bB <=> cC +dD }$$

$$ Q_c = [C]^c[D]^d/[A]^a[B]^b $$

Then,

If $Q_c >  K_c$, the reaction will proceed in the direction of reactants (reverse reaction).

If $Q_c <  K_c$, the reaction will proceed in the direction of the products (forward reaction).

If $Q_c = K_c$, the reaction mixture is already at equilibrium.

As illustrated above, the reaction quotient, $Q_ c$   will be useful in predicting the direction of reaction by comparing the values of $Q_ c$   and $K_c$  for a reaction. The same is depicted by figure, 7.7 below:

• If $ Q_ c <  K_c $, net reaction goes from left to right.

• If $Q_c > K_c$, net reaction goes from right to left.

• If $Q_c = K_c$, no net reaction occurs.

Example: The value of $K_c$ for the reaction 

2A $\rightleftharpoons$ B + C is 2 × 10$^{-3}$. At a given time, the composition of reaction mixture is [A] = [B] = [C] = 3 × 10$^{-4}$ M. In which direction the reaction will proceed?

Solution: 

For the reaction the reaction quotient $Q_c$ is given by, $Q_c = \frac{[B][C]}{[A]^2}$

as [A] = [B] = [C] = 3 × 10$^{-4}$ M 

$Q_c = \frac{(3 \times 10^{-4})(3 \times 10^{-4})}{(3 \times 10^{-4})^2} = 1$

as $Q_c > K_c$ so the reaction will proceed in the reverse direction.

7.6.3 Calculating Equilibrium Concentrations

In case of a problem in which we know the initial concentrations but do not know any of the equilibrium concentrations, the following three steps shall be followed:

Step 1. Write the balanced equation for the reaction.

Step 2. Under the balanced equation, make a table that lists for each substance involved in the reaction:

(a) its initial concentration,

(b) the change in concentration on reaching equilibrium, and

(c) the equilibrium concentration.

In constructing the table, “x” is defined as the concentration in mol L^-1 for one of the substances that reacts on going to equilibrium. Then, stoichiometry of the reaction is applied to determine the concentrations of the other substances in terms of x.

Step 3. Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If we are to solve a quadratic equation, we choose the mathematical solution that makes chemical sense.

Step 4. Calculate the equilibrium concentrations from the calculated value of x.

Step 5. Check your results by substituting them into the equilibrium equation.

Example:  13.8g of N$_2$O$_4$ was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium 

N$_2$O$_4$ (g) $\rightleftharpoons$ 2NO$_2$ (g) 

The total pressure at equilbrium was found to be 9.15 bar. Calculate $K_c$, $K_p$ and partial pressure at equilibrium.

Solution:

We know $pV = nRT$ 

Total volume $(V) = 1 \space L$ 

Molecular mass of N$_2$O$_4$ = 92 g 

Number of moles = 13.8g/92 g = 0.15 of the gas (n) 

Gas constant (R) = 0.083 bar L mol$^{-1}$K$^{-1}$ 

Temperature $(T) = 400$ K 

$pV = nRT$ 

$p \times 1$L = 0.15 mol $\times$ 0.083 bar L mol$^{-1}$K$^{-1}$ $\times$ 400 K 

$p = 4.98$ bar 

N$_2$O$_4$ $\rightleftharpoons$ 2NO$_2$ 

Initial pressure: 4.98 bar \hspace{1cm} 0 

At equilibrium: (4.98 – x) bar \hspace{1cm} 2x bar 

Hence, 

$p_{total}$ at equilibrium = $p_{N_2O_4} + p_{NO_2}$ 

9.15 = (4.98 – x) + 2x 

9.15 = 4.98 + x 

x = 9.15 – 4.98 = 4.17 bar 

Partial pressures at equilibrium are, 

$p_{N_2O_4} = 4.98 – 4.17 = 0.81$ bar 

$p_{NO_2} = 2x = 2 \times 4.17 = 8.34$ bar 

$$K_p = \frac{(p_{NO_2})^2}{p_{N_2O_4}}$$ 

$$K_p = \frac{(8.34)^2}{0.81} = 85.87$$ 

$$K_p = K_c(RT)^{\Delta n}$$ 

$$85.87 = K_c (0.083 \times 400)^1$$ 

$$K_c = 2.586 \approx 2.6$$     

Example: 3.00 mol of PCl$_5$ kept in 1L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate composition of the mixture at equilibrium. $K_c$ = 1.80 

Solution:

PCl$_5$ $\rightleftharpoons$ PCl$_3$ + Cl$_2$ 

Initial concentration: 3.0 \hspace{1cm} 0 \hspace{1cm} 0 

Let x mol per litre of PCl$_5$ be dissociated, 

At equilibrium: 

$(3-x) \hspace{1cm} x \hspace{1cm} x$

$$K_c = \frac{[PCl_3][Cl_2l]}{[PCl_5]}$$ 

$$1.8 = \frac{x^2}{(3-x)}$$ 

$x^2 + 1.8x – 5.4 = 0$

$x = [-1.8 \pm \sqrt{(1.8)^2 – 4(-5.4)}] / 2$

$x = [-1.8 \pm \sqrt{3.24 + 21.6}] / 2$

$x = [-1.8 \pm 4.98] / 2$

$x = [-1.8 + 4.98] / 2 = 1.59$ 

[PCl$_5$] = 3.0 – x = 3 – 1.59 = 1.41 M 

[PCl$_3$] = [Cl$_2$] = x = 1.59 M 

7.7 Relationship between Equilibrium Constant, K, Reaction Quotient, Q and Gibbs Energy, G

The value of $K_c$ for a reaction depends on the change in Gibbs energy, G as discussed:

• If G is negative, then the reaction is spontaneous and proceeds in the forward direction.
• If G is positive, then reaction is considered non-spontaneous and the reaction proceeds in reverse direction (where the products of the forward reaction will be converted to the reactants).
• If G is 0, reaction has achieved equilibrium. At this point, there is no longer any free energy left to drive the reaction in either direction.

The change in Gibbs energy and the composition of the reaction mixture of a reversible reaction is given by the equation:

$$ \Delta G = \Delta G^o + RT \space ln \space Q $$

where $\Delta G^o$ is the standard free energy change of the reaction and Q is the reaction quotient.

At equilibrium,,$\Delta G = 0 $ and $ Q = K_c$

$$ \therefore \Delta G^o = -RT \space ln \space K_c $$

$$ or,\Delta G^o = -2.303 \space RT \space log \space K_c$$

Note 1: $$ log \space K_c = \frac {\Delta G^o}{-2.303 RT}$$

$$ or \space K_c = e^{ \frac {-\Delta G^o}{2.303 \space RT}}$$

Note 2:

$$ ln \space K_c = -\frac {\Delta G^o}{RT}$$

$$ or \space K_c = e^{-\frac {\Delta G^o}{RT}}$$

Note 3: The temperature dependence of equilibrium constant is given by the expression

$$ log \frac {K_2}{K_1} = \frac {\Delta H}{2.303 R} \bigg[ \frac {T_2-T_1}{T_1T_2} \bigg] $$

Where, $K_1$ and $K_2$ are the equilibrium constants at temperatures, $T_1$ and $T_2$ respectively and;  is the enthalpy of reaction, assumed to be independent of temperature.

From equation, $ K = e^{-\Delta G^o/RT}$ , the spontaneity of a reaction can be interpreted in terms of the value $of ∆G^0$ as described:

• If of $ ∆G^0$ < 0, then – $ ∆G^0$ /RT is positive and $ e^{-\Delta G^o/RT} > 1$. Thus, K >1, which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.
• If $ ∆G^0$ > 0, then –$ ∆G^0$ /RT is negative, and$ e^{-\Delta G^o/RT} < 1$.. Thus, K < 1, which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.

Example:  The value of $\Delta G^{\ominus}$ for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of $K_c$ at 298 K. 

Solution:

$\Delta G^{\ominus} = 13.8$ kJ/mol = 13.8 × 10$^3$/mol 

Also, $\Delta G^{\ominus} = -RT \ln K_c$ 

Hence,  $\ln K_c = -13.8 \times 10^3 J/mol (8.314 J mol^{-1}K^{-1} \times 298 K)$

$$\ln K_c = -5.569$$ 

$$K_c = e^{-5.569}$$ 

$$K_c = 3.81 \times 10^{-3}$$

Questions from sections 7.6 and 7.7:

1. State the important features of equilibrium constant and list its applications.

2. Explain how the values of $Q_C$ and $K_c$ can be used to predict the direction of equilibrium of a reaction.

3. Explain stepwise, the procedure for calculating equilibrium concentration. Explain the relationship between Equilibrium Constant, K, Reaction Quotient, Q and Gibbs Energy, G with relevant equations.

4. Explain the significance of the equation, $ K = e^{-\Delta G^o/RT}$

7.8 Factors Affecting Equilibrium

Although equilibrium constant for a given reversible reaction is constant at constant temperature, equilibrium of the system gets altered when conditions like, pressure, volume, concentration and temperature of the system at equilibrium are altered.

