1st PUC Chemistry Question Bank Chapter 7 Equilibrium

Students can Download Chemistry Chapter 7 Equilibrium Questions and Answers, Notes Pdf, 1st PUC Chemistry Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams

Karnataka 1st PUC Chemistry Question Bank Chapter 7 Equilibrium

Question 1.
Define equilibrium state.
Answer:
It is a state of balance in a reversible reaction at which the rate of forward reaction is equal to the rate of backward reaction.
For example, at equilibrium, the rate of evaporation of water is equal to the rate of condensation.
H₂O(1) ⇌ H₂O(vap)

Question 2.
What is equilibrium mixture?
Answer:
Mixture of reactants and products in the equilibrium state is called an equilibrium mixture.

Question 3.
Chemical equilibrium is called dynamic equilibrium. Why?
Answer:
At equilibrium, there is no change in concentration of the reactants and products in the reaction mixture and the reaction continues in both forward and backward reaction with same rate. So, chemical equilibrium is called dynamic equilibrium.

Question 4.
What is ionic equilibrium?
Answer:
The equilibrium involving ions in aqueous solutions is called ionic equilibrium.

Equilibrium in Physical Processes

Question 1.
What is solid-liquid equilibrium? Give an example.
Answer:
The equilibrium in which the rate of melting of the solid is equal to the rate of freezing of the liquid is called solid-liquid equilibrium.
H₂O(S) ⇌ H₂O(l)

Question 2.
What is liquid – vapour equilibrium? Give an example.
Answer:
The equilibrium in which the rate of vapourisation of liquid is equal to rate of condensation of vapour is called liquid – vapour equilibrium.
H₂O(1) ⇌ H₂O(g)

Question 3.
What is solid-vapour equilibrium? Give an example.
Answer:
The equilibrium in which the rate of vapourisation of a solid is equal to rate of condensation of vapour into solid is called solid-vapour equilibrium.
Example: (i) Sublimation of iodine to iodine vapour and condensation of iodine vapours to solid iodine
I₂(solid) ⇌ I₂(vapour)
(ii) Camphor (solid) ⇌ Camphor (vapour)
(iii) NH₄CI (solid) ⇌ NH₄CI (vapour))

Question 4.
What is solid-solution equilibrium? Give an example.
Answer:
“The equilibrium in which the rate at which the solute particles goes into the solution phase is equal to rate at which solute particles in the solution gets into solid phase is called solid – solution equilibrium.”
Example: Sugar (s) ⇌ sugar solution

Question 5.
What is gas-solution equilibrium? Give an example.
Answer:
The equilibrium in which the rate at which molecules of a gas goes into the solution phase is equal to the rate at which the molecules of the gas in the solution phase going back to the gas phase is called gas solution equilibrium, example is soda water in air tight bottles.
CO₂(g) ⇌ CO₂(in solution)

Question 6.
State Henry’s law.
Answer:
Henry’s law states that “the mass of a gas dissolved in a given mass of a solvent is proportional to the pressure of the gas above the solvent”.

Question 7.
Give the general characteristics of equilibria involving physical processes.
Answer:

• Equilibrium is established only in closed system at constant temperature
• All measurable properties of the system remain constant.
• Both the opposing processes occur at the same rate
• When equilibrium is attained, it is characterised by constant value of one of its properties at a given temperature. This can be summarised as follows.

i. Solid-liquid equilibria: The amount of solid in liquid remains constant.
ii. Liquid- gas equilibria: Vapour pressure is constant at a given temperature
iii. Solid- solution equilibria: Solubility is constant at given temperature
iv. Gas- solution equilibria: the concentration of gas in a liquid is constant at a given temperature and pressure.

Equilibrium in Chemical Processes

Question 1.
What is chemical equilibrium?
Answer:
It is a state of balance in a reversible reaction in which the rate of forward reaction is equal to the rate of backward reaction.

Question 2.
Name the isotope used to show the dynamic nature of chemical equilibrium in the synthesis of ammonia by Haber’s process.
Answer:
Deuterium.

Question 3.
What is the indication obtained by the use of deuterium in the formation of ammonia by Haber’s process?
Answer:
Use of isotope (deuterium) in the formation of ammonia by Haber’s process clearly indicates that chemical reactions reach a state of dynamic equilibrium at which the rate of forward and reverse reactions are equal and there is no net change in composition.

Question 4.
Explain the attainment of chemical equilibrium for the reaction between H₂ and I₂ by using graph.
Answer:
The following graphical representation explain the attainment of chemical equilibrium for the reaction between H₂ and I₂.

It is clear from the figure that molar concentrations of H₂, I₂ and HI remain constant after the vertical line corresponding to time.

Law of Equilibrium and Equilibrium Constant

Question 1.
Write the equilibrium equation or equilibrium constant expression for the reversible reaction A + B ⇌ C + D according to law of mass action.
Answer:

where K_c is called equilibrium constant in terms of concentration. Equilibrium equation is called law of mass action.

Question 2.
State law of equilibrium or equilibrium law.
Answer:
Law of equilibrium states that “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.”

Question 3.
Define equilibrium constant.
Answer:
It is the ratio of the product of molar concentrations of the products to the product of molar concentrations of the reactants raised to the respective stoichiometric coefficients as mentioned in the balanced chemical equation.

Question 4.
Write the equilibrium constant expression for the general reaction aA + bB ⇌ cC + dD
Answer:

Question 5.
Wrire an expression for equilibrium constant of the reaction,
(a) H₂ + I₂ ⇌ 2HI
(b) 4NH₃ + 5O₂ ⇌ 4NO + 6H₂O
Answer:

Question 6.
Define equilibrium constant for the reverse reaction.
Answer:
Equilibrium constant for the reverse reaction is the reciprocal of equilibrium constant for the forward reaction.
Student’s illuminator
K_rev = \(\frac{1}{\mathrm{K}_{\mathrm{c}}}\)

Question 7.
Write the relation between equilibrium constant for the reactions:

Answer:

Question 8.
The following concentration were obtained for the formation of NH₃ from N₂ and H₂ at equilibrium at 500 K. [N₂] = 1.5 × 10^-2M, [H₂] = 3.0 × 10^-2M and [NH₃] = 1.2 × 10^-2M. Calculate the equilibrium constant.
Answer:
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), equilibrium constant (K_c) can be written as

Question 9.
At equilibrium, the concentration of N₂ = 3.0 × 10^-3 M, O₂ = 4.2 × 10^-3 M and NO = 2.8 × 10^-3 M in a sealed vessel at 800 K. What will be K_C for the reaction N₂(g) + O₂(g) ⇌ 2N0(g)?
Answer:

Question 10.
For the equilibrium reaction A + B ⇌ C + D. If the initial concentration of A and B is doubled then
what is the value of equilibrium constant?
Answer:
Equilibrium constant remains same.

Question 11.
K₁ and K₂ are equilibrium constants for reactions (a) and (b). Find the relation between them.
(a) N₂ + O₂ (g) ⇌ 2NO(g)
(b) NO(g) ⇌ \(\frac { 1 }{ 2 }\)N₂(g) + \(\frac { 1 }{ 2 }\)O₂(g)
Answer:

Homogeneous and Heterogeneous Equilibria

Question 1.
What is homogeneous equilibrium? Give an example.
Answer:
An equilibrium is said to be homogeneous if reactants and products are in the same phase.
Examples: N₂(g) + 3H_2(g) ⇌ 2NH₃(g)
CH₃COOCH₃(aq) + H₂O(I) ⇌ CH₃COOH(aq) + CH₃OH(aq)

Question 2.
What is heterogeneous equilibrium? Give an example.
Answer:
An equilibrium is said to be heterogeneous if reactants and products are in different phases
Examples:
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
NH₄HS(S) ⇌ NH₃(g) + H₂S(g)
NH₂CO₂NH₄(S) ⇌ 2NH₃(g) + CO₂(g)

Question 3.
Define equilibrium constant K_p for gaseous reactions.
Answer:
It is the ratio of the product of partial pressures of the products to the product of partial pressures of the reactants raised to the respective stoichiometric coefficients as mentioned in the balanced chemical equation.

