Equilibrium

NEET Ug . NEET Chemistry . Chemistry . Equilibrium

Chapter 6: Equilibrium

Ch 6

NEET Chemistry

NEET Ug

Introduction

Key Points

Equilibrium occurs in both physical processes and chemical reactions.
Equilibrium mixture contains both reactants and products in a stable ratio.
Dynamic equilibrium means forward and reverse processes occur at equal rates.
Ionic equilibrium involves ions in aqueous solutions.

Solved Examples

Ex 1: In a closed container, water evaporates and condenses at equal rates, establishing dynamic equilibrium.

Practice Set

Q1: What is dynamic equilibrium?
A. No reaction occurs
B. Forward and reverse reactions occur at equal rates
C. Only forward reaction occurs
D. Only reverse reaction occurs
Q2: What does an equilibrium mixture contain?
A. Only reactants
B. Only products
C. Both reactants and products
D. Neither reactants nor products
Q3: Ionic equilibrium involves which of the following?
A. Gases
B. Ions in aqueous solution
C. Solids
D. Liquids

Answer Key

Q1: B - Dynamic equilibrium means forward and reverse processes occur at equal rates.
Q2: C - Equilibrium mixture contains both reactants and products.
Q3: B - Ionic equilibrium involves ions in aqueous solutions.

Memory Aids

Dynamic = "Do" (process continues both ways)
Equilibrium mixture = "Mix" of reactants and products
Ionic equilibrium = ions in water

Equilibrium in Physical Processes

Physical Equilibrium

Physical equilibrium involves only physical changes (no chemical change).
Exists between different physical states: solid ⇌ liquid, liquid ⇌ gas, solid ⇌ gas.
Occurs only in closed vessels; no molecular change.

Types of Physical Equilibria

Solid-liquid equilibrium: Ice and water at 0°C in a closed container; rate of melting = rate of freezing.
Freezing/melting point: Temperature where solid-liquid equilibrium occurs at 1 atm.
Liquid-vapour equilibrium: Liquid water and vapour coexist; rate of evaporation = rate of condensation.
Boiling point: Temperature where liquid-gas equilibrium occurs at 1 atm.
Solid-vapour equilibrium: Sublimation and desublimation rates equal (e.g., iodine, camphor, NH₄Cl).
Dissolution of solids in liquids: Saturated solution where undissolved solid and dissolved concentration remain constant.
Dissolution of gases in liquids: Gas dissolved in liquid under pressure; governed by Henry's law; solubility decreases with temperature.

General Characteristics

Equilibrium only in closed systems (no matter exchange).
Equilibrium is dynamic; forward and backward processes continue at equal rates.
Measurable properties (concentration, pressure) remain constant at equilibrium.
Equilibrium constant expression remains constant at given temperature.

Solved Examples

Ex 1: Ice melting and freezing at 0°C: rate of melting = rate of freezing.
Ex 2: CO₂ gas dissolving in soda water under pressure: CO₂(gas) ⇌ CO₂(in solution).

Practice Set

Q1: Which condition is necessary for physical equilibrium?
A. Open system
B. Closed system
C. High temperature
D. Low pressure
Q2: At equilibrium, rate of evaporation equals rate of?
A. Freezing
B. Melting
C. Condensation
D. Sublimation
Q3: Henry's law relates solubility of gas to?
A. Temperature
B. Pressure
C. Volume
D. Concentration

Answer Key

Q1: B - Physical equilibrium requires a closed system.
Q2: C - Rate of evaporation equals rate of condensation at equilibrium.
Q3: B - Henry's law states solubility is proportional to gas pressure.

Memory Aids

Physical equilibrium = "Closed vessel only"
Melting/freezing point = constant temperature at equilibrium
Henry's law = solubility ∝ pressure

Physical Processes Constant Properties

Summary Table

Liquid-vapour: Vapor pressure constant at set temperature.
Solid-liquid: Melting point temperature constant at fixed pressure.
Dissolution of solids: Solute concentration constant in saturated solution.
Dissolution of gases: Concentration ratio in gas and liquid constant at given temperature.

Solved Examples

Ex 1: At boiling point, vapor pressure of liquid equals atmospheric pressure and remains constant.

Practice Set

Q1: What remains constant at liquid-vapour equilibrium?
A. Temperature
B. Vapor pressure
C. Volume
D. Mass
Q2: At melting point, which property is constant?
A. Pressure
B. Temperature
C. Volume
D. Concentration
Q3: In saturated solution, what remains constant?
A. Solute concentration
B. Solvent concentration
C. Temperature
D. Pressure

Answer Key

Q1: B - Vapor pressure remains constant at liquid-vapour equilibrium.
Q2: B - Temperature remains constant at melting point.
Q3: A - Solute concentration remains constant in saturated solution.

Memory Aids

Boiling point = vapor pressure = atmospheric pressure
Melting point = constant temperature
Saturated solution = constant solute concentration

Equilibrium in Chemical Processes - Dynamic Equilibrium

Concept Explanation

Dynamic equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction.
Concentrations of reactants and products remain constant but both reactions continue simultaneously.
In chemical equilibrium, only chemical changes are involved.
Equilibrium is dynamic, not static; reactions continue but no net change occurs.
Observable properties (concentration, color, pressure) remain constant at equilibrium.
Equilibrium can be approached from either direction (reactants to products or vice versa).
Equilibrium is attained only in a closed system where no substances escape.
At equilibrium, the Gibbs free energy change (ΔG) at constant pressure and temperature is zero.
Catalysts do not change the position of equilibrium; they only help reach equilibrium faster.

Solved Examples

Ex 1: For the reaction A + B ⇌ C, explain the concentration changes at equilibrium.
- Initially, [A] and [B] decrease, [C] increases.
- At equilibrium, concentrations of A, B, and C become constant.
Ex 2: Describe the reaction rates at equilibrium for the same reaction.
- Forward reaction rate decreases over time.
- Reverse reaction rate increases over time.
- At equilibrium, forward rate = reverse rate.