In order to decide what course the reaction adopts and to make a qualitative prediction about the effect of a change in conditions on equilibrium we use, Le Chatelier’s principle.

Definition box: Le Chatelier’s principle states that, change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change.

We shall now be discussing factors which can influence the equilibrium:

7.8.1 Effect of Concentration Change

When equilibrium is disturbed by the addition or removal of any reactant or product, then, according to Le Chatelier’s principle,

• The concentration stress of an added reactant/product is relieved by net reaction in the direction that consumes the added substance.
• The concentration stress of a removed reactant/product is relieved by net reaction in the direction that replenishes the removed substance. In other words, when the concentration of any of the reactants or products in a reaction at equilibrium is changed, the composition of the equilibrium mixture changes so as to minimize the effect of concentration changes.

Example box: Let us take the reaction, $$ \ce { H_2 (g)+ I_2  (g)<=> 2HI(g)}$$  If $H_2$ is added to the reaction mixture at equilibrium, then, the equilibrium of the reaction is disturbed. In order to restore the equilibrium state, by reducing $H_2$, the reaction proceeds in a direction wherein $H_2$  gets consumed. Hence, more of $H_2$  and $I_2$ react to form HI and finally the equilibrium shifts in right (forward) direction (Figure 7.8).   Figure 7.8 Effect of addition of $H_2$   on change of concentration for the reactants and products in the reaction, $H_2(g) + I_2(g) $\rightleftharpoons 2HI(g)$ This is in accordance with the Le Chatelier’s principle which implies that, in case of addition of a reactant or a product, a new equilibrium will be set up, in which the concentration of the reactant or a product would be less than what it was after the addition but more than what it was in the original mixture.

An experiment on Effect of Concentration:

• Let us consider the following reaction:

$$ \ce { Fe^{3+} (aq) + SCN^- (aq) <=> [Fe(SCN)]^{2+}(aq) }$$

$$ yellow \hspace{10mm} colourless \hspace{10mm} deep \space red $$

$$ K_c = \frac {[Fe(SCN)^{2+}(aq)]}{[Fe^{3+}(aq)][SCN^-(aq)]}$$

• A reddish colour appears on adding two drops of 0.002 M potassium thiocynate solution to 1 mL of 0.2 M iron(III) nitrate solution due to the formation of [Fe(SCN)]²⁺.
• The red colour becomes constant on attaining equilibrium.
• This equilibrium can be shifted in either forward or reverse directions by adding a reactant or a product respectively.
• Thus, the equilibrium can be shifted in the opposite direction by adding reagents that remove $Fe^{3+}$ or $SCN^–$ ions. [For example, oxalic acid ($H_2C_2O_4$), reacts with $Fe^{3+}$ ions to form the stable complex ion: $[Fe(C_2O_4)^3]^{3 –}$, thus decreasing the concentration of free $Fe^{3+}$(aq).]
• In accordance with the Le Chatelier’s principle, the removal of $Fe^{3+}$ is compensated by dissociation of $[Fe(SCN)]^{2+}$ to replenish the  $Fe^{3+}$ ions. Since the concentration of $[Fe(SCN)]^{2+}$  decreases during dissociation, the intensity of red colour decreases.
• Addition of aq. $HgCl_2$ also decreases red colour because $Hg^{2+}$ reacts with $SCN^–$ ions to form stable complex ion $[Hg(SCN)^4]^{2–}$.
• Removal of free $SCN^–$ (aq) shifts the equilibrium from right to left to replenish $SCN^–$  ions. Addition of potassium thiocyanate on the other hand increases the colour intensity of the solution as it shifts the equilibrium to right.

7.8.2 Effect of Pressure Change

• Pressure can be changed by changing the volume of reactants. This affects the yield of products in case of a gaseous reaction where the total number of moles of gaseous reactants and total number of moles of gaseous products are different.

Note box:   In applying Le Chatelier’s principle to a heterogeneous equation, the effect of pressure changes on solids and liquids can be ignored because, the volume (and concentration) of a solution or liquid is nearly independent of pressure.

• Let us consider the reaction,        

$$ \ce { CO(g) + 3H_2(g) <=> CH_4 (g) + H_2O(g) }$$

Here, 4 mol of gaseous reactants (CO + $3H_2$) become 2 mol of gaseous products ($CH_4 + H_2O$).

• Suppose, equilibrium mixture (for above reaction) kept in a cylinder fitted with a piston at constant temperature is compressed to one half of its original volume, then, total pressure will be doubled (according to pV = constant).
• The direction in which the reaction goes to re-establish equilibrium can be predicted by applying the Le Chatelier’s principle.  Here, since pressure has doubled, the equilibrium now shifts in the forward direction, a direction in which the number of moles of the gas or pressure decreases (we know pressure is proportional to moles of the gas).

7.8.3 Effect of Addition of Inert Gas

Case (i)

At Constant Volume: If an inert gas like argon is added to the system in equilibrium at constant volume, the pressure of the system increases as the number of moles of argon increases. But the number of moles of the reactants and products per unit volume (concentration) or their partial pressures remain the same. Therefore, the equilibrium remains constant.

Case (ii)

At Constant Pressure: If an inert gas is added to the system in equilibrium at constant pressure, the volume of the system increases. Therefore, the number of moles of reactants and products per unit volume decreases. The system opposes this. The equilibrium shifts in the direction in which the number of moles are increased. The reverse reaction is accompanied by an increase in the number of moles. Hence, the equilibrium shifts to the left.

7.8.4 Effect of Temperature Change

An increase in temperature favours endothermic reaction. For instance, higher temperature favours formation of nitric acid.

In general, the temperature dependence of the equilibrium constant depends on the sign of H for the reaction.

Production of ammonia according is an exothermic process.

$$ \ce { N_2(g) + 3H_2(g) <=> 2NH_3 (g) ; }$$

$$ \Delta H = – 92.38 \space kJ \space mol^{-1} $$

According to Le Chatelier’s principle, raising the temperature shifts the equilibrium to left and decreases the equilibrium concentration of ammonia. In other words, low temperature is favourable for high yield of ammonia.

Note box: In practice, very low temperatures slow down the reaction and thus a catalyst is used.

An experiment on Effect of Temperature

Concept box: Dimerise: Combination of a compound or a molecule with similar molecules to form another compound called, dimer.

• Effect of temperature on equilibrium can be demonstrated by taking $NO_2$ gas (brown in colour) which dimerises into $N_2O_4$ gas (colourless).
• $NO_2$ gas prepared by addition of Cu turnings to conc. $HNO_3$ is collected in two 5 mL test tubes (ensuring same intensity of colour of gas in each tube). The tubes are sealed with stoppers made of araldite.
• Three 250 mL beakers 1, 2 and 3 containing freezing mixture, water at room temperature and hot water (363K), respectively, are taken (Figure 7.9).

• Both the test tubes are placed in beaker 2 for 8-10 minutes.
• After this, one tube is placed in beaker 1 and the other in beaker 3.
• The effect of temperature on direction of reaction is depicted very well in this experiment.
• At low temperatures, as in beaker 1, the forward reaction of formation of $N_2O_4$ is preferred, as reaction is exothermic, and thus, intensity of brown colour due to $NO_2$ decreases.
• While in beaker 3, high temperature favours the reverse reaction of formation of $NO_2$ and thus, the brown colour intensifies.

Note box:   1. Effect of temperature can also be seen in an endothermic reaction, $$ \ce {[Co(H_2O)_6]^{3+}(aq) + 4Cl^-(aq) <=> [CoCl_4]^{2-}(aq) + 6H_2O(l)} $$ $$ pink \hspace{10mm} colourless \hspace{10mm} blue $$ At room temperature, the equilibrium mixture is blue due to $[CoCl_4]^{2–}$. When cooled in a freezing mixture, the colour of the mixture turns pink due to $[Co(H_2O)_6]^{3+}$. 2. Equilibrium constant decreases with increase in temperature. At low temperature rate decreases and it takes long time to reach at equilibrium, whereas high temperatures give satisfactory rates but poor yields.

7.8.5 Effect of a Catalyst

• A catalyst increases the rate of the chemical reaction by making available, a pathway which needs lower energy for the conversion of reactants to products.
• It increases the rate of forward and reverse reactions to the same extent. Hence, it does not affect the equilibrium. However, the equilibrium is reached more quickly.
• Catalyst lowers the activation energy for the forward and reverse reactions by exactly the same amount.
• Catalyst does not affect the equilibrium composition of a reaction mixture.

For example, in manufacture of sulphuric acid by contact process, $2SO_2(g) + O_2(g)  ⇌ 2SO_3(g); K_c = 1.7 \times 10^{26}$ though the value of K is suggestive of reaction going to completion, but practically, the oxidation of $SO_2$ to $SO_3$ is very slow. Thus, platinum or divanadium penta-oxide ($V_2O_5$) is used as catalyst to increase the rate of the reaction.

Note box: If a reaction has an exceedingly small K, a catalyst would be of little help.