Question 4.
Write the expression of equilibrium constant K_p for the gaseous reaction aA + bB ⇌ cC + dD.
Answer:

Question 5.
Write the expression of equilibrium constants K_c and K_p for the reaction H₂(g) + I₂(g) ⇌ 2HI(g).
Answer:

Question 6.
Write an expression for equilibrium constant for the heterogeneous equilibrium reaction,
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Answer:
K_P = p_CO2

Question 7.
Derive the relation between K_p and K_c.
Answer:
From the ideal gas equation;
P = \(\frac{\mathrm{nRT}}{\mathrm{V}}\) = [M]RT
Where [M] = molar concentration.
Now we can wiite: P_A = [A]RT, P_B = [B]RT, P_C = [C]RT and P_D = [D]RT
Thererfore,

Where Δn_g = (number of moles of gaseous products) – (number of moles of gaseous reactants).

Question 8.
Deduce the relationship between K_p and K_c for the following equilibrium.
Answer:
1. For the equilibrium H₂(g) + I₂(g), ⇌ 2HI(g)
Δn_g = n_p(g) – n_R(g) = 2 – 2 = 0
K_p = K_c(RT)^Δn_g
K_p = K_c (RT)°
∴ K_P = K_C

2. For the equilibrium, PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Δn_g = n_p(g) – n_R(g) = 2 – 1 = 1
K_p = K_c(RT)^Δn_g
∴ K_p = K_c × (RT) or K_p > K_c

3. For the equilibrium, N₂(g) + 3H₂(g), ⇌ 2NH₃(g)
Δn_g = n_p(g) – n_R(g) = 2 – 4 = 2
K_p = K_c(RT)^Δn_g
= K_c(RT)^-2
K_p = \(\frac{K_{c}}{(R T)^{2}}\) or K_c > K_p

4. For the equilibrium, CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Δn_g = n_p(g) – n_R(g) = 1 – 0 = 1
K_p = K_c(RT)^Δn_g
K_p = K_cRT or K_p > K_e
Note:
If, Δn_g = 0 then K_p = K_c
Δn_g > 0 (or + ve) then K_p > K_c
Δn_g < 0 (or – ve) then K_p < K_c

Question 9.
For the equilibrium 2NOCI(g) ⇌ 2NO(g) + Cl₂(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?
Answer:
We know that, K_p = K_c (RT)^Δn_g
For the above reaction, Δn = n_P(g) – n_R(g)
Δn = (2 + 1) -2 = 1
K_p = 3.75 × 10^-6 (0.0831 × 1069)
K_p = 0.000333

Question 10.
Find the value of K_c for each of the following equilibria from the value of K_p:
(a) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g),K_p = 1.8 × 10^-2 at500K
(b) CaCO₃ (S) ⇌ CaO(s) + CO₂(g),K_p = 167 at 1073K.
Answer:
(a) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
K_p = 1.8 × 10^-2
Δn_g = n_p(g) – n_R(g) = 3 – 2 = 1
K_p = K_c(RT)^Δn_g

(b) CaO(s) ⇌ CaO(s) + CO₂(g)
K_p = 167
Δn_g = 1 – 0 = 1
K_p = K_c(RT)^Δn_g

Question 11.
At 450 K, K_p = 2.0 × 10¹⁰ bar for the given reaction at equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). What is K_c at this temperature?
Answer:
For the given reaction,
Δn_g = n_p(g) – n_R(g) = 2 – 3 = 1
K_p = K_c(RT)^Δn_g
K_c = K_p(RT )^Δn_g
= (2.0 × 10¹⁰ bar)(0.0831 Lbar K^-1 mol^-1)^-(-1) (450 K)
= 7.48 × 10¹¹ Lmol^-1

Question 12.
For the following equilibrium, K_c = 6.3 × 10¹⁴ at 1000 K. NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g). Both the forward and reverse reactions in the equilibrium are elementary biomolecular reactions. What is K_c for the reverse reaction?
Answer:
For the reverse reaction

Question 13.

Answer:
4NO(g) + 6H₂O(g) → 4NH₃ (g) + 5O₂ (g)

Question 14.
One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation. H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g). Calculate the equilibrium constant for the reaction.
Answer:

Question 15.
A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 × 10^-1 mol L^-1. If value of K_c is 8.3 × 10^-3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?
Answer:

Question 16.
At 700 K, equilibrium constant for the reaction: H₂ (g) + I₂ (g) ⇌ 2HI(g) is 54.8. If 0.5 mol L^-1 of HI(g) is present at equilibrium at 700 K, what are the concentrations of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
Answer:

x = 0.068
[H₂] = 0.068 mol L^-1
[I₂] = 0.068 mol L^-1

Question 17.
K_p = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

Answer:

Question 18.
Deduce the units of equilibrium constant K_c and k_p for the following reactions.
(a) H₂ + I₂ ⇌ 2HI(g)
(b) N₂O₄(g) ⇌ 2NO₂(g)
Answer:

Question 19.
The value of K_p for the reaction, CO₂(g) + C(s) ⇌ 2CO(g) is 3.0 at 1000 K. If initially p_CO2 = 048 bar and P_co = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO₂
Answer:
For the reaction, let ‘x’ be the decrease in pressure of CO₂, then


K_p = (2x)² / (0.48 – x) = 3
4x² = 3(0.48 – x)
4x² =144 – x
4x² + 3x – 1.44= 0
a = 4, b = 3,c = -1.44

= [-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 = \(\frac{2 \cdot 66}{8}\) = 0.33
The equilibrium partial pressures are
Pco = 2(0.33) = 0.66bar
Pco₂ = 0.48 – x = 0.48 – 0.33 = 0.15 bar

Applications Equilibrium Constants

Question 1.
Summarise the important characteristic features of equilibrium constants.
Answer:

• Equilibrium constant is a constant value for a given equilibrium at constant temperature.
• The value of equilibrium constant of a given equilibrium does not change on changing pressure, F volume, concentrations of reactants and products and even by adding catalyst as long as temperature is constant.
• The value of equilibrium constant of a given equilibrium changes only when temperature is changed. Its value increases in case of endothermic reactions and decreases in case of exothermic reactions when temperature is increased.
• The value of equilibrium constant depends on its stoichiometry.
  Example A₂ + B₂ ⇌ 2AB
  Let K₁ be the equilibrium constant for the above equilibrium. Then the equilibrium constant K₂ for equilibrium is
  

Question 2.
Mention the three major applications of equilibrium constant.
Answer:
The major applications of equilibrium constant are

1. to predict the direction of the reaction.
2. to predict the extent to which a reaction occurs in a given direction
3. to calculate the equilibrium concentrations.