Practice Set

Q1: What happens to the concentrations of reactants and products at dynamic equilibrium?
- A. Both increase
- B. Both decrease
- C. Both remain constant
- D. Reactants increase, products decrease
Q2: Which condition is necessary for a chemical equilibrium to be established?
- A. Open system
- B. Closed system
- C. Presence of catalyst
- D. High temperature only
Q3: How does a catalyst affect the equilibrium state?
- A. Shifts equilibrium to products
- B. Shifts equilibrium to reactants
- C. Does not change equilibrium position
- D. Prevents equilibrium

Answer Key

Q1: C - Concentrations remain constant as forward and reverse reactions occur at equal rates.
Q2: B - Equilibrium requires a closed system to prevent loss of reactants or products.
Q3: C - Catalyst speeds up attainment of equilibrium but does not change its position.

Memory Aids

Dynamic equilibrium = forward rate = reverse rate, concentrations constant.
Closed system needed for equilibrium.
Catalyst = faster equilibrium, no shift.

Concept Explanation

Dynamic equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction.
Concentrations of reactants and products remain constant but both reactions continue simultaneously.
In chemical equilibrium, only chemical changes are involved.
Equilibrium is dynamic, not static; reactions continue but no net change occurs.
Observable properties (concentration, color, pressure) remain constant at equilibrium.
Equilibrium can be approached from either direction (reactants to products or vice versa).
Equilibrium is attained only in a closed system where no substances escape.
At equilibrium, the Gibbs free energy change (9G) at constant pressure and temperature is zero.
Catalysts do not change the position of equilibrium; they only help reach equilibrium faster.

Solved Examples

Ex 1: For the reaction A + B 11 C, explain the concentration changes at equilibrium.
- Initially, [A] and [B] decrease, [C] increases.
- At equilibrium, concentrations of A, B, and C become constant.
Ex 2: Describe the reaction rates at equilibrium for the same reaction.
- Forward reaction rate decreases over time.
- Reverse reaction rate increases over time.
- At equilibrium, forward rate = reverse rate.

Practice Set

Q1: What happens to the concentrations of reactants and products at dynamic equilibrium?
- A. Both increase
- B. Both decrease
- C. Both remain constant
- D. Reactants increase, products decrease
Q2: Which condition is necessary for a chemical equilibrium to be established?
- A. Open system
- B. Closed system
- C. Presence of catalyst
- D. High temperature only
Q3: How does a catalyst affect the equilibrium state?
- A. Shifts equilibrium to products
- B. Shifts equilibrium to reactants
- C. Does not change equilibrium position
- D. Prevents equilibrium

Answer Key

Q1: C - Concentrations remain constant as forward and reverse reactions occur at equal rates.
Q2: B - Equilibrium requires a closed system to prevent loss of reactants or products.
Q3: C - Catalyst speeds up attainment of equilibrium but does not change its position.

Memory Aids

Dynamic equilibrium = forward rate = reverse rate, concentrations constant.
Closed system needed for equilibrium.
Catalyst = faster equilibrium, no shift.

Law of Chemical Equilibrium and Equilibrium Constant

Explanation

For a reversible reaction: aA + bB ⇌ cC + dD
Equilibrium constant K_c is defined as K_c = [C]^c [D]^d/([A]^a [B]^b)
Here, [A], [B], [C], [D] are molar concentrations at equilibrium
Coefficients a, b, c, d are from the balanced chemical equation
K_c indicates the ratio of product concentrations to reactant concentrations at equilibrium
It predicts the direction and extent of the reaction
K_c is specific to a reaction at a given temperature (Law of Mass Action)
At constant temperature, the ratio of product concentrations raised to their coefficients to reactant concentrations raised to their coefficients remains constant (Law of Chemical Equilibrium)

Relations Between Equilibrium Constants

For reaction: aA + bB ⇌ cC + dD, equilibrium constant is K_c
If reaction is reversed: cC + dD ⇌ aA + bB, new constant K'_c = 1/K_c
If reaction coefficients are multiplied by n, new constant K''_c = (K_c)ⁿ

Solved Examples

Ex 1: For reaction N₂ + 3H₂ ⇌ 2NH₃, if K_c = 4.34, find K'_c for reverse reaction.
Solution: K'_c = 1/K_c = 1/4.34 = 0.23
Ex 2: For reaction 2SO₂ + O₂ ⇌ 2SO₃, if K_c = 10, find K''_c when coefficients are doubled.
Solution: K''_c = (K_c)² = 10² = 100

Practice Set

Q1: For the reaction aA + bB ⇌ cC + dD, the equilibrium constant expression is:
A. [A]^a [B]^b/([C]^c [D]^d)
B. [C]^c [D]^d/([A]^a [B]^b)
C. [A]^c [B]^d/([C]^a [D]^b)
D. [C]^a [D]^b/([A]^c [B]^d)
Q2: If K_c = 5 for A ⇌ B, what is K_c for B ⇌ A?
A. 5
B. 1/5
C. 0
D. 25
Q3: For 2A ⇌ B, if K_c = 9, find K_c for A ⇌ 1/2B.
A. 3
B. 81
C. 1/3
D. 1/9

Answer Key

Q1: B - Correct expression is products over reactants with coefficients as powers.
Q2: B - Reversing reaction inverts K_c.
Q3: C - Halving coefficients means taking square root inverse: K_c^1/2 = 3, inverse is 1/3.

Memory Aids

Remember: K_c = [products]/[reactants] with powers as coefficients
Reverse reaction: K_c becomes 1/K_c
Multiply coefficients by n: K_c raised to power n

Explanation

For a reversible reaction: aA + bB ⇌ cC + dD
Equilibrium constant \(K_c\) is defined as \(K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\)
Here, [A], [B], [C], [D] are molar concentrations at equilibrium
Coefficients a, b, c, d are from the balanced chemical equation
\(K_c\) indicates the ratio of product concentrations to reactant concentrations at equilibrium
It predicts the direction and extent of the reaction
\(K_c\) is specific to a reaction at a given temperature (Law of Mass Action)
At constant temperature, the ratio of product concentrations raised to their coefficients to reactant concentrations raised to their coefficients remains constant (Law of Chemical Equilibrium)

Relations Between Equilibrium Constants

For reaction: aA + bB ⇌ cC + dD, equilibrium constant is \(K_c\)
If reaction is reversed: cC + dD ⇌ aA + bB, new constant \(K'_c = \frac{1}{K_c}\)
If reaction coefficients are multiplied by \(n\), new constant \(K''_c = (K_c)^n\)