Questions from section 7.8:

1. State Le Chatelier’s principle.

2. Explain the effect of concentration change as a consequence of Le Chatelier’s principle.

3. Explain the experiment that demonstrates the effect of concentration change.

4. Explain the effect of pressure change in a reaction according to Le Chatelier’s principle.

5. Explain the effect of addition of inert gas on the equilibrium of a reaction.

6. Explain the effect of temperature change as a consequence of Le Chatelier’s principle.

7. Explain the experiment demonstrating the effect of temperature change on equilibrium of a reaction.

7.9 Ionic Equilibrium in Solution

Electrolytes and Non-electrolytes

Electrolytes are substances which conduct electricity either in molten state or in their aqueous solutions.

Some of the examples include: NaCl, KCl, $CuSO_4, Ag NO_3$ and $H_2SO_4$.

Non-electrolytes are substances which do not conduct electricity either in molten state or in their aqueous solutions.

Some of the examples are: Glucose, Sucrose, Benzene and Ethanol.

Strong and Weak Electrolytes:

Strong Electrolytes are those electrolytes which dissociate almost completely in aqueous solutions even at moderate concentrations.

Examples include: Strong acids like, HCl, $HNO_3, H_­2SO_4$; Strong bases like, NaOH, KOH, $Ba(OH)_2$ and; almost all water soluble salts.

Weak Electrolytes are those electrolytes which dissociate only partially in aqueous solutions even at moderate concentrations.

Examples include: Weak acids like, $H_3PO_4, CH_3COOH$, HCOOH, HCN; Weak bases like, $NH_3, CH_3NH_2$ and salts like, $ZnCl_2, AlCl_3, Hg_2Cl_2$.

Ionic equilibrium: In weak electrolytes, equilibrium is established between ions and the unionized molecules. This type of equilibrium involving ions in aqueous solution is called ionic equilibrium.

Questions from section 7.9:

1. What are electrolytes?

2. What are non-electrolytes?

3. What are weak electrolytes and strong electrolytes?

4. What do we mean by ionic equilibrium?

7.10 Acids, Bases and Salts

Some common properties of acids:

1. Acids are sour to taste.
2. They turn blue litmus red.
3. Dilute acids react with active metals to liberate hydrogen.
4. They decompose carbonates and bicarbonates, liberating $CO_2$ with effervescence.
5. They neutralize bases to form salt and water.

Some common properties of bases:

1. Bases are bitter to taste.
2. Their aqueous solutions are soapy to touch.
3. They turn red litmus blue.
4. They neutralize acids to form salts.

7.10.1 Arrhenius Concept of Acids and Bases

According to Arrhenius theory, acids are substances that dissociates in water to give hydrogen ions $H^+$ (aq) and bases are substances that produce hydroxyl ions $OH^–$(aq).

I. The ionization of an acid HX (aq) can be represented by the following equations:

$$ HX (aq) \rightarrow H^+(aq) + X^-(aq) $$

$$or$$

$$ HX(aq) + H_2O (l) \rightarrow H_3O^+ (aq) + X^-(aq)$$

II. Similarly, a base molecule like MOH ionizes in aqueous solution according to the equation:

$$ MOH(aq) → M^+(aq) + OH^–(aq) $$

7.10.2 The Brönsted-Lowry Acids and Bases

According to Brönsted-Lowry theory, “acid is a substance that is capable of donating a hydrogen ion $H^+$ and bases are substances capable of accepting a hydrogen ion, $H^+$”.

In other words, acids are proton donors and bases are proton acceptors.

• Let us consider the example of dissolution of $NH_3$ in $H_2O$ represented by the following equation:

• The basic solution is formed due to the presence of hydroxyl ions.
• In this reaction, water molecule acts as proton donor and is thus called, Lowry-Brönsted acid and; ammonia molecule acts as proton acceptor and is thus, called Lowry-Brönsted base.
• In the reverse reaction, $H^+$ is transferred from $NH_4^+$ to $OH^–$. In this case, $H^+$  acts as a Bronsted acid while $OH^–$.  acted as a Brönsted base.
• The acid-base pair that differs only by one proton is called a conjugate acid-base pair. Therefore, OH^– is called the conjugate base of an acid $H_2O$ and $NH_4^+$  is called conjugate acid of the base $NH_3^+$.

Note box: If Brönsted acid is a strong acid then its conjugate base is a weak base and vice versa. Also, conjugate acid has one extra proton and each conjugate base has one less proton.

• Let us now consider the example of ionization of hydrochloric acid in water. HCl (aq) acts as an acid by donating a proton to $H_2O$ molecule which acts as a base.

• It can be seen in the above equation that, water acts as a base because it accepts the proton.
• The species $H_3O^+$ is produced when water accepts a proton from HCl.
• Therefore, $Cl^–$ is a conjugate base of HCl and HCl is the conjugate acid of base $Cl^–$.

Example: What will be the conjugate bases for the following Brønsted acids: HF, H$_2$SO$_4$ and HCO$_3^-$? 

Solution:

The conjugate bases should have one proton less in each case and therefore the corresponding conjugate bases are: F$^-$, HSO$_4^-$ and CO$_3^{2-}$ respectively.

7.10.3 Lewis Acids and Bases

Lewis base: An acid is an electron-pair acceptor while a base is an electron-pair donor.

For example: BF₃ does not have a proton but still acts as an acid and reacts with $NH_3^+$  by accepting its lone pair of electrons. The reaction can be represented by,

Other examples of Lewis acids and bases are listed below:

Lewis Bases:

(i) Neutral species having at least one pair of electrons. Examples include:

(ii) Negatively charged species, that is, anions. Examples include: Cl^–, OH^– and CN^–.

Lewis Acids:

(i) Neutral molecules with electron deficient central atom.

(ii) Cations like, $Na^+, Ca^{2+}$ and $K^+$ have very little tendency to accept electrons. On the other hand, cations like, $H^+, Ag^+$ have a greater tendency to accept electrons and therefore act as Lewis acids.

(iii) Molecules containing central atom with vacant d-orbitals

Examples: $ SiF_4, SnCl_4, PF_5, AlCl_3$

(iv) Molecules containing atoms with dissimilar electronegativity’s by multiple bonds.

Example: $ CO_2, SO_2$

Acid base reactions based on Lewis Theory:

Example: Classify the following species into Lewisacids and Lewis bases and show howthese act as such:(a) $HO^-$ (b) $F^-$ (c) $H^+$ (d) $BCl_3$

Solution:

(a) Hydroxyl ion is a Lewis base as it can donate an electron lone pair (:$OH^–$).

(b) Flouride ion acts as a Lewis base as it can donate any one of its four electron lone pairs.

(c) A proton is a Lewis acid as it can accept a lone pair of electrons from bases like hydroxyl ion and fluoride ion.

(d) $BCl_3$ acts as a Lewis acid as it can accept a lone pair of electrons from species like ammonia or amine molecules.

Merit of Lewis theory:

It explains all acidic and basic nature on the basis of transfer or gain of electrons accompanied by loss or donation of electron pair.

Questions from section 7.10:

1. State the properties of acids and bases.

2. Explain Arrhenius Concept of Acids and Bases with illustration.

3. Explain Brönsted and Lowry’s theory of Acids and Bases.

4. What is a Lewis acid and a Lewis base? Also state the merits of Lewis theory of acids and bases.

7.11 Ionization of Acids and Bases

• Arrhenius concept of acids and bases are useful in studying ionization of acids and bases since ionization occurs mainly in aqueous medium. According to the concept,
• Strong acids like perchloric acid ($HClO_4$), hydrochloric acid (HCl), hydrobromic acid (HBr), hyrdoiodic acid (HI), nitric acid $(HNO_3)$ and sulphuric acid $(H_2SO_4)$ are termed strong because they are almost completely dissociated into their constituent ions in an aqueous medium, thereby acting as proton (or $H^+$) donors.
• Strong bases like lithium hydroxide (LiOH), sodium hydroxide (NaOH), potassium hydroxide (KOH), caesium hydroxide (CsOH) and barium hydroxide $Ba(OH)_2$ are called strong bases as they get almost completely dissociated into ions in an aqueous medium giving hydroxyl ions, $OH^–$.
• Thus, according to Arrhenius concept strong acids and bases are able to completely dissociate and produce $H_3O^+$ and $OH^–$ ions respectively in the medium.

7.11.1 The Ionization Constant of Water and its Ionic Product

Water dissociates to a small extent as follows:

$$ \ce { H_2O(l) + H_2O(l) <=> H_3O^+(aq) + OH^-(aq) } $$

$$ acid \hspace{10mm} base  \hspace{10mm} conjugate \space acid \hspace{10mm} conjugate \space base $$

The dissociation constant is represented by,

$$ \ce { K = [H3O+] [OH–] / [H2O] …..}(7.26) $$   

The concentration of water is omitted from the denominator as water is a pure liquid and its concentration remains constant. [$H_2O$] is incorporated within the equilibrium constant to give a new constant, $K_w$, which is called the ionic product of water.

$$ \ce { Kw = [H+][OH–]…. (7.27) } $$

The concentration of H+ has been found out experimentally as 1.0 × 10$^{–7}$ M at 298 K. And, as dissociation of water produces equal number of $H^+$ and $OH^–$ ions, the concentration of hydroxyl ions, [$OH^–] = [H^+]$ = 1.0 × 10$^{–7}$ M. Thus, the value of $K_w$ at 298K,

$$ \ce { Kw = [H3O+][OH–] = (1 × 10–7)^2 = 1 × 10^{–14} M^2….(7.28) }$$

The value of $K_w$ is temperature dependent as it is an equilibrium constant.