Question 3.
How do you predict the extent of reaction in terms of K_c and K_p values?
Answer:
The numerical value of the equilibrium constant for a reaction indicates the extent of the reaction. The magnitude of K_c or K_p is directly proportional to the concentrations of the products and inversely proportional to the concentrations of the reactants. This implies that a high value of K indicates the high concentration of products and vice versa.
Case i: If K_c is high (> 10³) products predominate over reactants. Therefore, high value of K_c favours the forward reaction.
Case ii: If K_c is low (< 10^-3) reactants predominates over reactants products. Therefore, low value of K_c favours the backward reaction.
Case iii: K_c = 1, both reactants and products are present at equilibrium.

Question 4.
Define reaction quotient.
Answer:
The reaction quotient Q is the ratio of the product of concentrations of the products to that of the reactants.

where concentrations or partial pressures are not equilibrium concentrations or partial pressures.

Question 5.
How do you predict the direction of reaction interms of equilibrium constant and reaction quotient (Q_c)?
Answer:
Consider the following reaction:
aA + bB ⇌ cC + dD
The reaction quotient is given by

a. If Q_c > K_c, the reaction will proceed in the reverse direction i.e. in the direction of reactants.
b. If Qc_c < K_c, the reaction will proceed in the forward direction i.e. in the direction of products.
c. If Q_c = K_c, the reaction mixture is at equilibrium.
Hence we can conclude that a reaction has the tendency to form products if Q < K_c and to form reactants if Q > K_c.

Question 6.
The value of Kc for the reaction 2A ⇌ 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 will the reaction proceed?
Answer:
For the given reaction, the reaction quotient Q_c is given by,

as [A] = [B] = [C] = 3 × 10^-4 M
Q_c = \(\frac{\left(3 \times 10^{-4}\right)\left(3 \times 10^{-4}\right)}{\left(3 \times 10^{-4}\right)^{2}}=1\)
as Q_c > K_c so the reaction will proceed in the reverse direction.

Question 7.
A mixture of 1.57 mol of N₂,1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 1.7 × 10². Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Answer:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Since, Q_c ≠ K_c, the reaction mixture is not at equilibrium.
Since, Q_c > K_c, the net reaction will be in the backward direction.

Question 8.
In which case the yield of the product will be more
A ⇌ B K₁ =1010
A ⇌ Y K₂ = 106
Answer:
Since K₁ > K₂, yield of the product is more in first reaction.

Question 9.
Equilibrium constant, K_c for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g). at 500 K is 0.061. At a particular time, the analysis shows that composition of the reaction mixture of 3.0 mol L^-1 N₂,2.0 mol L^-1 H₂and 0.5 mol L^-1 NH₃. Is the reaction at equilibrium? If not in which direction would the reaction tend to proceed to reach equilibrium?
Answer:

Since Q_c is not equal to K_c, the reaction is not at equilibrium
Since Q_c, is less than K_c, the reaction would tend to proceed in forward direction to attain equilibrium.

Relation between Equilibrium Constant(K), Reaction Quotient(Q) and Gibbs Energy(G)

Question 1.
Write the relation between ΔG, ΔG° and reaction quotient (Q).
Answer:
ΔG = ΔG° + RT ln Q ΔG = ΔG° + 2.303 RT log Q

Question 2.
When does reaction quotient is equal to equilibrium constant?
Answer:
At equilibrium ΔG = 0, Q = K_c

Question 3.
Write the relation between ΔG° and K_c.
Answer:
ΔG° = -2 . 303 RT log K_c or K = e \(\frac{-\Delta G^{0}}{R T}\)

Question 4.
Give the importance of ΔG° and equilibrium constant K.
Answer:
ΔG°< 0, then K_c > 1, process is spontaneous.
ΔG° > 0, then K_c < 1, process is non spontaneous.
ΔG° = 0 then K_c = 1, process is at equilibrium.

Question 5.
The value of ΔG° for the phosphorylation of glucose in glycolysis is 13.8 kJ mol^-1. Find the value of K_c at 298 K.
Answer:
ΔG° = -2-303 RT log K_c or ΔG° = 2.303 RT log \(\frac{1}{K_{c}}\)
13.8 × 1000 = -2.303 × 8.314 × 298 log k_c
log\(\frac{1}{K_{c}}\) = \(\frac{13 \cdot 8 \times 1000}{2 \cdot 303 \times 8 \cdot 314 \times 298}\)
log\(\frac{1}{K_{c}}\) = 2.418
\(\frac{1}{K_{c}}\) = Antilog (2.4185)
= 2.623 × 10²
∴ K_c = \(\frac{1}{2 \cdot 623 \times 10^{2}}\) = 3.814 × 10^-3

Question 6.
Hydrolysis of sucrose gives, sucrose + H₂O ⇌ Glucose + fructose. Equilibrium constant K_c for the reaction is 2 × 10³ at 300 K. Calculate ΔG° at 300 K.
Answer:
ΔG° = -2.303 RT log K_c
= -2.303 × 8 . 314 × 300 log 2 × 10¹³
= -2-303 × 8.314 × 300 × 13.3010
= -7.64 × 10⁴ J mol^-1

Factors Affecting Equilibria

Question 1.
What are the factors affecting the equilibria?
Answer:

• Temperature
• Pressure
• Concentration of reactant and products.

Question 2.
Name the principle used to study the effect of concentration, temperature and pressure on equilibria.
Answer:
Lechatelier’s principle.

Question 3.
State Lechatelier’s principle.
Answer:
It states that “if a constraint (change in temperature, pressure, concentration) is applied to a system at equilibrium, the system itself adjust to nullify the effect of the applied constraint”.

Question 4.
Discuss the effect of concentration on equilibria according to Lechatelier’s principle?
Answer:
According to Lechatelier’s principle, “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 change.”

• When the concentration of reactant(s) is increased, the system tries to reduce their concentration by favouring the forward reaction.
• When the concentration of product(s) is increased, the system tries to reduce their concentration by favouring the backward reaction.
• When the concentration of reactant(s) is decreased, the system tries to increase their concentration by favouring the backward reaction.
• When the concentration of product(s) is decreased, the system tries to increase their concentration by favouring the forward reaction.

Question 5.
What is the effect of change in concentration of H₂,I₂ and HI on H₂(g) + I₂(g) ⇌ 2HI.
Answer:
In the equilibrium reaction,
H₂ (g) + I₂ (g) ⇌ 2HI(g)
If H₂ is added to the above equilibrium, [H₂] in the equilibrium increases. According to Le Chatelier’s principle, the equilibrium opposes this and to reduce the [H₂], favours forward reaction.
In case HI is added to the above equilibrium, in order to oppose this as per Le-Chatelier’s principle, backward reaction is favoured.

Question 6.
Describe the experimental study of the effect of concentration on the equilibrium
Fe³⁺ + SCN^– ⇌ [Fe(SCN)]²⁺
Answer:
When 2 drops of 0.002 M potassium thiocyanate solution is added to 10 ml of 0.2 M ferric nitrate solution, the following equilibrium is attained.

Red colour solution is formed, for [Fe(SCN)]²; is red in colour. The intensity of the red colour remains same with time indicating the attainment of the equilibrium. If we add a drop of 0.002 M Fe³⁺ ion solution to the above equilibrium, to oppose the increase in concentration of Fe³⁺, according to Le Chatelier’s principle forward reaction is favoured. The equilibrium is shifted to right which is indicated by the increase in the intensity of red colour.
Similarly the addition of SCN^– ions to the equilibrium shifts the equilibrium to right.