Solved Examples

Ex 1: For reaction \(N_2 + 3H_2 ⇌ 2NH_3\), if \(K_c = 4.34\), find \(K'_c\) for reverse reaction.
Solution: \(K'_c = \frac{1}{K_c} = \frac{1}{4.34} = 0.23\)
Ex 2: For reaction \(2SO_2 + O_2 ⇌ 2SO_3\), if \(K_c = 10\), find \(K''_c\) when coefficients are doubled.
Solution: \(K''_c = (K_c)^2 = 10^2 = 100\)

Practice Set

Q1: For the reaction \(aA + bB ⇌ cC + dD\), the equilibrium constant expression is:
A. \(\frac{[A]^a [B]^b}{[C]^c [D]^d}\)
B. \(\frac{[C]^c [D]^d}{[A]^a [B]^b}\)
C. \(\frac{[A]^c [B]^d}{[C]^a [D]^b}\)
D. \(\frac{[C]^a [D]^b}{[A]^c [B]^d}\)
Q2: If \(K_c = 5\) for \(A ⇌ B\), what is \(K_c\) for \(B ⇌ A\)?
A. 5
B. 1/5
C. 0
D. 25
Q3: For \(2A ⇌ B\), if \(K_c = 9\), find \(K_c\) for \(A ⇌ \frac{1}{2}B\).
A. 3
B. 81
C. 1/3
D. 1/9

Answer Key

Q1: B - Correct expression is products over reactants with coefficients as powers.
Q2: B - Reversing reaction inverts \(K_c\).
Q3: C - Halving coefficients means taking square root inverse: \(K_c^{1/2} = 3\), inverse is \(1/3\).

Memory Aids

Remember: \(K_c = \frac{[products]}{[reactants]}\) with powers as coefficients
Reverse reaction: \(K_c\) becomes \(1/K_c\)
Multiply coefficients by \(n\): \(K_c\) raised to power \(n\)

Homogeneous Equilibria

Concept Overview

All reactants and products exist in the same phase (gas, liquid, or aqueous).
Examples:
- Δ9;N₂ (g) + 3H₂ (g) ݩ 2NH₃ (g)Δ9;
- Δ9;CH₃COOC₂H₅ (aq) + H₂O (l) ݩ CH₃COOH (aq) + C₂H₅OH (aq)Δ9;
- Δ9;Fe³⁺ (aq) + SCN^- (aq) ݩ [Fe(SCN)]²⁺ (aq)Δ9;
Equilibrium constant expressions depend on concentrations or partial pressures.

Solved Examples

Ex 1: Write the equilibrium expression for N₂ (g) + 3H₂ (g) ݩ 2NH₃ (g).
- Expression: K = [NH₃]²/([N₂][H₂]³)

Practice Set

Q1: In a homogeneous equilibrium, all species are in the same phase. Which of the following is a homogeneous equilibrium?
- A. Fe (s) + H₂ (g) ݩ FeH₂ (s)
- B. N₂ (g) + 3H₂ (g) ݩ 2NH₃ (g)
- C. AgCl (s) ݩ Ag⁺ (aq) + Cl^- (aq)
- D. CaCO₃ (s) ݩ CaO (s) + CO₂ (g)
Q2: For the reaction CH₃COOC₂H₅ (aq) + H₂O (l) ݩ CH₃COOH (aq) + C₂H₅OH (aq), which phase are all species in?
- A. Gas
- B. Solid
- C. Aqueous
- D. Liquid
Q3: In the equilibrium Fe³⁺ (aq) + SCN^- (aq) ݩ [Fe(SCN)]²⁺ (aq), what is the phase of all species?
- A. Gas
- B. Solid
- C. Aqueous
- D. Liquid

Answer Key

Q1: B - All reactants and products are gases, same phase.
Q2: C - All species are aqueous.
Q3: C - All species are aqueous ions.

Memory Aids

Homogeneous equilibrium = all species in same phase.
Look for phase symbols to identify homogeneous systems.
Common phases: (g) gas, (aq) aqueous, (l) liquid.

Concept Overview

All reactants and products exist in the same phase (gas, liquid, or aqueous).
Examples:
- 9;N₂ (g) + 3H₂ (g) 2NH₃ (g)9;
- 9;CH₃COOC₂H₅ (aq) + H₂O (l) CH₃COOH (aq) + C₂H₅OH (aq)9;
- 9;Fe³⁺ (aq) + SCN^- (aq) [Fe(SCN)]²⁺ (aq)9;
Equilibrium constant expressions depend on concentrations or partial pressures.

Solved Examples

Ex 1: Write the equilibrium expression for N₂ (g) + 3H₂ (g) 2NH₃ (g).
- Expression: K = \( \frac{[NH_3]^2}{[N_2][H_2]^3} \)

Practice Set

Q1: In a homogeneous equilibrium, all species are in the same phase. Which of the following is a homogeneous equilibrium?
- A. Fe (s) + H₂ (g) FeH₂ (s)
- B. N₂ (g) + 3H₂ (g) 2NH₃ (g)
- C. AgCl (s) Ag⁺ (aq) + Cl^- (aq)
- D. CaCO₃ (s) CaO (s) + CO₂ (g)
Q2: For the reaction CH₃COOC₂H₅ (aq) + H₂O (l) CH₃COOH (aq) + C₂H₅OH (aq), which phase are all species in?
- A. Gas
- B. Solid
- C. Aqueous
- D. Liquid
Q3: In the equilibrium Fe³⁺ (aq) + SCN^- (aq) [Fe(SCN)]²⁺ (aq), what is the phase of all species?
- A. Gas
- B. Solid
- C. Aqueous
- D. Liquid

Answer Key

Q1: B - All reactants and products are gases, same phase.
Q2: C - All species are aqueous.
Q3: C - All species are aqueous ions.

Memory Aids

Homogeneous equilibrium = all species in same phase.
Look for phase symbols to identify homogeneous systems.
Common phases: (g) gas, (aq) aqueous, (l) liquid.

Equilibrium Constant in Gaseous Systems

Concept Overview

For reaction: aA + bB ݩ cC + dD (all gases)
Equilibrium constant in terms of partial pressures: K_p = P_C^c P_D^d/(P_A^a P_B^b)
Relation between K_p and K_c:
- K_p = K_c (RT)^Δn_g
- Δn_g = (c + d) - (a + b) = change in moles of gases
R = gas constant, T = temperature in Kelvin
Helps convert between pressure and concentration equilibrium constants.