The density of pure water is 1000 g / L and its molar mass is 18.0 g /mol. From this the molarity of pure water can be given as,

$[H_2O]$ = (1000 g /L)(1 mol/18.0 g) = 55.55 M.

Therefore, the ratio of dissociated water to that of undissociated water can be given as:

$10^{–7}$ / (55.55) = 1.8 × $10^{–9}$ or ~ 2 in $10^{–9}$ (thus, equilibrium lies mainly towards undissociated water)

We can distinguish acidic, neutral and basic aqueous solutions by the relative values of the $H_3O^+$ and $OH^–$ concentrations:

$$ Acidic: [H_3O^+] > [OH^–] $$

$$ Neutral: [H3O^+] = [OH^– ] $$

$$ Basic : [H3O^+] < [OH^–]$$

7.11.2 The pH Scale

Definition box: pH of a solution is defined as the negative logarithm to the base 10 of hydronium ion (or hydrogen ion) concentration in mol $L^{-1}$.

Note box: Sorensen introduced a scale to express the concentration of hydronium ions known as the, pH scale. pH means, the power of hydrogen derived from the French expression, puissance d’ hydrogen.

$$ pH = -log_{10}[H_3O^+]$$

$$or \space pH = -log_{10}[H^+]$$

For Example,

(i) pH of water at 298 K

$ \ce { [H_3O]^+ = 10^{-7} mol \space L^{-1}}$

$ pH = – log_{10} [H_3O^+]$

$ = -log_{10}10^{-7}$

$=-[-7 \space log_{10} 10]$

$= 7$

(ii) pH of 0.1 M HCl

$ \ce { HCl + H_2O -> H_3O^+ + Cl^- } $

$0.1 M  \quad 0.1 M$

$ [H_3O^+]=0.1 \space M$

$= 10^{-1} M$

$pH= -log_{10}[H_3O^+]$

$ = -log_{10}10^{-1}$

$ =1$

(ii) pH of 0.1 M NaOH

$  NaOH \Rightarrow Na^+ + OH^- $

$1 M \quad 1 M$

$ [OH^-]=1 \space M$

$[H_3O^+][OH^-]=1 \space M$$

$ \therefore [H_3O^+] \times 1 =10^{-14}$$

$[H_3O^+] =10^{-14}$$

$pH= -log_{10}[H_3O^+]$$

$ = -log_{10}10^{-14}$$

$ =14$

(iii) pH of 1M HCl

$ \ce { HCl + H_2O -> H_3O^+ + Cl^- } $

$1 M \quad 1 M$

$ [H_3O^+]=1 \space M = 10^0M$

$ pH = -log_{10}[H_3O^+]$

$=-log_{10}10^o$

$= 0$

Therefore, we can summarise that

Acidic solutions have pH < 7

Neutral solutions have pH = 7

Basic solutions have pH > 7

$pOH$ of a solution is defined as the negative logarithm to the base 10 of the hydroxyl ion concentration in mol $L^{-1}$

$$ pOH=-log_{10}[OH^-]$$

To prove $\textbf{pH + pOH = 14}$

We know that

$$ [H_3O^+][OH^-]=K_w$$

$$[H_3O^+][OH^-]=10^{-14} at \space 298 \space K$$

Taking log,

$$ log[H_3O^+] + log[OH^-] = log 10^{-14} = 14$$

Multipyling throughout by -1,

$$-log[H_3O^+]-log[OH^-] = 14$$

$$ \therefore pH + pOH =14$$

The pH of a solution can be found roughly with the help of pH paper that imparts different colours in solutions of different pH. The pH in the range of 1-14 can be determined with an accuracy of _~0.5 using a pH paper (shown in figure 7.10).

Note box: 1. pH + pOH = pKw 2. A change in [H+] by a factor of 10 changes the pH by one unit. When [H+] changes by a factor of 100, the value of pH changes by 2 units. Hence, a change in temperature on pH is negligible.

Example: The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10$^{-3}$ M. what is its pH ? 

Solution:

pH = – log[3.8 × 10$^{-3}$] 

= – {log[3.8] + log[10$^{-3}$]} 

= – {(0.58) + (– 3.0)} = – {– 2.42} = 2.42 

Therefore, the pH of the soft drink is 2.42 and it can be inferred that it is acidic. 

Example: Calculate pH of a 1.0 × 10$^{-8}$ M solution of HCl. 

Solution:

2H$_2$O (l) $\rightleftharpoons$ H$_3$O$^+$ (aq) + OH$^-$ (aq) 

$K_w$ = [OH$^-$][H$_3$O$^+$] 

= 10$^{-14}$ 

Let, x = [OH$^-$] = [H$_3$O$^+$] from H$_2$O. The H$_3$O$^+$ concentration is generated (i) from the ionization of HCl dissolved i.e., 

HCl(aq) + H$_2$O(l) $\rightleftharpoons$ H$_3$O$^+$ (aq) + Cl$^-$ (aq), 

and (ii) from ionization of H$_2$O. In these very dilute solutions, both sources of H$_3$O$^+$ must be considered: 

[H$_3$O$^+$] = 10$^{-8}$ + x 

$K_w$ = (10$^{-8}$ + x)(x) = 10$^{-14}$ 

or x$^2$ + 10$^{-8}$x – 10$^{-14}$ = 0 

[OH$^-$] = x = 9.5 × 10$^{-8}$ 

So, pOH = 7.02 and pH = 6.98 

7.11.3 Ionization Constants of Weak Acids

Let us consider a weak acid HA that is partially ionized in the aqueous solution as expressed:

$$ \ce { HA(aq) + H_2O (l) <=> H_3O^+ (aq) + A^-(aq)}$$

Applying the law of mass action,

$$ K = \frac {[H_3O^+][A^-]}{[HA][H_2O]}$$

Since water is present in large excess,

$$ K_a = \frac {[H_3O^+][A^-]}{[HA]}$$

where $K_a$ is the ionisation constant of the acid.

Higher the value of $K_a$ larger is the concentration of $[H_3O^+]$ and therefore stronger is the acid.

If C is the concentration of the acid and $\alpha $ is the degree of ionisation,then

$$ \ce { HA(aq) + H_2O (l) <=> H_3O^+ (aq) + A^-(aq)}$$

Initial conc.       C            0             0

Eqlbm.conc.      $(1-\alpha)$    $\alpha C $   $\alpha C$

$$ K_a = \frac {[H_3O^+][A^-]}{[HA]}$$

$$ = \frac {\alpha C \times \alpha C}{(1-\alpha)C}$$

$$ K_a = \frac { \alpha^2 C}{1-\alpha } ….(1)$$

Equation (1) represents Ostwald’s dilution law.For a week acid $\alpha$ is very small and may be neglected compared to 1

$$ 1- \alpha \approx 1$$

$$ \therefore K_a = \alpha^2 C $$

$$ \alpha = \sqrt { \frac {K_a}{C} } ….(2)$$

$$[H_3O^+] = \alpha C = \sqrt { \frac {K_a}{C}} \times C $$

$$ \therefore [H_3O^+] = \sqrt { K_a \times C} ….(3)$$

From equation (3),pH of the solution can  be calculated.

Note 1 : $$pK_a = -log K_a$$

Larger the value of $pK_a$ weaker is the acid

Note 2 : Comparison  of strengths of acids :

$$ For \space acid_1,[H^+]_1 = \sqrt {K_{a1}C}$$

$$ For \space acid_2,[H^+]_2 = \sqrt {K_{a2}C}$$

$$ \therefore \frac {[H^+]_1}{[H^+]_2} = \sqrt { \frac {K_{a1}}{K_{a2}}}$$

Knowing the ionization constant, $K_a$ of an acid and its initial concentration, c, it is possible to calculate the equilibrium concentration of all species, the degree of ionization of the acid and the pH of the solution.

A step-wise approach can be adopted to evaluate the pH of the weak electrolyte as follows:

Step 1: The species present before dissociation are identified as Brönsted-Lowry acids or bases.

Step 2: Balanced equations for all possible reactions that is, with a species acting both as acid as well as base are written.

Step 3: The reaction with the higher $K_a$ is identified as the primary reaction and the other is a subsidiary reaction.

Step 4. In a tabular form, the following values for each of the species in the primary reaction is listed:

(a) Initial concentration, c.

(b) Change in concentration on proceeding to equilibrium in terms of α, degree of ionization.

(c) Equilibrium concentration.