If we add a drop of solution of [Fe(SCN)]²⁺, to the equilibrium, its concentration increases. To oppose this, according to Le Chatelier’s principle the equilibrium will be shifted to left. This shift is indicated by change in colour of the equilibrium mixture. On addition of [Fe(SCN)]²⁺ initially red colour intensifies but later decreases slightly and then remains same. In case a drop of solution of oxalic acid is added, Fe³⁺ combines with oxalate ion to form ferric oxalate ion [Fe(C₂O₄)₃]^3-. As a result the concentration of Fe³⁺ ion decreases. According to Le Chatelier’s principle, to oppose this, backward reaction is favoured and the equilibrium is shifted to left. This shift is indicated by the decrease in the intensity of red colour.

Question 7.
What is the effect of pressure on equilibria according to Lechatelier’s principle?
Answer:
According to Lechatliers principle,
If R(g) = P(g), pressure has no effect on equilibrium. (Δn_g = 0)
If R(g) > P(g), high pressure favours favours the forward reaction and viceversa.
If R(g) < P(g), high pressure favours favours the bachward reaction and viceversa.
Where R = number of moles of reactants and P = number of moles of products.

Question 8.
Explain the effect of pressure on the equilibrium. CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)
Answer:
In the above equilibrium forward reaction involves decrease in number of moles of gaseous products and hence increase in pressure favours forward reaction and equilibrium will be shifted to right. Obviously decrease in pressures favours backward reaction and equilibrium shifts to left.

Question 9.
The equilibrium remains undisturbed by the addition of inert gas at constant volume. Why?
Answer:
It is because the addition of inert gas at constant volume does not change the partial pressures or the molar concentrations of the substance involved in the reaction.

Question 10.
What s the effect of change in temperature on equilibrium?
Answer:
In endothermic reactions, high temperature favours forward reaction and low temperature favours the backward reaction. In exothermic reaction, high temperature favours backward reaction and low temperature favours forward reaction.

Question 11.
What is the effect of temperature on the equilibrium. 2NO₂(g) ⇌ N₂O₄(g) ΔH = -57.2 kj
Answer:
2NO₂(g) ⇌ N₂O₄(g) ΔH = -57.2 kj
When temperature is increased, according to Le-Chatelier’s principle, to oppose the increase in temperature, decomposition of N₂O₄ is favoured and equilibrium is shifted to left.

Question 12.
What is the effect of catalyst on equilibrium?
Answer:
There is no effect of catalyst on the position of chemical equilibrium. But catalyst shortens the time to attain equilibrium quickly by increasing the rate of both forward and backward reaction to the same extent by decreasing the activation energy.

Ionic Equilibrium in Solution

Question 1
What is ionic equilibrium?
Answer:
The equilibrium involving ions in aqueous solutions is called ionic equilibrium.

Question 2.
What are electrolytes? Give examples.
Answer:
Compounds that conduct electricity either in solution state or in fused state are called electrolytes.
Example: NaCl, HCl, NaOH, etc.

Question 3.
What are nonelectrolytes?
Answer:
Compounds that do not conduct electricity either in solution state or in fused state are called non electrolytes.

Question 4.
What are strong electrolytes? Give examples.
Answer:
Electrolytes in which dissociation proceeds to almost completion are called strong electrolytes.
Examples’.
1. All salts solutions like NaCl, KCl, MgSO₄ etc., are strong electrolytes.
2. All strong acids such a< HCl, H₂SO₄, HNO₃ etc., are strong electrolytes.
3. All strong bases such at NaOH and KOH etc., are strong electrolytes.

Question 5.
What are weak electrolytes? Give examples.
Answer:
Electrolytes in which dissociation is partial i.e., does not proceeds to completion are called weak electrolytes:
Examples:
1. Weak acids such as H₂CO₃, H₃PO₄, CH₃COOH, H – COOH
2. Weak bases such Al(OH)₃, NH₄OH, Ca(OH)₂, CH₃ – NH₂etc.,

Acids, Bases and Salts

Question 1.
Define acid and bases based on Arrhenius concept with examples.
Answer:
Ac cording to Arrhenius theory
“An acid is a substance which produces hydrogen ion in the aqueous solution”
Examples: HCl (g) + aq → H⁺(aq) + Cl^–(aq)
HCl, HNO₃ HNO₃(l) + aq → H⁺(aq) + NO^–₃ (aq)

A base is a substance, which produces hydroxyl ions (OH^–) in the aqueous solution
Examples: NaOH(s) + aq → Na⁺(aq) + OH^–(aq)
NaOH, KOH KOH(s) + aq → K⁺(aq) + OH^–(aq)

Question 2.
Explain Bronsted – Lowry theory of acids and bases.
Answer:
According to this theory “Acid is a substance, which has a tendency to donate a proton “Base is a substance, which has a tendency to accept a proton.
In short, “acid is a proton donor and base is proton acceptor”.
An acid after donating the proton becomes base, which is called conjugate base of the acid.
Examples

A base after accepting the proton behaves as an acid, which is called conjugate acid of the base.
Examples:

Question 3.
What is hydronium ion?
Answer:
Protonated water or hydrated hydrogen ion is called hydronium ion or oxonium ion. It is represented by H₃O⁺

Question 4.
What is acid base conjugate pair? Illustrate with an example.
Answer:
“A pair of an acid and a base that differs by only one proton is called conjugate acid base pair”. For example,

Here NH⁺₄ and NH₃ is a conjugate acid base pair.
H₂O and OH^– is a conjugate acid-base pair.
The conjugate base of a strong acid is weak where as the conjugate base of a weak acid is strong.
Since HCl is strong acid, so Cl^– is weak conjugate base of HCl
Since acetic acid weak acid, CH₃COO^– is the strong conjugate base of CH₃COOH
Note:
(1) CH₃COO^– is a stronger base than Cl^–
(2) Similarly, the conjugate acid of a strong base is weak where as the conjugate acid of a weak base is strong.

Question 5.
What will be the conjugate bases for the following Bronsted acids: HF, H₂SO₄ and HCOO^–?
Answer:
The conjugate bases should have one proton less in each case and therefore the corresponding conjugate bases are: F^– HSO^–₄ and CO^2-₃ respectively.

Question 6.
Write the conjugate acids for the following Bronsted bases: NH^–₂, NH₃ and HCOO^–
Answer:
The conjugate acid should have one extra proton in each case and therefore the corresponding conjugate acids are: NH₃, NH⁺₄ and HCOOH respectively.

Question 7.
The species: H₂O, HCO^–₃, HSO^–₄ and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and conjugate base.
Answer:
The answer is given in the following table.
Note: (i) To get conjugate acid, add H⁺
Example: H₂O + H⁺ → H₃O⁺ conjugate acid.

(ii) To get conjugate base remove H⁺
Example: H₂O → H⁺ + OH^– conjugate base.

Question 8.
Define acid and base according to Lewis theory with examples.
Answer:
According to Lewis, “Acid is an electron pair acceptor and base is an electron pair donor”
Examples for Lewis bases:
NH₃, H₂O, R – OH, F^–, CN^–, H^– etc.
Examples for Lewi’s acids
1) Neutral molecules such as BF₃, AlCl₃, FeCl₃ etc.,
2) Simple cations such as Fe³⁺, Ag⁺, H⁺ etc.,

Question 9.
BF₃ is a Lewis acid and NH₃ is a Lewis base. Explain.
Answer:
In BF₃ boron is electron deficient. It is ready to accept a pair of electron from any Lewis base to form a coordinate bond.
In NH₃ nitrogen is electron rich, it is ready to give a pair of electron to any Lewis acid to form co-ordinate bond resulting in the formation of adduct.
F₃B + : NH₃ → [F₃B ← NH₃]

Question 10.
Classify the following species into Lewis acids and Lewis bases and show how these act as such: (a) HO^– (b) F^– (c) H⁺ (d) BCl₃.
Answer:
(a) Hydroxyl ion (HO^–) is a Lewis base as it can donate an electron lone pair (:OH^–).
(b) Flouride ion (F^–) acts as a Lewis base as it can donate any one of its four electron lone pairs.
(c) A proton (H⁺) is a Lewis acid as it can accept a lone pair of electrons from bases like hydroxyl ion and fluoride ion.
(d) BCl₃ acts as a Lewis acid as it can accept a lone pair of electrons from species like ammonia or amine molecules.