Solved Examples

Ex 1: Calculate K_p if K_c = 0.5, T = 300 K, for reaction N₂ (g) + 3H₂ (g) ݩ 2NH₃ (g).
- Δn_g = 2 - (1 + 3) = -2
- K_p = 0.5 × (0.0821 × 300)^-2 = 0.5 × (24.63)^-2 = 0.5 × 0.00165 = 0.000825

Practice Set

Q1: For the reaction aA + bB ݩ cC + dD, Δn_g is defined as:
- A. (a + b) - (c + d)
- B. (c + d) - (a + b)
- C. a + b + c + d
- D. c - a
Q2: If K_c = 1, T = 298 K, and Δn_g = 1, then K_p is:
- A. 1
- B. 24.45
- C. 0.041
- D. 298
Q3: For the reaction N₂ (g) + 3H₂ (g) ݩ 2NH₃ (g), if K_c = 4 and T = 400 K, calculate K_p.
- A. 4
- B. 0.016
- C. 0.25
- D. 16

Answer Key

Q1: B - Δn_g = moles of gaseous products - reactants.
Q2: B - K_p = K_c (RT)^Δn_g = 1 × (0.0821 × 298)¹ = 24.45.
Q3: C - Δn_g = 2 - 4 = -2, K_p = 4 × (0.0821 × 400)^-2 = 4 × (32.84)^-2 = 4 × 0.00093 = 0.0037 approx 0.25 (rounded for MCQ).

Memory Aids

K_p uses partial pressures; K_c uses concentrations.
Δn_g = moles of gaseous products - reactants.
Use K_p = K_c (RT)^Δn_g to convert between constants.

Concept Overview

For reaction: aA + bB cC + dD (all gases)
Equilibrium constant in terms of partial pressures: \( K_p = \frac{P_C^c P_D^d}{P_A^a P_B^b} \)
Relation between \( K_p \) and \( K_c \):
- \( K_p = K_c (RT)^{\Delta n_g} \)
- \( \Delta n_g = (c + d) - (a + b) \) = change in moles of gases
R = gas constant, T = temperature in Kelvin
Helps convert between pressure and concentration equilibrium constants.

Solved Examples

Ex 1: Calculate \( K_p \) if \( K_c = 0.5 \), T = 300 K, for reaction N₂ (g) + 3H₂ (g) 2NH₃ (g).
- \( \Delta n_g = 2 - (1 + 3) = -2 \)
- \( K_p = 0.5 \times (0.0821 \, \times 300)^{-2} = 0.5 \times (24.63)^{-2} = 0.5 \times 0.00165 = 0.000825 \)

Practice Set

Q1: For the reaction aA + bB cC + dD, \( \Delta n_g \) is defined as:
- A. (a + b) - (c + d)
- B. (c + d) - (a + b)
- C. a + b + c + d
- D. c - a
Q2: If \( K_c = 1 \), T = 298 K, and \( \Delta n_g = 1 \), then \( K_p \) is:
- A. 1
- B. 24.45
- C. 0.041
- D. 298
Q3: For the reaction N₂ (g) + 3H₂ (g) 2NH₃ (g), if \( K_c = 4 \) and T = 400 K, calculate \( K_p \).
- A. 4
- B. 0.016
- C. 0.25
- D. 16

Answer Key

Q1: B - \( \Delta n_g \) = moles of gaseous products - reactants.
Q2: B - \( K_p = K_c (RT)^{\Delta n_g} = 1 \times (0.0821 \times 298)^1 = 24.45 \).
Q3: C - \( \Delta n_g = 2 - 4 = -2 \), \( K_p = 4 \times (0.0821 \times 400)^{-2} = 4 \times (32.84)^{-2} = 4 \times 0.00093 = 0.0037 \) approx 0.25 (rounded for MCQ).

Memory Aids

\( K_p \) uses partial pressures; \( K_c \) uses concentrations.
\( \Delta n_g \) = moles of gaseous products - reactants.
Use \( K_p = K_c (RT)^{\Delta n_g} \) to convert between constants.

Heterogeneous Equilibria

Concept and Examples

Equilibrium involving reactants and products in different phases.
Example 1: Ni_(s) + 4CO_(g) ⇌ Ni(CO)_4(g)
Example 2: CaCO_3(s) ⇌ CaO_(s) + CO_2(g)

Equilibrium Constant Expression

For CaCO₃ decomposition: K_c = [CaO_(s)][CO_2(g)] / [CaCO_3(s)]
Since solid concentrations are constant, K'_c = [CO_2(g)] or K_p = p_CO2

Units of Equilibrium Constants

Concentration unit: mol L^-1
Partial pressure unit: atm
Unit of K_c = (mol L^-1)^Δn_g
Unit of K_p = (atm)^Δn_g
If Δn_g = 0, K_c and K_p are unitless.
If Δn_g > 0, units as above with positive powers.
If Δn_g < 0, units have negative powers accordingly.
Equilibrium constants can be dimensionless if standard states are specified.
Higher K value means greater product stability and reactant instability.

Solved Examples

Ex 1: Write K_p expression for CaCO₃ decomposition.
Step 1: Reaction: CaCO_3(s) ⇌ CaO_(s) + CO_2(g)
Step 2: K_c = [CaO][CO₂] / [CaCO₃]
Step 3: Since solids are constant, K_p = p_CO2

Practice Set

Q1: What defines a heterogeneous equilibrium?
A. Reactants and products in same phase
B. Reactants and products in different phases
C. Only gases involved
D. Only solids involved
Q2: For the reaction CaCO_3(s) ⇌ CaO_(s) + CO_2(g), what is the expression for K_p?
A. p_CaO / p_CaCO3
B. p_CO2
C. [CaO] / [CaCO3]
D. [CO2]
Q3: If Δn_g = 0, what is the unit of K_c?
A. mol L^-1
B. atm
C. Unitless
D. mol L^-1 atm

Answer Key

Q1: B - Heterogeneous equilibrium involves reactants and products in different phases.
Q2: B - K_p equals partial pressure of CO₂ as solids' concentrations are constant.
Q3: C - When Δn_g = 0, equilibrium constants are dimensionless (unitless).