Step 5: Equilibrium concentrations are substituted into equilibrium constant equation for principal reaction solved for α.

Step 6: The concentration of species in principal reaction is calculated.

Step 7: Finally, pH is calculated as: pH = – $log[H_3O^+$]

The above mentioned methodology has been elucidated in the following examples:

Example 1:  The ionization constant of HF is 3.2 × 10$^{-4}$. Calculate the degree of dissociation of HF in its 0.02 M solution. Calculate the concentration of all species present (H$_3$O$^+$, F$^-$ and HF) in the solution and its pH. 

Solution: The following proton transfer reactions are possible: 

1) HF + H$_2$O $\rightleftharpoons$ H$_3$O$^+$ + F$^- \hspace{1cm} K_a = 3.2 \times 10^{-4}$ 

2) H$_2$O + H$_2$O $\rightleftharpoons$ H$_3$O$^+$ + OH$^- \hspace{1cm} K_w = 1.0 \times 10^{-14}$ 

As $K_a \gg K_w$, [1] is the principle reaction. 

HF + H$_2$O $\rightleftharpoons$ H$_3$O$^+$ + F$^-$ 

Initial concentration (M) $\hspace{1cm}$ 0.02 $\hspace{1.2cm}$ 0 $\hspace{1.2cm}$ 0 

Change (M) $\hspace{3.2cm}$ –0.02$\alpha$ $\hspace{1cm}$ +0.02$\alpha$ $\hspace{1cm}$ +0.02$\alpha$ 

Equilibrium concentration (M) $\hspace{0.6cm}$ 0.02 – 0.02$\alpha$ $\hspace{0.6cm}$ 0.02$\alpha$ $\hspace{0.6cm}$ 0.02$\alpha$ 

Substituting equilibrium concentrations in the equilibrium reaction for principal reaction gives: 

$$K_a = \frac{(0.02\alpha)^2}{(0.02 – 0.02\alpha)}$$ 

$$= \frac{0.02 \alpha^2}{1 – \alpha} = 3.2 \times 10^{-4}$$ 

We obtain the following quadratic equation: 

$$\alpha^2 + 1.6 \times 10^{-2}\alpha – 1.6 \times 10^{-2} = 0$$ 

The quadratic equation in $\alpha$ can be solved and the two values of the roots are: 

$$\alpha = +0.12 \ \text{and} \ -0.12$$ 

The negative root is not acceptable and hence, 

$$\alpha = 0.12$$ 

This means that the degree of ionization, $\alpha = 0.12$, then equilibrium concentrations of other species viz., HF, F$^-$ and H$_3$O$^+$ are given by: 

[H$_3$O$^+$] = [F$^-$] = c$\alpha$ = 0.02 × 0.12 

= 2.4 × 10$^{-3}$ M 

[HF] = c(1 – $\alpha$) = 0.02 (1 – 0.12) 

= 17.6 × 10$^{-3}$ M 

pH = –log[H$^+$] = –log(2.4 × 10$^{-3}$) = 2.62 

Example 2: The pH of 0.1M monobasic acid is 4.50. Calculate the concentration of species $H^+$,

$A^–$ and $HA$ at equilibrium. Also, determine the value of $K_a$ and $Pk_a$ of the monobasic acid.

Solution:

pH = – log [H$^+$] 

Therefore, [H$^+$] = 10$^{-pH}$ = 10$^{-4.50}$ 

= 3.16 × 10$^{-5}$ 

[H$^+$] = [A$^-$] = 3.16 × 10$^{-5}$ 

Thus, 

$$K_a = \frac{[H^+][A^-]}{[HA]}$$ 

[HA]$_{eqlbm}$ = 0.1 – (3.16 × 10$^{-5}$) ≈ 0.1 

$$K_a = \frac{(3.16 \times 10^{-5})^2}{0.1} = 1.0 \times 10^{-8}$$ 

$$pK_a = -\log(10^{-8}) = 8$$ 

Alternatively, “Percent dissociation” is another useful method for measure of strength of a weak acid and is given as: Percent dissociation 

$$= \frac{[HA]_{dissociated}}{[HA]_{initial}} \times 100\% \hspace{1cm} (7.32)$$ 

Example 3: Calculate the pH of 0.08M solution of hypochlorous acid, $HOCl$. The ionization constant of the acid is 2.5 × 10$^{–5}$. Determine the percent dissociation of HOCl.

Solution:

HOCl(aq) + H$_2$O (l) $\rightleftharpoons$ H$_3$O$^+$(aq) + ClO$^-$(aq) 

Initial concentration (M) 

0.08 $\hspace{2cm}$ 0 $\hspace{2cm}$ 0 

Change to reach equilibrium concentration (M) 

–x $\hspace{2cm}$ +x $\hspace{2cm}$ +x 

equilibrium concentration (M) 

0.08 – x $\hspace{1.7cm} \times \hspace{2cm}$ x 

$$K_a = \{[H_3O^+][ClO^-] / [HOCl]\}$$ 

$$= x^2 / (0.08 – x)$$ 

As x << 0.08, therefore 0.08 – x ≃ 0.08 

$$x^2 / 0.08 = 2.5 \times 10^{-5}$$ 

$$x^2 = 2.0 \times 10^{-6}, \ \text{thus,} \ x = 1.41 \times 10^{-3}$$ 

[H$^+$] = 1.41 × 10$^{-3}$ M. 

Therefore, 

Percent dissociation 

= \{[HOCl]$_{dissociated}$ / [HOCl]$_{initial}$\} × 100 

= 1.41 × 10$^{-3}$ × 10$^2$ / 0.08 = 1.76 %. 

pH = –log(1.41 × 10$^{-3}$) = 2.85. 

7.11.4 Ionization of Weak Bases

The ionization of base BOH can be represented by equation:

$$ \ce { BOH (aq) <=> B^+ (aq) + OH^-(aq)}$$

$$ K_b = \frac {[B^+][OH^-]}{[BOH]}$$

Higher the value of $K_b$,larger is the concentration of $OH^-$ ions and therefore stronger is the base.

If C is the concentration of the base and $\alpha$ is the degree of ionisation,

$ \ce { BOH (aq) <=> B^+ (aq) + OH^-(aq)}$

Initial conc.  C $\quad$  0            0

Eqlbm conc.   $(1-\alpha)C$           $\alpha C $     $\alpha C $

$$K_b = \frac {[B^+][OH^-]}{[BOH]}$$

$$ = \frac { \alpha C \times \alpha C }{(1-\alpha )C}$$

$$ K_b = \frac {\alpha^2 C}{1-\alpha}….(1)$$

Equation (1) represents Ostwald’s dilution law

For weak base ,$1-\alpha \approx 1$

$$ \therefore K_b = \alpha^2 C….(2)$$

$$ \alpha = \sqrt { \frac {K_b}{C}}….(3)$$

$$[OH^-] = \alpha C = \sqrt { \frac {K_b}{C}} \times C$$

$$ [OH^-] = \sqrt { K_b \times C} ….(4)$$

Using equation (4),pOH cam be calculated.The pH of the solution can be calculated using the relation pH=14- pOH

Example: The pH of 0.004M hydrazine solution is 9.7. Calculate its ionization constant $K_b$ and $pK_b$.

Solution:

NH$_2$NH$_2$ + H$_2$O $\rightleftharpoons$ NH$_2$NH$_3^+$ + OH$^-$ 

From the pH we can calculate the hydrogen ion concentration. Knowing hydrogen ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have: 

[H$^+$] = antilog (–pH) 

= antilog (–9.7) = 1.67 × 10$^{-10}$ 

[OH$^-$] = $K_w$ / [H$^+$] = 1 × 10$^{-14}$ / 1.67 × 10$^{-10}$ 

= 5.98 × 10$^{-5}$ 

The concentration of the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to 0.004M. 

Thus, 

$K_b = \frac{[NH_2NH_3^+][OH^-]}{[NH_2NH_2]}$

$= (5.98 \times 10^{-5})^2 / 0.004 \space = \space 8.96 \times 10^{-7}$ 

$pK_b \space = \space –logK_b \space = \space –log(8.96 \times 10^{-7}) \space = \space 6.04.$

7.11.5 Relation between $K_a$(strength of an acid) and $K_b$ (strength of a base)

Let us consider the dissociation of a weak acid,

$$ \ce {HA(aq) + H_2O(l) <=> H_3O^+ (aq) + A^- (aq)}$$

$$ K_a = \frac {[H_3O^+][A^-]}{[HA]}….(1)$$

The conjugate base interacts with water as follows:

$$ \ce { A^-(aq) + H_2O(l) <=> HA(aq) + OH^- (aq) }$$

$$ K_b = \frac {[HA][OH^-]}{[A^-]}….(2)$$

Multiplying equation (1) and (2),we get

$$ k_a \times k_b = \frac {[H_3O^+][A^-]}{[HA]} \times \frac {[HA][OH^-]}{[A^-]} $$

$$ =[H_3O^+][OH^-]$$

$$=K_w$$

$$Thus,K_a \times K_b = K_w$$

if we take negative logarithm on both sides of the equation, then pK values of the conjugate acid and base are related to each other by the equation:

$$pK_a + pK_b = pK_w = 14 (at \space 298K)$$

Example: Determine the degree of ionization and pH of a 0.05M of ammonia solution. The ionization constant of ammonia can be taken from Table 7.7. Also, calculate the ionization constant of the conjugate acid of ammonia.