 Ionisation of Acids and Bases

Question 1.
Define ionic product of water. What its value at 298 K.
Answer:
It is defined as the product of molar concentration of hydrogen ion and hydroxyl ion.
K_W = [H⁺][OH^–] or K_w = [H₃O⁺][OH^–]
At 298 K, K_w = 10^-14 M²
∴ [H⁺][OH^–] = 10^-14 at 298 K

Question 2.
What is the hydrogen ion concentration in pure water at 298 K?
Answer:
[H⁺] = 10^-7 M at 298 K.

Question 3.
What is the hydroxyl ion concentration in pure water at 298 K?
Answer:
[OH^–] = 10^-7 M at 298 K

Question 4.
What is the molarity of pure water if density of water is 1000 g/L?
Ans.
55.5 M

In acidic solution, [H₃O⁺] > [OH^–] or [H⁺] > [OH^–]
In neutral solution, [H₃O⁺] = [OH^–] or [H⁺] = [OH^–]
In basic solution, [H₃O⁺] < [OH^–] or [H⁺] < [OH^–]
In acidic solution, [H⁺] is expressed as [H₃O⁺].
At 298 K, it follows that ;

• In pure water, [H⁺] = 10^-7 moles/dm³, [OH^–] = 10^-7 moles/dm³
• [H⁺] = [OH^–] = 10^-7 moles/dm³, the solution is neutral.
• [H⁺] > 10^-7 moles/dm³, the solution is acidic.
• [H⁺] < 10^-7 moles/dm³, the solution is basic.
• At constant temperature if [H⁺] is increased [OH^–] decreases and vice versa.
• If [H⁺] is increased from 10^-7 to 10^-3, the OH^– decreases from 10^-7 to 10^-11 such that [H⁺][OH^–] = 10^-3 × 10^-11 = 10^-14.

Question 5.
Define pH scale.
Answer:
pH scale is a negative logarithmic scale used to express hydronium ion concentration in molarity. pH scale has range from 0 to 14.

Question 6.
Define pH of a solution.
Answer:
It is defined as the negative logarithm to base ten of the molar concentration of hydrogen ion.
i.e.,

Question 7.
Define pOH of a solution.
Answer:
It is defined as the negative logarithm to the base ten of the molar concentration of hydroxyl ion.

Question 8.
Define pK_w.
Answer:
It is defined as negative logarithm to the base ten of ionic product of water (kw).
pK_w = -log₁₀K_w

Question 9.
Prove that pH + pOH = 14 = pK_w at 298 K.
Answer:
We know that [H⁺] [OH^–] = KW = 10^-14
Taking negative log on both sides
-log₁₀[H⁺] +(- log₁₀[OH^–]) = -log₁₀K_w = -log₁₀10^-14
pH + pOH = 14 at 298 K
Note: At 298 K it follows that;

• pH = pOH = 7, the solution is neutral.
• pH > 7, the solution is basic.
• pH < 7, the solution is acidic.
• pH changes by one unit corresponds to a 10 fold change in the concentration of hydrogen ions and vice versa.
• Higher the concentration of the hydrogen ions lower will be the pH value and vice versa.

Question 10.
Give the different methods to determine the pH of a solution.
Answer:
1. pH of a solution can be measured using several methods. pH of a solution can be determined using pH papers. A strip of pH paper is dipped in the solution whose pH is to be determined. The paper gets a particular colour which is compared with the colours of the standard colour chart and matching colour is identified. The pH of the solution will be equal to that of matching colour in the colour chart. By this method pH of a solution can be determined with an accuracy of 0.5 pH.

2. pH of a solution can also be determined colourimetrically using buffer solutions of known pH and universal indicator.

3. pH of a solution is also determined using pH meters with an accuracy of 0.001. In this method pH is determined by measuring the electrode potential of the hydrogen electrode or glass electrode dipped in, the given solution.

Question 11.
Define dissociation constant of acid (Ka).
Answer:
It is defined as “the ratio of product of molar concentration of dissociated ions to the molar concentration of undissociated molecules at a given temperature.”
Let us consider a weak acid, which dissociates partially in aqueous solution as
CH₃COO ⇌ CH₃COO^– + H⁺
Applying Law of mass action,

Where K_a is called dissociation constant of a weak acid.

Question 12.
Define dissociation constant of base (K_b).
Answer:
“It is defined as the ratio of product of molar concentration of dissociated ions to the molar concentration of undissociated base molecules”
Let us consider a weak base which dissociates partially in aqueous solution as,
NH₄OH ⇌ + NH₄⁺ + OH^–

Where K_b is called dissociation constant of a weak base.

Question 13.
Write the relation between K_a and K_b.
Answer:
K_a × K_b = 10^-14 or K_a × K_b = K_w

Question 14.
Give the importance of K_a and K_b.
Answer:
At a given temperature, K_a is a measure of the strength of the acid. Larger the value of K_a, the stronger is the acid and vice versa.
Similarly, K_b is the measure of the strength of the base. Larger the value of K_b, stronger is the base and vice versa.

Question 15.
The dissociation constants of formic acid and acetic acid rate 1.8 × 10^-4 and 1.74 × 10^-5 respectively at 298 K. Which of them is stronger acid?
Answer:
Formic acid.

Question 16.
Define pK_a.
Answer:
“pK_a is defined as negative logarithm to the base ten of dissociation constant of a weak acid”.
i.e. pK_a = -log₁₀K_a

Question 17.
Define pK_b.
Answer:
“pK_b is defined as negative logarithm to the base ten of dissociation constant of a weak base”.
pK_b = -log₁₀K_b

Question 18.
Give the significance of pK_a and pK_b.
Answer:
pK_a values of different acids give their relative acidic strength. Lower the value of pK_a stronger is the acid. Example, If weak acids HA, HB, and HC have pK_a values 3, 4 and 5 respectively, their relative strengths will be HA > HB > HC.
pK_b values of different bases give their relative basic strengths. Lower the pK_b value stronger is the base. Example, if pK_b values of weak bases XOH, YOH and ZOH are respectively 3, 4, and 5, then their relative strengths will be XOH > YOH > ZOH.

Question 19.
Prove that pK_a + pK_b = 14.
Answer:
We know that K_a × K_b = 10^-14
Take negative log₁₀ on both sides
-logK_a + (-log K_b) = -log10^-14
PK_a + pK_b = 14
or pK_a + pK_b = pK.

Question 20.
Derive the relation between dissociation constant of weak acid (Ka) and degree of dissociation (a). Or Derive Ostwald’s dilution law for a weak electrolyte.
Answer:
Consider a weak acid HX that is partially ionized in the aqueous solution. Let ‘a’ be the extent of dissociation. The equilibrium can be expressed as

This is called Ostwald’s dilution for a weak electrolyte.

Question 21.
What is degree of dissociation?
Answer:
It is amount of solute dissociation into ions.