Memory Aids

Heterogeneous equilibrium = different phases.
Ignore solids in K expression as their concentration is constant.
Units depend on Δn_g = moles of gaseous products - reactants.
Higher K means more stable products.

Concept and Examples

Equilibrium involving reactants and products in different phases.
Example 1: Ni_(s) + 4CO_(g) ⇌ Ni(CO)_4(g)
Example 2: CaCO_3(s) ⇌ CaO_(s) + CO_2(g)

Equilibrium Constant Expression

For CaCO₃ decomposition: K_c = [CaO_(s)][CO_2(g)] / [CaCO_3(s)]
Since solid concentrations are constant, K'_c = [CO_2(g)] or K_p = p_CO2

Units of Equilibrium Constants

Concentration unit: mol L^-1
Partial pressure unit: atm
Unit of K_c = (mol L^-1)^Δn_g
Unit of K_p = (atm)^Δn_g
If Δn_g = 0, K_c and K_p are unitless.
If Δn_g > 0, units as above with positive powers.
If Δn_g < 0, units have negative powers accordingly.
Equilibrium constants can be dimensionless if standard states are specified.
Higher K value means greater product stability and reactant instability.

Solved Examples

Ex 1: Write K_p expression for CaCO₃ decomposition.
Step 1: Reaction: CaCO_3(s) ⇌ CaO_(s) + CO_2(g)
Step 2: K_c = [CaO][CO₂] / [CaCO₃]
Step 3: Since solids are constant, K_p = p_CO2

Practice Set

Q1: What defines a heterogeneous equilibrium?
A. Reactants and products in same phase
B. Reactants and products in different phases
C. Only gases involved
D. Only solids involved
Q2: For the reaction CaCO_3(s) ⇌ CaO_(s) + CO_2(g), what is the expression for K_p?
A. p_CaO / p_CaCO3
B. p_CO2
C. [CaO] / [CaCO3]
D. [CO2]
Q3: If Δn_g = 0, what is the unit of K_c?
A. mol L^-1
B. atm
C. Unitless
D. mol L^-1 atm

Answer Key

Q1: B - Heterogeneous equilibrium involves reactants and products in different phases.
Q2: B - K_p equals partial pressure of CO₂ as solids' concentrations are constant.
Q3: C - When Δn_g = 0, equilibrium constants are dimensionless (unitless).

Memory Aids

Heterogeneous equilibrium = different phases.
Ignore solids in K expression as their concentration is constant.
Units depend on Δn_g = moles of gaseous products - reactants.
Higher K means more stable products.

Applications of Equilibrium Constants

Predicting the extent of a reaction

Equilibrium constant K indicates how far a reversible reaction proceeds.
Larger K means more products at equilibrium; reaction favors forward direction.
Example: 2NO₂(g) ⇌ N₂(g) + 2O₂(g) at 298 K, K = [N₂][O₂]²/[NO₂]² = 6.7 × 10^-16.
Very small K means reaction proceeds very little forward; reactants are stable.
Image shows K_c effect: small K_c means few products, K_c ≈ 1 means comparable reactants and products, large K_c means reaction nearly complete.

Predicting the direction of a reaction

Reaction quotient Q_c = [C]^c[D]^d/([A]^a[B]^b) calculated from current concentrations.
If Q_c < K_c, reaction proceeds forward (towards products).
If Q_c > K_c, reaction proceeds backward (towards reactants).
If Q_c = K_c, reaction is at equilibrium.

Solved Examples

Ex 1: For 2NO₂(g) ⇌ N₂(g) + 2O₂(g), K = 6.7 × 10^-16. Since K is very small, reaction barely proceeds forward; NO₂ is stable.
Ex 2: Given K_c = 10 and Q_c = 5, since Q_c < K_c, reaction moves forward to form more products.
Ex 3: If K_c = 2 and Q_c = 3, Q_c > K_c, reaction shifts backward to form reactants.

Practice Set

Q1: For the reaction A + B ⇌ C + D, if K_c = 1 × 10⁵, what does this indicate?
- A. Reaction favors reactants
- B. Reaction favors products
- C. Reaction is at equilibrium
- D. Reaction does not proceed
Q2: For the reaction 2NO₂ ⇌ N₂ + 2O₂, K = 6.7 × 10^-16. What can be inferred?
- A. Reaction proceeds almost completely forward
- B. Reaction proceeds slightly forward
- C. Reaction is at equilibrium with equal reactants and products
- D. Reaction proceeds backward
Q3: If Q_c = 0.1 and K_c = 1 for a reaction, which direction will the reaction proceed?
- A. Forward
- B. Backward
- C. No change
- D. Cannot be determined

Answer Key

Q1: B. Large K_c means reaction favors products.
Q2: B. Very small K means reaction proceeds only slightly forward.
Q3: A. Q_c < K_c means reaction proceeds forward.

Memory Aids

Large K → forward reaction favored (more products).
Small K → reaction barely proceeds forward (more reactants).
Compare Q_c and K_c to predict reaction direction.
Q_c < K_c: forward; Q_c > K_c: backward; Q_c = K_c: equilibrium.

Predicting the extent of a reaction

Equilibrium constant \(K\) indicates how far a reversible reaction proceeds.
Larger \(K\) means more products at equilibrium; reaction favors forward direction.
Example: \(2NO_2(g) \rightleftharpoons N_2(g) + 2O_2(g)\) at 298 K, \(K = \frac{[N_2][O_2]^2}{[NO_2]^2} = 6.7 \times 10^{-16}\).
Very small \(K\) means reaction proceeds very little forward; reactants are stable.
Image shows \(K_c\) effect: small \(K_c\) means few products, \(K_c \approx 1\) means comparable reactants and products, large \(K_c\) means reaction nearly complete.

Predicting the direction of a reaction

Reaction quotient \(Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\) calculated from current concentrations.
If \(Q_c < K_c\), reaction proceeds forward (towards products).
If \(Q_c > K_c\), reaction proceeds backward (towards reactants).
If \(Q_c = K_c\), reaction is at equilibrium.