Solution:

The ionization of $NH_3$ in water is represented by equation:

NH$_3$ + H$_2$O $\rightleftharpoons$ NH$_4^+$ + OH$^-$ 

We use equation (7.33) to calculate hydroxyl ion concentration, 

[OH$^-$] = c α = 0.05 α 

$K_b$ = 0.05 α$^2$ / (1 – α) 

The value of α is small, therefore the quadratic equation can be simplified by neglecting α in comparison to 1 in the denominator on right hand side of the equation, 

Thus, 

$K_b = c \alpha^2$  or  $\alpha = \sqrt{(1.77 \times 10^{-5} / 0.05)}$ 

= 0.018. 

[OH$^-$] = c α = 0.05 × 0.018 = 9.4 × 10$^{-4}$ M. 

[H$^+$] = $K_w$ / [OH$^-$] = 10$^{-14}$ / (9.4 × 10$^{-4}$) 

= 1.06 × 10$^{-11}$ 

pH = –log(1.06 × 10$^{-11}$) = 10.97. 

Now, using the relation for conjugate acid-base pair, 

$K_a \times K_b = K_w$ 

using the value of $K_b$ of NH$_3$ from Table 7.7. 

We can determine the concentration of conjugate acid NH$_4^+$ 

$K_a = K_w / K_b = 10^{-14} / 1.77 \times 10^{-5}$ 

= 5.64 × 10$^{-10}$.

7.11.6 Di- and Polybasic Acids and Di- and Polyacidic Bases

Some of the acids like oxalic acid, sulphuric acid and phosphoric acids which have more than one ionizable proton per molecule of the acid are known as polybasic or polyprotic acids.

The ionization reactions for example, for a dibasic acid $H_2X$, are represented by the equations:

$$ \ce { H_2X(aq) <=> H^+(aq) + HX^-(aq)}$$

$$ \ce { HX^-(aq) <=> H^+(aq) + X^{2-} (aq)}$$

And the  corresponding  equilibrium constants  are given  below :

$$ K_{a1} = {[H^+][HX^-]}/[H_2X]$$ and

$$ K_{a2} = {[H^+][X^{2-}]}/[HX^-]$$

Where, $k_{a1}$ and $k_{a2}$ are called the first and second ionization constants respectively of the acid $H_2X$. Similarly, for tribasic acids like $H_3PO_4$, we have three ionization constants.

The values of the ionization constants for some common polyprotic acids are given in Table 7.8:

Note box: From the table, we see that, higher order ionization constants ($ k_{a1} , k_{a2})$ are smaller than the lower order ionization constant, $(k_{a1})$ of a polyprotic acid. The reason for this is that, it is more difficult to remove a positively charged proton from a negative ion due to electrostatic forces. This can be seen in the case of removing a proton from the uncharged $H_2CO_3$ as compared from a negatively charged $HCO_3^–$. Similarly, it is more difficult to remove a proton from a doubly charged $HPO_4^{2–}$ anion as compared to $H_2PO_4^–$.

7.11.7 Factors Affecting Acid Strength

• Strength of the H-A bond—the weaker the bond, the lesser the energy required to break it. Hence, the acid is strong.
• The polarity of the H-A bond—as the H-A bond becomes more polar, the proton tends to leave the molecule more easily, making it a strong acid.
• Atomic size of A – As the atom becomes larger, the bond gets weaker. Consequently, acid strength increases.
• While comparing elements in the same group of the periodic table, H-A bond strength is a more important factor in determining acidity than its polar nature.

Common Ion Effect

Definition box: Common Ion Effect is defined as the shift in equilibrium on adding a substance that provides more of an ionic species already present in the dissociation equilibrium.

Consider an example of acetic acid dissociation equilibrium represented as:

$$ \ce { CH_3COOH (aq) <=> H^+(aq) +CH_3COO^- (aq)}$$

$$ or \space  \ce { HAc(aq) <=> H^+(aq) +Ac^-(aq)}$$

$$ K_a = [H^+][Ac^-]/[HAc]$$

From above equation, we can say that, addition of acetate ions to an acetic acid solution results in decrease in concentration of hydrogen ions, $[H^+]$. Also, if $H^+$ ions are added from an external source, then the equilibrium moves in the direction of undissociated acetic acid that is, in a direction of reducing the concentration of hydrogen ions, $[H^+]$.. This phenomenon is an example of Common Ion Effect. Thus, we can say that common ion effect is a phenomenon based on the Le Chatelier’s principle.

We shall now evaluate the pH of solution resulting from addition of 0.05M acetate ion to 0.05M acetic acid solution. On dissociation of acetic acid,

 $$ HAc(aq) ⇌ H^+(aq) + Ac^–(aq) $$

Initial concentration (M)

$$ 0.05 \hspace{10mm} 0 \hspace{10mm} 0.05$$

Let x be the extent of ionization of acetic

acid.

Change in concentration (M)

$$ –x \hspace{10mm} +x \hspace{10mm} +x $$

Equilibrium concentration (M)

$$ 0.05-x \hspace{10mm} x \hspace{10mm} 0.05+x $$

Therefore,

$$K_a= [H^+][Ac^– ]/[H Ac] = {(0.05+x)(x)}/(0.05-x)$$

As $K_a$ is small for a very weak acid, x<<0.05.

Hence, (0.05 + x) ≈ (0.05 – x) ≈ 0.05

Thus,

$$ 1.8 \times 10^{–5} = (x) (0.05 + x) / (0.05 – x) $$

$$ = x(0.05) / (0.05) = x = [H^+] = 1.8 × 10^{–5}M $$

$$ pH = – log(1.8 \times 10^{–5}) = 4.74 $$

Example: Calculate the pH of a 0.10M ammonia solution. Calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.10M HCl. The dissociation constant of ammonia, $K_b = 1.77 \times 10^{-5}$ 

Solution:

NH$_3$ + H$_2$O $\rightarrow$ NH$_4^+$ + OH$^-$ 

$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = 1.77 \times 10^{-5}$ 

Before neutralization, 

[NH$_4^+$] = [OH$^-$] = x 

[NH$_3$] = 0.10 – x ≃ 0.10 

$\frac{x^2}{0.10} = 1.77 \times 10^{-5}$

Thus, x = 1.33 × 10$^{-3}$ = [OH$^-$] 

Therefore, [H$^+$] = $K_w$ / [OH$^-$] = 10$^{-14}$ / (1.33 × 10$^{-3}$) = 7.51 × 10$^{-12}$ 

pH = –log(7.5 × 10$^{-12}$) = 11.12 

On addition of 25 mL of 0.1M HCl solution (i.e., 2.5 mmol of HCl) to 50 mL of 0.1M ammonia solution (i.e., 5 mmol of NH$_3$), 2.5 mmol of ammonia molecules are neutralized. The resulting 75 mL solution contains the remaining unneutralized 2.5 mmol of NH$_3$ molecules and 2.5 mmol of NH$_4^+$. 

NH$_3$ + HCl $\rightarrow$ NH$_4^+$ + Cl$^-$ 

2.5 $\hspace{1cm}$ 2.5 $\hspace{1cm}$ 0 $\hspace{1cm}$ 0 

At equilibrium 

$0 \hspace{1cm} 0 \hspace{1cm} 2.5 \hspace{1cm} 2.5 $

The resulting 75 mL of solution contains 2.5 mmol of NH$_4^+$ ions (i.e., 0.033 M) and 2.5 mmol (i.e., 0.033 M) of unneutralised NH$_3$ molecules. This NH$_3$ exists in the following equilibrium: 

NH$_4$OH $\rightleftharpoons$ NH$_4^+$ + OH$^-$ 

0.033M – y \hspace{1cm} y \hspace{1cm} y 

where, y = [OH$^-$] = [NH$_4^+$] 

The final 75 mL solution after neutralisation already contains 2.5 m mol NH$_4^+$ ions (i.e. 0.033M), thus total concentration of NH$_4^+$ ions is given as: 

[NH$_4^+$] = 0.033 + y 

As y is small, [NH$_4$OH] ≃ 0.033 M and [NH$_4^+$] ≃ 0.033M. 

We know, 

$$K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]}$$ 

= y(0.033)/(0.033) = 1.77 × 10$^{-5}$ 

Thus, y = 1.77 × 10$^{-5}$ = [OH$^-$] 

[H$^+$] = 10$^{-14}$ / 1.77 × 10$^{-5}$ = 0.56 × 10$^{-9}$ 

Hence, pH = 9.24 

7.11.9 Hydrolysis of Salts and the pH of their Solutions

Definition box: Salt Hydrolysis: The process of interaction between water and cations or anions or both of salts is called hydrolysis.