Question 22.
Write the relation between dissociation constant of weak base (Kb) and degree of dissociation (a).
Answer:
K_b = \(\frac{c \alpha^{2}}{(1-\alpha)}\)

Question 23.
Calculate the pH of 0.001 M HCl.
Answer:

[H⁺] = 0.001M = 10^-3M
PH = =log[H⁺]
= – log 10^-3 = -(-3 log 10)
pH = 3

Question 24.
Calculate the pH of 0.01 MH₂SO₄ by assuming complete ionisation
Answer:

[H⁺] = 0.02 M = 2 × 10^-2
pH = -log [H⁺]
= -1og 2 × 10^-2 = [log2 + log10^-2]
= -log2-(-2 log10) = -log2 + 2
= -0.3010 + 2
pH = 1 . 6990

Question 25.
Calculate the pH of 0.02 M Ba (OH)₂ solution by assuming complete ionisation.
Answer:

∴ [OH^–] = 0.04 M = 4 × 10^-2M
pOH = -log[OH^–]
= -log[4 × 10^-2] = -[log4 + log10^-2]
=-log4-(-21ogl0) = -0.6021 + 2
pOH = 1.3979
We know that pH + pOH = 14
pH = 14-1.3979
pH = 12.6021

Question 26.
Calculate the pH of 0.1 M CH3COOH. The dissociation constant of acetic acid is 1.8 × 10^-5.
Answer:
Acetic acid is weak acid dissociated partially as
CH3COOH ⇌ CH₃COO^– + H⁺
[H⁺] = \(\sqrt{\mathrm{K}_{\mathrm{a}} \times \mathrm{c}}=\sqrt{1 \cdot 8 \times 10^{-5} \times 0 \cdot 1}\)
= 1.341 × 10^-3M
pH = -log[H⁺]
pH = -log 1.341 × 10^-3 = 3 – 0.1587
= 2.8413

Question 27.
Calculate the pH of 0.002 M acetic acid if it is 2.3% ionised at this solution.
Answer:
[H⁺] = cα

[H⁺] = cα
∴ [H⁺] = 0.002 × 0.023 = 4.6 × 10^-5M
PH = -log[H⁺]
= -log4.6 x 10^-5
= -[log 4 . 6 + log 10^-5 ] = -0 . 6628 – (-5 log 10)
= -0.6628 + 5 = 4.3372

Question 28.
Calculate the pH of 0.1 M NH₃ solution. The ionisation constant Kb for NH₃ is 1.8 × 10 s at 298 K.
Answer:
NH₃ + H₂O ⇌ NH⁺₄ + OH^–

pOH = -log[OH^–]
= -log[1.341 × 10^-3]
= -log 1.341 + log10^-3
= -1og1 . 341 – (-3 log 10) = -0 . 1587 + 3
pOH = 2.8413
We know that pH = 14 – pOH
= 14 – 2.8413
=11.1587

Question 29.
The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3 M. What is its pH?
Answer:
pH = -log[H⁺]
= -log(3.8 × 10^-3)
= -log 3.8 + 3 = 3-0.5798 = 2.4202 = 2.42

Question 30.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer:
pH = -log[H⁺]
or log[H⁺] = -pH = -3.76 = \(\overline{4}\).24
∴ [H⁺] = Anti log 4̄.24 = 1.738 × 1c^-4 M

Question 31.
It has been found that the pH of a 0.01 M solution of monobasic organic acid is 4.15. Calculate the concentration of the anion, the ionisation constant of the acid and its pKa.
Answer:
HA ⇌ H⁺ + A^–
pH = -log[H⁺]
4-15 = -log[H⁺] log[H⁺] = -4.15
[H⁺] = antilog (\(\overline{5}\).85) = 7.08 ×10^-5M
= 7.08 × 10^-5M
[A^– ] = [H⁺] = 7.08 ×10^-5 M

= \(\frac{\left(7.08 \times 10^{-5}\right)\left(7.08 \times 10^{-5}\right)}{10^{-3}}=5.01 \times 10^{-7}\)
pK_a = – log K_a
= – log (5.01 × 10^-7) = 7 – 0.6998 = 6.302

Question 32.
The pH of 0.005 M codeine (C₁₅H₂₁NO₃) solution is 9.95. Calculate the ionisation constant and pK_b.
Answer:
Codeine is a base, its ionisation may be represented as
B ⇌ H₂O BH⁺ + OH^–
pH = 9.95 ∴ pOH = 14 – 9.95 = 4.05
pOH = -log[OH^–]
-log[OH^–] = 4.05
or log[OH^–] = -4.05 = \(\overline{5}\).95
[OH^–] = 8.91 × 10^-5 M

\(=\frac{\left(8.91 \times 10^{-5}\right)^{2}}{5 \times 10^{-3}}=1.588 \times 10^{-6}\)
pK_b = – log_b
pK_b = -log(1.588 × 10 ^-6) = 6 – 0.1987 = 5.7991

Question 33.
The dissociation constan.t of formic acid 1.77 × 10^-4 at 298 K. Calculate its pK_a and pK_b of its conjugate base.
pK_a = -logK_a
= – log 1.77 × 10 ^-4
= -[log1.77 + log10 ^-4]
= – log 1. 77 + 4
= -0.2480 + 4
= 3.752
We know that pK_a + pK_b = 14
pK_b = 14 – 3.752
pK_b =10.248

Question 34.
The pK_a of acetic acid is 4.7447. Calculate Its K_a and K_b of its conjugate base.
Answer:
pK_a = -logK_a
-logK_a = 4.7447
IogK_a = -4.7447
logK = \(\overline{5}\).2553
K_a = Antilog (\(\overline{5}\).2553)
K_a = 1.800 × 10^-5
We know that _a × K_b =10
K_b = \(=\frac{1 \times 10^{-14}}{1 \cdot 8 \times 10^{-5}}\) = 0.555 × 10^-9
K_b= 5.55 × 10^-10

Question 35.
Determine the degree of ionisation and pH of a 0.05 M of ammonia solution. The ionisation constant of ammonia is 1.77 × 10^-5. Also, calculate the ionisation constant of the conjugate acid of ammonia.
Answer:

= 0.01881 moles
∴ [OH^–] = 0.05 × 0.018
[OH^–] = 9.4 × 10^-4M

pH = -log[H⁺]
PH = – log(1.06 × 10^-11) = 10.97
Now, using the relation for conjugate acid-base pair K_a x K_b = K_w
Using the value of K_b of NH₃ is 1.77 × 10^-5
We can determine the concentration of conjugate acid NH⁺₄

Question 36.
The pH of 0.004 M hydrazine solution is 9.7. Calculate its ionisation constant K_b and pK_b.
Answer:
NH₂ NH₂ + H₂O ⇌ NH_b NH⁺3 + OH^–
pH = 1og[H⁺]
[H⁺]= Antilog (-pH)
= Anti1og(-9 . 7)
= Antilog \((\overline{10} \cdot 3) \) = 1.995 × 10^-5 M

The concentration of the corresponding hydrazinium ion ¡s 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 0.004 M.