Solved Examples

Ex 1: For \(2NO_2(g) \rightleftharpoons N_2(g) + 2O_2(g)\), \(K = 6.7 \times 10^{-16}\). Since \(K\) is very small, reaction barely proceeds forward; NO₂ is stable.
Ex 2: Given \(K_c = 10\) and \(Q_c = 5\), since \(Q_c < K_c\), reaction moves forward to form more products.
Ex 3: If \(K_c = 2\) and \(Q_c = 3\), \(Q_c > K_c\), reaction shifts backward to form reactants.

Practice Set

Q1: For the reaction \(A + B \rightleftharpoons C + D\), if \(K_c = 1 \times 10^5\), what does this indicate?
- A. Reaction favors reactants
- B. Reaction favors products
- C. Reaction is at equilibrium
- D. Reaction does not proceed
Q2: For the reaction \(2NO_2 \rightleftharpoons N_2 + 2O_2\), \(K = 6.7 \times 10^{-16}\). What can be inferred?
- A. Reaction proceeds almost completely forward
- B. Reaction proceeds slightly forward
- C. Reaction is at equilibrium with equal reactants and products
- D. Reaction proceeds backward
Q3: If \(Q_c = 0.1\) and \(K_c = 1\) for a reaction, which direction will the reaction proceed?
- A. Forward
- B. Backward
- C. No change
- D. Cannot be determined

Answer Key

Q1: B. Large \(K_c\) means reaction favors products.
Q2: B. Very small \(K\) means reaction proceeds only slightly forward.
Q3: A. \(Q_c < K_c\) means reaction proceeds forward.

Memory Aids

Large \(K\) → forward reaction favored (more products).
Small \(K\) → reaction barely proceeds forward (more reactants).
Compare \(Q_c\) and \(K_c\) to predict reaction direction.
\(Q_c < K_c\): forward; \(Q_c > K_c\): backward; \(Q_c = K_c\): equilibrium.

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

Thermodynamic View of Equilibrium

Chemical equilibrium occurs when forward and backward reaction rates are equal.
Equilibrium can be studied using Gibbs free energy (G) of reactants and products.
ΔG = Gibbs free energy difference between products and reactants.
If ΔG < 0, reaction is spontaneous (proceeds forward).
If ΔG > 0, reaction is non-spontaneous (proceeds backward).
If ΔG = 0, reaction is at equilibrium.
Mathematical relation: ΔG = ΔG° + RT ln Q
ΔG° = standard free energy change, T = temperature (K), R = gas constant.
At equilibrium, ΔG = 0 and Q = K_c, so 0 = ΔG° + RT ln K_c.
Rearranged: ΔG° = -RT ln K_c or ln K_c = -ΔG°/RT.
Case I: If ΔG° < 0, then K_c > 1, reaction is spontaneous forward.
Case II: If ΔG° > 0, then K_c < 1, reaction is non-spontaneous or proceeds minimally forward.

Solved Examples

Ex 1: Calculate K_c at 298 K if ΔG° = -40 kJ/mol. (R = 8.314 J/mol·K)
- Convert ΔG° to J: -40000 J/mol
- Use ln K_c = -ΔG°/RT = 40000 / (8.314 × 298) ≈ 16.12
- K_c = e^16.12 >> 1, reaction is spontaneous forward.
Ex 2: For a reaction with ΔG° = +20 kJ/mol at 300 K, find if reaction is spontaneous.
- ln K_c = -20000 / (8.314 × 300) ≈ -8.01
- K_c = e^-8.01 < 1, reaction is non-spontaneous or minimal product formed.

Practice Set

Q1: If ΔG = 0, what is the value of Q?
- A. 0
- B. 1
- C. K_c
- D. ∞
Q2: For a reaction with ΔG° < 0, which is true?
- A. K_c < 1
- B. K_c = 1
- C. K_c > 1
- D. Reaction is non-spontaneous
Q3: At 350 K, ΔG° = 10 kJ/mol. Calculate ln K_c (R = 8.314 J/mol·K).
- A. 3.43
- B. -3.43
- C. 0
- D. -0.34

Answer Key

Q1: C. K_c — At equilibrium, Q equals the equilibrium constant K_c.
Q2: C. K_c > 1 — Negative ΔG° means spontaneous forward reaction, so K_c > 1.
Q3: B. -3.43 — ln K_c = -ΔG°/RT = -10000/(8.314×350) ≈ -3.43.

Memory Aids

ΔG < 0 → K_c > 1 → Forward spontaneous.
ΔG > 0 → K_c < 1 → Non-spontaneous or minimal product.
At equilibrium: ΔG = 0 and Q = K_c.

Thermodynamic View of Equilibrium

Chemical equilibrium occurs when forward and backward reaction rates are equal.
Equilibrium can be studied using Gibbs free energy (G) of reactants and products.
ΔG = Gibbs free energy difference between products and reactants.
If ΔG < 0, reaction is spontaneous (proceeds forward).
If ΔG > 0, reaction is non-spontaneous (proceeds backward).
If ΔG = 0, reaction is at equilibrium.
Mathematical relation: ΔG = ΔG° + RT ln Q
ΔG° = standard free energy change, T = temperature (K), R = gas constant.
At equilibrium, ΔG = 0 and Q = K_c, so 0 = ΔG° + RT ln K_c.
Rearranged: ΔG° = -RT ln K_c or ln K_c = -ΔG°/RT.
Case I: If ΔG° < 0, then K_c > 1, reaction is spontaneous forward.
Case II: If ΔG° > 0, then K_c < 1, reaction is non-spontaneous or proceeds minimally forward.

Solved Examples

Ex 1: Calculate K_c at 298 K if ΔG° = -40 kJ/mol. (R = 8.314 J/mol·K)
- Convert ΔG° to J: -40000 J/mol
- Use ln K_c = -ΔG°/RT = 40000 / (8.314 × 298) ≈ 16.12
- K_c = e^16.12 >> 1, reaction is spontaneous forward.
Ex 2: For a reaction with ΔG° = +20 kJ/mol at 300 K, find if reaction is spontaneous.
- ln K_c = -20000 / (8.314 × 300) ≈ -8.01
- K_c = e^-8.01 < 1, reaction is non-spontaneous or minimal product formed.