The cations (like, $Na^+, K^+, Ca^{2+}, Ba^{2+}$) of strong bases and anions (like, $Cl^–, Br^–, NO3^–, ClO_4^–$) of strong acids get hydrated but do not hydrolyse, and therefore the solutions of salts formed from strong acids and bases are neutral that is, their pH is 7.

However, the following types of salts undergo hydrolysis:

(i) salts of weak acid and strong base like, $CH_3COONa$.

(ii) salts of strong acid and weak base like, $NH_4Cl$, and

(iii) salts of weak acid and weak base, like, $CH_3COONH_4$.

Hydrolysis of a salt of weak acid and a strong base:

As in first case, $CH_3COONa$ being a salt of weak acid, $CH_3COOH$ and strong base, NaOH gets completely ionised in aqueous solution as depicted:

$$ CH_3COONa(aq) \rightarrow CH_3COO^-(aq) + Na^+(aq)$$

Acetate ion thus formed undergoes hydrolysis in water to give acetic acid and $OH^–$ ions as depicted:

$$ \ce { CH_3COO^-(aq) + H_2O(l) <=> CH_3COOH(aq)+ OH^-(aq)}$$

Acetic acid being a weak acid ($K_a$ = 1.8 × $10^{–5}$) remains unionised in solution. This results in increase of $OH^–$ ion concentration in the solution making it alkaline (with pH>7).

Similarly, $NH_4Cl$ formed from weak base, $NH_4OH$ and strong acid, HCl, in water dissociates completely.

$$ \ce { NH_4Cl(aq) -> NH_4^+ (aq) + Cl^-(aq) } $$

Ammonium ions undergo hydrolysis with water to form $NH_4OH$ and $H^+$ ions

$$ \ce { NH^+_4(aq) + H_2O (l) <=> NH_4OH(aq) + H^+(aq) } $$

Ammonium hydroxide is a weak base $(K_b=1.77 \times 10^{-5})$ and therefore remains almost unionised in solution. This results in increased of $H^+$ ion concentration in solution making the solution acidic. Thus, the pH of $NH_4Cl$ solution in water is less than 7 Consider the hydrolysis of $CH_3COONH_4$ salt formed from weak acid and weak base. The ions formed undergo hydrolysis as follow:

$$ \ce { CH_3COO^- + NH_4^+ + H_2O <=> CH_3COOH + NH_4OH } $$

$CH_3COOH$ and $NH_4OH$, also remain into partially dissociated form:

$$ \ce { CH_3COOH <=> CH_3COO^- + H^+ }$$

$$ \ce { NH_4OH <=> NH_4^+ + OH^- }$$

$$ \ce { H_2O <=> H^+ + OH^- }$$

Without going into detailed calculation, it can be said that degree of hydrolysis is independent of concentration of solution, and pH of such solutions is determined by their pK values:

$$ pH = 7 + 1/2 (pK_a -pK_b)…..(7.38)$$

The pH of solution can be greater than 7, if the difference is positive and it will be less than 7, if the difference is negative.

Example: The p$K_a$ of acetic acid and p$K_b$ of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution. 

Solution:

pH = 7 + ½ [pK$_a$ – pK$_b$] 

= 7 + ½ [4.76 – 4.75] 

= 7 + ½ [0.01] = 7 + 0.005 = 7.005 

7.12 Buffer Solutions

Definition box: Salt Hydrolysis: The process of interaction between water and cations or anions or both of salts is called hydrolysis.

Buffer Solutions of known pH can be prepared from the knowledge of $pK_a$ of the acid or $pK_b$ of base and by controlling the ratio of the salt and acid or salt and base.

Applications of buffer solutions:

• pH controlled buffer solutions is very important in many chemical and biochemical processes.
• Many medical and cosmetic formulations are Buffer Solutions which can be kept and administered at a particular pH.

Types of buffer solutions:

Acidic buffer: A buffer solution containing a large amount of a weak acid and its salt, with a strong base, is termed as an acidic buffer.

For example: A mixture of ethanoic acid and sodium ethanoate.

Basic buffer: A buffer solution containing relatively large amounts of a weak base and its salt, with a strong acid, is termed as a basic buffer.

For example: Ammonium acetate solution is a neutral buffer.

Note box: A mixture of acetic acid and sodium acetate acts as buffer solution around pH of 4.75 and a mixture of ammonium chloride and ammonium hydroxide acts as a buffer around pH of 9.25.

Questions from sections 7.11 and 7.12:

1. How can we differentiate between an acidic, basic and a neutral aqueous solution?  

2. Define pH for a solution. Explain the stepwise approach to evaluate the pH of a weak electrolyte.

3. Obtain the relation between $K_a$ and $K_b$.

4. What are polybasic acids?

5. List the factors affecting the strength of an acid.

6. Explain the common ion effect with an example.

7. What is Salt-hydrolysis? Which are the types of salts that undergo salt hydrolysis?

8. What are buffer solutions? Explain its types.

9. State the applications of buffer solutions.

7.13 Solubility Equilibria of Sparingly Soluble Salts

Solubility of ionic solids in water depends on nature of salts. For example, calcium chloride is so soluble that they are hygroscopic in nature (that is, they even absorb water vapour from atmosphere); while others like, lithium fluoride have so little solubility that they are commonly termed as insoluble.

The solubility of ionic solids in water depends on:

• lattice enthalpy of the salt;
• the solvation enthalpy (or enthalpy of hydration) of the ions in a solution.

Solubility of ionic solids:

• For an ionic solid to dissolve in a solvent, the strong forces of attraction between its ions (lattice enthalpy) are overcome by the ion-solvent interactions during the process of solvation of ions.
• The amount of solvation enthalpy depends on the nature of the solvent. In case of a non-polar (covalent) solvent, solvation enthalpy is small and hence, not sufficient to overcome lattice enthalpy of the salt. Consequently, the salt does not dissolve in non-polar solvent.
• In general, for a salt to be able to dissolve in a particular solvent, its solvation enthalpy must be greater than its lattice enthalpy so that the latter may be overcome by former.
• Each salt has its characteristic solubility which depends on temperature.

We classify salts on the basis of their solubility in the following three categories:

7.13.1 Solubility Product Constant

Let us now consider a solid like barium sulphate in contact with its saturated aqueous solution.

The equilibrium between the undissolved solid barium sulphate and the ions in a saturated solution can be represented by the equation:

$$ \ce { BaSO_4(s) ->[{Saturated \space Solution}][{in water}]Ba^{2+}(aq) + SO^{2-}_4(aq)},$$

The equilibrium constant is given by the equation:

$$K = {[Ba^{2+}][SO_4^{2–}]} / [BaSO_4]$$

For a pure solid substance the concentration remains constant and we can write

$$k_{sp} = K[BaSO_4]=[Ba^{2+}][SO^{2-}_4]……(7.39)$$

Where, $K_{sp}$ is the solubility product constant or simply solubility product.

The Solubility Product of a salt can thus be defined at a given temperature as the product of the molar concentrations of its ions in saturated solution, with each concentration term raised to the power of its stoichiometric coefficient in the balanced chemical equation.

Relationship between the solubility product and solubility:

(i) AB type of electrolytes

Examples: Li F,Ag Cl,$BaSO_4,BaCrO_4,CaCO_3$ etc.

$$ \ce { AB(s) + aq <=> A^+(aq) + B^-(aq)}$$

Conc. At eqlbm           S          S

(mol $L^{-1}) $

$$ K_{sp} = [A^+][B^-]$$

$$ = S \times S $$

$$ K_{sp} = S^2 $$

Where S is the solubility of the salt in mol $L^{-1}$

(ii) $A_2B$ type

Examples:$Ag_2CrO_4,Ag_2S,Cu_2S,Ag_2SO_4$ etc.

$$ \ce { A_2B (s) + aq <=> 2A^+ (aq) + B^- (aq)}$$

Eqlbm. Conc.                 2S        S

(mol $l^{-1}$)

$$ K_{sp} = [A^+]^2[B^-]$$

$$=(2S)^2 \times S $$

$$ K_{sp} = 4S^3$$

Thus, for a solid salt of the general formula:  $M_x^{p+},X_y^{q-}$ with molar solubility S, in equilibrium with its saturated solution may be represented by the equation:

$$ \ce { M_xX_y(s) <=> xM^{p+}(aq) +yX^{q-}(aq)}$$

(where x × $p^+$ = y × $q^–$)

And its solubility product constant is given by:

$$ K_{sp} =[M^{p+}]^x[X^{q-}]^y=(xS)^x(yS)^y $$

$$ (7.40) \hspace{10mm} =x^x.y^y.S^{(x+y)}$$

$$ S^{(x+y)}=K_{sp}/x^x.y^y $$

$$S=(K_{sp}/x^x.y^y)^{1/x+y}……(7.41)$$

The solubility product constants of a number of common salts at 298K are given in Table 7.9 below:

Name of the Salt	Formula	Ksp
Silver Bromide	AgBr	5.0 × 10⁻¹³
Silver Carbonate	Ag₂CO₃	8.1 × 10⁻¹²
Silver Chromate	Ag₂CrO₄	1.1 × 10⁻¹²
Silver Chloride	AgCl	1.8 × 10⁻¹⁰
Silver Iodide	AgI	8.3 × 10⁻¹⁷
Silver Sulphate	Ag₂SO₄	1.4 × 10⁻⁵
Aluminium Hydroxide	Al(OH)₃	1.3 × 10⁻³³
Barium Chromate	BaCrO₄	1.2 × 10⁻¹⁰
Barium Fluoride	BaF₂	1.0 × 10⁻⁶
Barium Sulphate	BaSO₄	1.1 × 10⁻¹⁰
Calcium Carbonate	CaCO₃	2.8 × 10⁻⁹
Calcium Fluoride	CaF₂	5.3 × 10⁻⁹
Calcium Hydroxide	Ca(OH)₂	5.5 × 10⁻⁶
Calcium Oxalate	CaC₂O₄	4.0 × 10⁻⁹
Calcium Sulphate	CaSO₄	9.1 × 10⁻⁶
Cadmium Hydroxide	Cd(OH)₂	2.5 × 10⁻¹⁴
Cadmium Sulphide	CdS	8.0 × 10⁻²⁷
Chromic Hydroxide	Cr(OH)₃	6.3 × 10⁻³¹
Cuprous Bromide	CuBr	5.3 × 10⁻⁹
Cupric Carbonate	CuCO₃	1.4 × 10⁻¹⁰
Cuprous Chloride	CuCl	1.7 × 10⁻⁶
Cupric Hydroxide	Cu(OH)₂	2.2 × 10⁻²⁰
Cuprous Iodide	CuI	1.1 × 10⁻¹²
Cupric Sulphide	CuS	6.3 × 10⁻³⁶
Ferrous Carbonate	FeCO₃	3.2 × 10-11
Ferrous Hydroxide	Fe(OH)₂	8.0 × 10-16
Ferric Hydroxide	Fe(OH)₃	1.0 × 10⁻³⁸
Ferrous Sulphide	FeS	6.3 × 10-18
Mercurous Bromide	Hg₂Br₂	5.6 × 10⁻²³
Mercurous Chloride	Hg₂Cl₂	1.3 × 10⁻¹⁸
Mercurous Iodide	Hg₂I₂	4.5 × 10⁻²⁹
Mercurous Sulphate	Hg₂SO₄	7.4 × 10⁻⁷
Mercuric Sulphide	HgS	4.0 × 10⁻⁵³
Magnesium Carbonate	MgCO₃	3.5 × 10⁻⁸
Magnesium Fluoride	MgF₂	6.5 × 10⁻⁹
Magnesium Hydroxide	Mg(OH)₂	1.8 × 10⁻¹¹
Magnesium Oxalate	MgC₂O₄	7.0 × 10⁻⁷
Manganese Carbonate	MnCO₃	1.8 × 10⁻¹¹
Manganese Sulphide	MnS	2.5 × 10⁻¹³
Nickel Hydroxide	Ni(OH)₂	2.0 × 10⁻¹⁵
Nickel Sulphide	NiS	4.7 × 10⁻⁵
Lead Bromide	PbBr₂	4.0 × 10⁻⁵
Lead Carbonate	PbCO₃	7.4 × 10⁻¹⁴
Lead Chloride	PbCl₂	1.6 × 10⁻⁵
Lead Fluoride	PbF₂	7.7 × 10⁻⁸
Lead Hydroxide	Pb(OH)₂	1.2 × 10⁻¹⁵
Lead Iodide	PbI₂	7.1 × 10⁻⁹
Lead Sulphate	PbSO₄	1.6 × 10⁻⁸
Lead Sulphide	PbS	8.0 × 10⁻²⁸
Stannous Hydroxide	Sn(OH)₂	1.4 × 10⁻²⁸
Stannous Sulphide	SnS	1.0 × 10⁻²⁵
Strontium Carbonate	SrCO₃	1.1 × 10⁻¹⁰
Strontium Fluoride	SrF₂	2.5 × 10⁻⁹
Strontium Sulphate	SrSO₄	3.2 × 10⁻⁷
Thallous Bromide	TlBr	3.4 × 10⁻⁶
Thallous Chloride	TlCl	1.7 × 10⁻⁴
Thallous Iodide	TlI	6.5 × 10⁻⁸
Zinc Carbonate	ZnCO₃	1.4 × 10⁻¹¹
Zinc Hydroxide	Zn(OH)₂	1.0 × 10⁻¹⁵
Zinc Sulphide	ZnS	1.6 × 10⁻²⁴

Example: Calculate the solubility of A$_2$X$_3$ in pure water, assuming that neither kind of ion reacts with water. The solubility product of A$_2$X$_3$, $K_{sp}$ = 1.1 × 10$^{-23}$. 

Solution:

A$_2$X$_3$ → 2A$^{3+}$ + 3X$^{2-}$ 

$K_{sp}$ = [A$^{3+}$]$^2$ [X$^{2-}$]$^3$ = 1.1 × 10$^{-23}$ 

If S = solubility of A$_2$X$_3$, then 

[A$^{3+}$] = 2S; $\hspace{0.5cm}$ [X$^{2-}$] = 3S 

therefore, $K_{sp}$ = (2S)$^2$(3S)$^3$ = 108S$^5$ 

= 1.1 × 10$^{-23}$ 

thus, S$^5$ = 1 × 10$^{-25}$ 

S = 1.0 × 10$^{-5}$ mol/L. 

7.13.2 Common Ion Effect on Solubility of Ionic Salts

Concept box: Gravimetric analysis is a technique through which the amount of an ion being analyzed can be determined through the measurement of mass.

Applications of Common ion effect:

• When the concentration of one of the ions is decreased, more salt will dissolve to increase the concentration of both the ions until, $K_{sp} = Q_{sp}$. For instance, if we take a saturated solution of sodium chloride and pass HCl gas through it, then sodium chloride is precipitated due to increased concentration (activity) of chloride ion available from the dissociation of HCl. Sodium chloride thus obtained is of very high purity and we can get rid of impurities like sodium and magnesium sulphates.
• The common ion effect is also used for almost complete precipitation of a particular ion when it is sparingly soluble salt, with very low value of solubility product for gravimetric estimation. Thus, we can precipitate silver ion as silver chloride, ferric ion as its hydroxide (or hydrated ferric oxide) and barium ion as its sulphate for quantitative estimations.

Example: Calculate the molar solubility of Ni(OH)$_2$ in 0.10 M NaOH. The ionic product of Ni(OH)$_2$ is 2.0 × 10$^{-15}$. 

Solution:

Let the solubility of Ni(OH)$_2$ be equal to S. Dissolution of S mol/L of Ni(OH)$_2$ provides 

S mol/L of Ni$^{2+}$ and 2S mol/L of OH$^-$, but the total concentration of OH$^-$ = (0.10 + 2S) mol/L because the solution already contains 0.10 mol/L of OH$^-$ from NaOH. 

$K_{sp}$ = 2.0 × 10$^{-15}$ = [Ni$^{2+}$] [OH$^-$]$^2$ 

= (S)(0.10 + 2S)$^2$ 

As $K_{sp}$ is small, 2S << 0.10, 

thus, (0.10 + 2S) ≈ 0.10 

Hence, 

2.0 × 10$^{-15}$ = S (0.10)$^2$ 

S = 2.0 × 10$^{-13}$ M = [Ni$^{2+}$] 

The solubility of salts of weak acids like phosphates increases at lower pH. This is because, at lower pH, the concentration of the anion decreases due to its protonation. This in turn increases the solubility of the salt so that $K_{sp} = Q_{sp}$.

We have to satisfy two equilibria simultaneously that is,

$$ K_{sp}=[M^+][X^-],$$

$$ \ce {HX(aq) <=> H^+(aq) +  X^-(aq);}$$

$$K_a = \frac {[H^+(aq)][X^-(aq)]}{[HX(aq)]}$$

$$ [X^–] / [HX] = K_a / [H^+] $$

Taking inverse of both side and adding 1 we get

$$ \frac {[HX]}{[X^-]} + 1 = \frac {[H^+]}{[K_a]}+1$$

$$ \frac {[HX]+[H^-]}{[X^-]} = \frac {[H^+] +K_a}{K_a}$$

Now, again taking inverse, we get

$[X^–] / \{[X^–] + [HX]\} = f = K_a / (K_a + [H^+])$ and it

can be seen that ‘f ’ decreases as pH decreases. If S is the solubility of the salt at a given pH then

$ K_{sp} =[S][fS]=S^2\{K_a / (K_a+[H^+])\}$ and

$$ S = \{ K_{sp}([H^+] + K_a )/K_a \}^{1/2}…..(7.42)$$

Thus solubility, S increases with increase in $[H^⁺]$ or decrease in pH.

Questions from section 7.13:

1. Explain the factors upon which the solubility of ionic solids depends.

2. Establish the relationship between solubility product and solubility.

3. Explain the applications of Common Ion Effect.