Question 37.
Calculate pH of a 1.0 × 10^-8M solution of HCl.
Answer:
2H₂O(l) ⇌ H₃O⁺(aq) + 0H^–(aq)
K_W = [OH^–][H₃O⁺] = 10^-14
The H₃O⁺ concentration is generated (i) from the ionisation of HCl dissolved and (ii) from ionisation of H₂O. In these very dilute solutions, both sources of H₃O⁺ must be considered.
(i) [H₃O⁺] = 10^-8M
(ii) [H₃O⁺] = 10^-7M
∴ [H₃O⁺] = 10^-8 + 10^-7
= 1.1 × 10^-7M
pH = -log[H₃O⁺]
= -log 1.1 × 10^-7
= -[log 1.1 + log10^-7 ]
= -0.0414 + 7 = +6.958

Question 38.
The pH of 0.1 M 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.
Answer:
pH = -log[H⁺]
[H⁺] = Anti log (-pH)
= Antilog(-4.5)
= Antilog \(\overline{5}\).5
[H+] = 3.162 × 10^-5M = [A^–]

Thus

K_a = 1 × 10^-8
pK_a = -logK_a
PK_a = -log(10^-8) = 8

Question 39.
The degree of ionisation of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid.
Answer:
[H⁺] = cα
= 0.1 × 0.132
= 0.0132 = 1.32 × 10^-2 M
pH = -log 1.32 × 10^-2 = 1.88
K_a = cα²
= 0.1 × (0.132)²
K_a = 1.74 × 10^-3

Question 40.
What is dibasic acid? Give an example.
Answer:
An acid containing two ionisable protons per molecule is called diabasic acid.
Example: H₂SO₄ → 2H⁺ + SO₄

Question 41.
What are poly basic acids?
Answer:
Acids containing more than two ionisable protons per molecule are called poly basic acids.
Example: H₃PO₄ (Tri basic acid)
H₃PO₄ → 3H⁺ + PO^3-₄

Question 42.
Give an expression of first and second ionisation constants (K_a1 and K_a1) of the acid H₂X.
Answer:

Question 43.
On which factors does the acid strength depends?
Answer:
Acid strength mainly depends on
(a) bond strength (H – A) (b) polarity of HA bond

Question 44.
What is common ion effect? Give an example.
Answer:
“The suppression of degree of dissociation of weak electrolyte by adding strong electrolyte having common ion is called common ion effect”.
Example: Consider a weak electrolyte, acetic acid its dissociation at equilibrium is,
CH₃COOH ⇌ CH₃COO^– + H⁺
If little of sodium acetate be added to the solution. Sodium acetate being a strong electrolyte dissociates completely as,
CH₃COONa → CH₃COO^–+ Na⁺
So on addition of sodium acetate, [CH₃COO^–] in the equilibrium increases. According to Le-Chatelier principle, to oppose this increase in concentration of CH₃COO^– ions the equilibrium favours backward reaction and shifts to left. As a result [H⁺] decreases and [CH₃COOH] increases. Thus the degree of dissociation of CH₃COOH is suppressed by the addition of CH₃COONa.
Similarly the degree of dissociation of NH₄OH is suppressed by the addition of NH₄Cl.

Question 45.
What is hydrolysis of salts? What type of salts undergo hydrolysis?
Answer:
The process of interaction between hydrogen ion and hydroxyl ion of water and anion and cation of salt to give either acidic or basic solution is called hydrolysis.
Only the salts of
(a) Weak acid and strong base (example: CH₃COONa)
(b) Strong acid and weak base (example: NH₄CI)
(c) Weak acid and weak base (example: CH₃COOCH₄) undergo hydrolysis.

Question 46.
Salts of strong acid and strong base solutions are neutral. Why?
Answer:
Because they do not undergo hydrolysis.

Question 47.
Salts of weak acid and strong base solutions are basic. Explain with an example.
Answer:
Sodium acetate (CH₃COONa) is a salt of weak acid and strong base. It undergoes ionisation in water to form acetate ion and Na+ ion. Aeetate ion further undergoes hydrolysis to form more OH^– ion concentration in solution making it basic.

The ionic form is,
CH₃COO^– + Na⁺ + H₂O ⇌ CH₃COOH + Na⁺ + OH^–
∴ Solution containing OH” ions. So solution is basic and the pH > 7 and overall solution is basic.
So, the pH of salt of weak acid and strong base solutions is greater than 7.

Question 48.
Salts of strong acid and weak base solutions are acidic. Explain with an example.
Answer:
NH₄CI is a salt of weak base and strong acid salt which undergoes dissociation in water to form NH⁺₄ and Cl^–.
NH₄Cl + aq → NH⁺₄ (aq) + Cl^–(aq)
Further NH⁺₄ ions undergo hydrolysis with water to form NH₄OH and H⁺ ions. So H⁺ ions in the solution increases making the solution acidic and pH < 7.
NH⁺₄ (aq) + H₂O(1) ⇌ NH₄OH(aq) + H⁺(aq)

Question 49.
Hydrolysis of salt of weak acid and weak base solutions are either acidic or basic. Explain with an example.
Answer:
Consider the hydrolysis of CH₃COONH₄ salt formed from weak acid and weak base. The ions formed ’ undergo hydrolysis as follow:
CH₃COO +NH⁺₄+H₂O ⇌ CH₃COOH + NH₄OH
CH₃COOH and NH₄OH, also remain into partially dissociated form:
CH₃COOH ⇌ CH₃COO^– + H⁺
NH₄OH ⇌ NH⁺ + OH^–
H₂O ⇌ 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 + \(\frac { 1 }{ 2 }\)(pK_a – pK_b)
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.

Question 50.
The pK_a of acetic acid and pK_b of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution.
Answer:
pH = 7 + \(\frac { 1 }{ 2 }\)[pK_a – pK_b]
= 7 + \(\frac { 1 }{ 2 }\)[4.76 – 4.75]
= 7 + \(\frac { 1 }{ 2 }\)[0 . 01] = 7 + 0 . 005 = 7 . 005

Buffer Solutions

Question 1.
What are buffer solutions?
Answer:
A solution which resist the change in pH with the addition of small amounts of an acid or alkali is called buffer solution.

Question 2.
How do you prepare buffer solutions?
Answer:
Buffer solutions are prepared from the knowledge of pK_a of acid and pK_b of base and by controlling the ratio of the salt and acid or salt and base.

Question 3.
Give an example for acidic buffer.
Answer:
Mixture of acetic acid and sodium acetate. (CH₃COOH + CH₃COONa)

Question 4.
Give an example of basic buffer.
Answer:
Mixture of ammonium hydroxide and ammonium chloride. (NH₄OH + NH₄Cl)

Question 5.
Give an example for neutral buffer,
Answer:
Ammonium acetate

Question 6.
Derive Henderson – Hasserbalch equation for acidic buffer.
Answer:
Consider an acidic buffer solution consisting of a weak acid HA and its highly ionised salt, BA.
HA + H₂O ⇌ H₃O⁺ + A^–
BA → B⁺ + A^–
As the acid is very weak, it is only slightly ionized, the concentration of the free acid (unionised acid) at equilibrium is almost equal to the concentration of the acid present originally, i.e., [HA] ≈ [acid] and the salt BA is dissociated completely, concentration [A^–] is equal to the molar concentration of the salt i.e., [BA]] = [salt].
By applying law of mass action to the equation 1
Dissociation constant of the weak acid,

Taking logarithms

Mulitiply by -ve sign throughout,

By definitions

This is known as the Henderson’s equation for pH of an acidic buffer solution.

Question 7.
Write the Henderson – Hasselbalch equation for basic buffer.
Answer:

Solubility Equilibria of Sparingly Soluble Salts

Question 1.
Define solubility product (K_sp).
Answer:
“The solubility product of a sparingly soluble salt is defined as the product of the molar concentration of ions in a saturated solution, the concentration terms being raised to their appropriate power equal to the number of ions produced by one molecule at given temperature”.