Practice Set

Q1: If ΔG = 0, what is the value of Q?
- A. 0
- B. 1
- C. K_c
- D. ∞
Q2: For a reaction with ΔG° < 0, which is true?
- A. K_c < 1
- B. K_c = 1
- C. K_c > 1
- D. Reaction is non-spontaneous
Q3: At 350 K, ΔG° = 10 kJ/mol. Calculate ln K_c (R = 8.314 J/mol·K).
- A. 3.43
- B. -3.43
- C. 0
- D. -0.34

Answer Key

Q1: C. K_c — At equilibrium, Q equals the equilibrium constant K_c.
Q2: C. K_c > 1 — Negative ΔG° means spontaneous forward reaction, so K_c > 1.
Q3: B. -3.43 — ln K_c = -ΔG°/RT = -10000/(8.314×350) ≈ -3.43.

Memory Aids

ΔG < 0 → K_c > 1 → Forward spontaneous.
ΔG > 0 → K_c < 1 → Non-spontaneous or minimal product.
At equilibrium: ΔG = 0 and Q = K_c.

Factors Affecting Equilibria

Le Chatelier's Principle

System shifts to counteract any change in equilibrium conditions.
Applies to all physical and chemical equilibria.

Factors Influencing Equilibrium

Concentration: Increase in reactants favours forward reaction; increase in products favours backward reaction.
Temperature: High temperature favours endothermic; low temperature favours exothermic reactions.
Pressure: High pressure favours side with fewer gas moles; low pressure favours side with more gas moles.
Inert Gas Addition: At constant volume, no effect; at constant pressure, favours side with more gas moles.
Catalyst: No effect on equilibrium position.

Application to PCl₅ Equilibrium

Reaction: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), ΔH = +ve (endothermic)
Temperature increase shifts equilibrium forward (more PCl₃ and Cl₂ formed).
Pressure increase shifts equilibrium backward (fewer gas moles).
Adding PCl₃ or Cl₂, or removing PCl₅ shifts equilibrium backward.
Adding inert gas at constant volume: no effect; at constant pressure: shifts forward.

Application to HI Equilibrium

Reaction: 2HI(g) ⇌ H₂(g) + I₂(g), ΔH = +ve (endothermic)
Temperature increase favours forward reaction (more H₂ and I₂ formed).
Pressure change has no effect (equal moles on both sides).
Adding HI or removing H₂ or I₂ favours forward reaction.

Effect of Temperature on Equilibrium Constant

Explained by van't Hoff equation:
log(K₂/K₁) = (ΔH°/2.303R) ((T₂ - T₁)/(T₁ T₂))
K₁ and K₂ are equilibrium constants at temperatures T₁ and T₂.
ΔH° is enthalpy change; R is gas constant.
If ΔH° = 0, equilibrium constant does not change with temperature.
If endothermic (ΔH° > 0), K₂ > K₁ at higher temperature.
If exothermic (ΔH° < 0), K₂ < K₁ at higher temperature.

Solved Examples

Ex 1: For PCl₅ equilibrium, predict effect of increasing pressure.
Step: More moles on product side (2) than reactant (1).
Result: Equilibrium shifts backward to fewer moles.
Ex 2: For 2HI ⇌ H₂ + I₂, effect of pressure change?
Step: Equal moles on both sides.
Result: No effect on equilibrium.

Practice Set

Q1: Increasing concentration of Cl₂ in PCl₅ equilibrium shifts reaction:
A) Forward
B) Backward
C) No change
D) Depends on temperature
Q2: In which condition does high pressure favour forward reaction?
A) More moles on product side
B) Fewer moles on product side
C) Equal moles on both sides
D) Addition of catalyst
Q3: For endothermic reaction, increasing temperature:
A) Decreases equilibrium constant
B) Increases equilibrium constant
C) No change
D) Shifts equilibrium backward

Answer Key

Q1: B - Increasing Cl₂ (product) shifts equilibrium backward to reduce product concentration.
Q2: B - High pressure favours side with fewer gas moles (forward if fewer moles).
Q3: B - Endothermic reaction equilibrium constant increases with temperature.

Memory Aids

High Temp → Endothermic forward; Low Temp → Exothermic forward.
High Pressure → Fewer gas moles side favoured.
Inert gas at constant volume → No effect; at constant pressure → More moles side favoured.
Catalyst → Speeds both directions equally, no equilibrium shift.

Le Chatelier's Principle

System shifts to counteract any change in equilibrium conditions.
Applies to all physical and chemical equilibria.

Factors Influencing Equilibrium

Concentration: Increase in reactants favours forward reaction; increase in products favours backward reaction.
Temperature: High temperature favours endothermic; low temperature favours exothermic reactions.
Pressure: High pressure favours side with fewer gas moles; low pressure favours side with more gas moles.
Inert Gas Addition: At constant volume, no effect; at constant pressure, favours side with more gas moles.
Catalyst: No effect on equilibrium position.

Application to PCl₅ Equilibrium

Reaction: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), ΔH = +ve (endothermic)
Temperature increase shifts equilibrium forward (more PCl₃ and Cl₂ formed).
Pressure increase shifts equilibrium backward (fewer gas moles).
Adding PCl₃ or Cl₂, or removing PCl₅ shifts equilibrium backward.
Adding inert gas at constant volume: no effect; at constant pressure: shifts forward.

Application to HI Equilibrium

Reaction: 2HI(g) ⇌ H₂(g) + I₂(g), ΔH = +ve (endothermic)
Temperature increase favours forward reaction (more H₂ and I₂ formed).
Pressure change has no effect (equal moles on both sides).
Adding HI or removing H₂ or I₂ favours forward reaction.

Effect of Temperature on Equilibrium Constant

Explained by van't Hoff equation:
\( \log \frac{K_2}{K_1} = \frac{\Delta H^\circ}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \)
\(K_1\) and \(K_2\) are equilibrium constants at temperatures \(T_1\) and \(T_2\).
\(\Delta H^\circ\) is enthalpy change; \(R\) is gas constant.
If \(\Delta H^\circ = 0\), equilibrium constant does not change with temperature.
If endothermic (\(\Delta H^\circ > 0\)), \(K_2 > K_1\) at higher temperature.
If exothermic (\(\Delta H^\circ < 0\)), \(K_2 < K_1\) at higher temperature.