Question 2.
Write the solubility product expression for BaSO₄.
Answer:
BaSO₄ v ⇌ Ba²⁺(aq) + SO^2-₄(aq)
K_sp = [Ba²⁺] [SO^2-₄]

Question 3.
Derive the relation between solubility and solubility product for Zirconium phosphate [Zr₃(PO₄)₄]
Answer:
Zr₃(PO₄)₄ ⇌ 3Zr⁴⁺ + 4PO^3-₄
K_sp = [Zr⁴⁺]³ [PO^3-₄]⁴
If S is the solubility of each ion in the solution, then
[Zr⁴⁺] = 3S and [PO^3-₄] = 4S .
= (3S)³ (4S)⁴
K_Sp = 6912S⁷
S = \(\left(\frac{K_{s p}}{6912}\right)^{1 / 7}\)

Question 4.
Write solubility product expression for AB type salt.
Answer:
Example: AgCl, AgNO₃, BaSO₄, CaSO₄ etc.
AB ⇌ A⁺ + B^–
K_sp = [A⁺]¹[B^–]¹
K_sp = (S)(S) = S²
∴ K_sp = S²
∴ S = \(\sqrt{\mathrm{K}_{\mathrm{sp}}}\)
where S = solubility

Question 5.
Write solubility product expression for AB₂ type salt.
Answert:
Example. BaCl₂, PbCl₂
AB₂ ⇌ A²⁺ + 2B^–
K_SP = [A]²⁺ + [B^–]²
= (s)(2s)²
K_sp = 4s³
∴ S = \(\sqrt[3]{\frac{\mathrm{K}_{\mathrm{SP}}}{4}}\)

Question 6.
Derive the relation between solubility and solubility product for the general solid salt of the type M^p+_x X^q-_y
Answer:
M^p+_x X^q-_y (s) ⇌ xM^p+ + yX^q-
where X x ^p+ = y x ^q-
If S is the solubility of each ion in the solution, then [M^p+] = xS and [X^q-] = YS
∴ K_sp = [M^p+]^x[X^q-]^y
K_sp = (xS)^x (yS)^y
K_sp = x^x S^x × y^yS^y
K_sp = x^xy^y × S^x+y

Question 7.
Calculate the solubility of A₂X₃ in pure water assuming that neither kind of ion reacts with water. The solubility product of A₂X₃(Ksp) = 1.1 × 10^-23.
Answer:
A₂X₃ ⇌ 2A³⁺ + 3X^2-
If S is the solubility of salt, then in solution
[A³⁺] = 2S [X^2-] = 3S
∴ K_sp = [A³⁺]²[X^2-]³
= (2S)² × (3S)³
K_sp = 108 S³

Question 8.
The values of K of two sparingly soluble salts Ni(OH)₂ and AgCN are 2.0 × 10^-15 and 6 × 10^-17 respectively. Which salt is more soluble? Explain.
Answer:
AgCN ⇌ Ag⁺ +CN^–
K_sp = [Ag⁺ ] [CN^– ] = 6 × 10^-17
Ni(OH)₂ ⇌ Ni²⁺+ 2OH^–
K_sp = [Ni²⁺][OH^–]² = 2 × 10^-15
Let [Ag⁺ ] = S₁, then [CN^–] = S₁
Let [Ni²⁺] = S₂, then [OH^–] = 2S₂
S₁² = 6 × 10^-17, S₁= 7.8 × 10^-9
(S₂)(2S₂)² = 2 × 10^-15, S2 = 0.58 × 10^-4
Ni(OH)₂ is more soluble than AgCN.

Question 9.
A saturated solution of silver chloride contains 1.46 × 10^-3 g dm^-3 at 291K. What is the solubility product of AgCI at this 1 emperature?
Answer:
Gram molecular mass o. AgCI = 107.8 + 35.5 = 143.3g
The solubility of AgCI = 1.46 × 10^-3 g dm^-3
= \(\frac{1.46 \times 10^{-3}}{143.3}\) = 1.02 × 10^-5 mol dm^-3
At equilibrium, AgCl_(s) ⇌ Ag⁺ (aq) + Cl^– (aq)
K_sp = S²
= (1.02 × 10^-5)² = 1.04 × 10^-10 mol² dm^-6

Question 10.
The solubility product of AgCI at a particular temperature is 1.08 × 10^-10 mol² dm^-6. Calculate its solubility in 0.01M HCl.
Answer:
Note:
In the presence of a common ion with the concentration of ‘c’ mol dm^-3, the solubility ‘S’ of the sparingly
soluble salt is generally given by \(\frac{K_{s p}}{c}\).
Let ‘S’ be the solubility of AgCI in water.

Solubility product of AgCl, K_sp = [Ag⁺] [Cl^–]
1.08 × 10^-10 = (S)(S + 0.01)
Assuming S to be small when compared to 0.01
S + 0.01 ≈ 0.01
∴ 1.08 × 10^-10 = (S)(0.01)
Solubility (S) = \(\frac{1.08 \times 10^{-10}}{0.01}\) = 1.8 × 10^-8 mol dm^-3

Question 11.
Solubility product constant of Pbl₂ in water at a given temperature is 3.4 × 10^-11. Calculate its solubility in grams per dm³.
Answer:
Let S be the solubility of CaF₂ in water. It is a AB₂ type ternary salt.

S = (8.5 × 10¹⁰)^1/3
Take log on both sides
log S = log (8 . 5 × 10^-10)^1/3
= \(\frac { 1 }{ 3 }\) (log 8.5-10) = \(\frac { 1 }{ 3 }\) (0.7079-10) = \(\frac { 1 }{ 3 }\) (-9.2921)
logS = -3.0973
S = Anti log (-3.0973)
S = Antilog = \(\vec{4}\).9027
S = 7.993 × 10^-4
Solubility in gram per dm³ = 7.993 × 10^-4 × 461.19
= 0 . 3686 g dm^-3

Question 12.
The solubility product of AgCl is 2.8 × 10^-10 at 298 K. Calculate the solubility of AgCl in (i) pure water (ii) 0.1M AgNO₃ solution and (iii) 0.1M HCl solution.
Answer:

Question 13.
The solubility product of BaSO₄ at 298K is 2 × 10^-5 mol²dm^-6. Predict whether BaSO₄ will precipitate when (i) equal volumes of 10^-3 M solution of Ba²⁺ ions and 10^-3 M solution of SO^2-₄ ions are mixed, (ii) equal volumes of 2 × 10^-2M solution of [Ba²⁺] ions and SO^2-₄ ions are mixed.
Answer:
i. After mixing the two solutions in equal volumes, ionic concentrations become half of their original values.
10~3
[Ba²⁺] = [SO^2-₄] = \(\frac{10^{-3}}{2}\) = 5 × 10^-4mol dm^-3
∴ Ionic Product of BaSO₄ = [Ba²⁺] [SO^2-₄]
= (5 × 10^-4) (5 × 10^-4) = 2.5 × 10^-7 mol² dm^-6 < K_sp of BaSO₄
Since the ionic product is less than the solubility product value, BaSO₄ will not precipitate.

ii. After mixing, [Ba²⁺] = [S0^2-₄] = \(\frac{2 \times 10^{-2}}{2}\) = 1 × 10^-2 mol dm^-3
∴ Ionic Product of BaS04 = [Ba²⁺] [SO^2-₄] = [1 × 10^-2] [1 × 10^-2]
= 1 × 10^-4 mol² dm^-6 > K_sp of BaSO₄
Since, ionic product value exceeded the solubility product value, BaSO₄ will precipitate.