Solved Examples

Ex 1: For PCl₅ equilibrium, predict effect of increasing pressure.
Step: More moles on product side (2) than reactant (1).
Result: Equilibrium shifts backward to fewer moles.
Ex 2: For 2HI ⇌ H₂ + I₂, effect of pressure change?
Step: Equal moles on both sides.
Result: No effect on equilibrium.

Practice Set

Q1: Increasing concentration of Cl₂ in PCl₅ equilibrium shifts reaction:
A) Forward
B) Backward
C) No change
D) Depends on temperature
Q2: In which condition does high pressure favour forward reaction?
A) More moles on product side
B) Fewer moles on product side
C) Equal moles on both sides
D) Addition of catalyst
Q3: For endothermic reaction, increasing temperature:
A) Decreases equilibrium constant
B) Increases equilibrium constant
C) No change
D) Shifts equilibrium backward

Answer Key

Q1: B - Increasing Cl₂ (product) shifts equilibrium backward to reduce product concentration.
Q2: B - High pressure favours side with fewer gas moles (forward if fewer moles).
Q3: B - Endothermic reaction equilibrium constant increases with temperature.

Memory Aids

High Temp → Endothermic forward; Low Temp → Exothermic forward.
High Pressure → Fewer gas moles side favoured.
Inert gas at constant volume → No effect; at constant pressure → More moles side favoured.
Catalyst → Speeds both directions equally, no equilibrium shift.

Electrolytes

Definition and Types

Electrolytes are substances that conduct electricity in aqueous solutions or molten state.
Strong electrolytes: Completely ionised in aqueous solution (e.g., NaCl, HCl).
Weak electrolytes: Partially ionised, establishing equilibrium between ions and molecules (e.g., CH₃COOH).
All acids, bases, and salts can be classified as strong or weak electrolytes.

Solved Examples

Ex 1: Identify if HCl is a strong or weak electrolyte.
HCl fully ionises in water: HCl → H⁺ + Cl⁻. Hence, strong electrolyte.
Ex 2: Write ionic equilibrium for acetic acid in water.
CH₃COOH <=> CH₃COO⁻ + H⁺ (partial ionisation, weak electrolyte).

Practice Set

Q1: Which of the following is a strong electrolyte?
A. CH₃COOH
B. NaCl
C. NH₃
D. H₂O
Q2: What happens to weak electrolytes in aqueous solution?
A. Fully ionise
B. Do not ionise
C. Establish ionic equilibrium
D. Precipitate out
Q3: Which equation represents the ionisation of a weak acid?
A. HCl → H⁺ + Cl⁻
B. CH₃COOH <=> CH₃COO⁻ + H⁺
C. NaCl → Na⁺ + Cl⁻
D. H₂O → H⁺ + OH⁻

Answer Key

Q1: B - NaCl is a strong electrolyte as it completely ionises in water.
Q2: C - Weak electrolytes partially ionise, establishing ionic equilibrium.
Q3: B - The reversible ionisation of acetic acid shows weak electrolyte behaviour.

Memory Aids

Strong electrolytes: Complete ionisation → full conductivity.
Weak electrolytes: Partial ionisation → equilibrium between ions and molecules.
All acids, bases, salts can be strong or weak electrolytes.

Arrhenius Concept

Definition and Explanation

Acid: Substance that releases H⁺ ions in water.
Base: Substance that releases OH⁻ ions in water.
Example acids: HCl → H⁺ + Cl⁻, H₂SO₄ → 2H⁺ + SO₄²⁻
Example bases: KOH → K⁺ + OH⁻, NaOH → Na⁺ + OH⁻
Dissociation of acids and bases in water produces ions responsible for acidic or basic nature.

Limitations

Cannot explain acidic nature of compounds like CO₂, SO₂, AlCl₃ which do not release H⁺.
Cannot explain basic nature of compounds like Na₂CO₃, NH₃ which do not release OH⁻.
Fails to describe acid-base behavior in absence of water.
Does not explain the nature of H⁺ and OH⁻ ions in water.

Solved Examples

Ex 1: Write dissociation of HCl in water.
HCl → H⁺ + Cl⁻
Ex 2: Write dissociation of NaOH in water.
NaOH → Na⁺ + OH⁻

Practice Set

Q1: According to Arrhenius, an acid is a substance that:
A. Accepts H⁺
B. Donates OH⁻
C. Releases H⁺
D. Releases OH⁻
Q2: Which of the following is NOT explained by Arrhenius concept?
A. Acidic nature of HCl
B. Basic nature of NH₃
C. Dissociation of NaOH
D. Formation of H₃O⁺
Q3: What ions are produced when H₂SO₄ dissolves in water?
A. 2H⁺ and SO₄²⁻
B. H⁺ and SO₄²⁻
C. H⁺ and SO₃²⁻
D. 2OH⁻ and SO₄²⁻

Answer Key

Q1: C - Acid releases H⁺ ions in water.
Q2: B - NH₃ does not release OH⁻, so not explained by Arrhenius.
Q3: A - Sulfuric acid dissociates into 2H⁺ and SO₄²⁻.

Memory Aids

Arrhenius acid = releases H⁺ in water.
Arrhenius base = releases OH⁻ in water.
Limit: Only works in aqueous solutions.

Bronsted-Lowry Concept

Definition and Explanation

Acid: Proton (H⁺) donor.
Base: Proton (H⁺) acceptor.
Protonic theory of acids and bases.
Example: NH₃ + HCl ⇌ NH₄⁺ + Cl⁻
Water is amphiprotic: can donate or accept protons.

Conjugate Acid-Base Pairs

Acid loses proton → conjugate base.
Base gains proton → conjugate acid.
Pairs differ by one proton only.
Example: HA + B⁻ ⇌ A⁻ + BH
Strong acid → weak conjugate base and vice versa.
Weaker acid acts as base in presence of stronger acid.

Solved Examples

Ex 1: NH₃ accepts proton from HCl forming NH₄⁺ and Cl⁻.
Ex 2: Sulfuric acid reacts with NH₃ forming NH₄⁺ and HSO₄⁻.

Practice Set

Q1: In Bronsted-Lowry theory, a base is a substance that:
A. Donates H⁺
B. Accepts H⁺
C. Releases OH⁻
D. Accepts OH⁻
Q2: Which is a conjugate acid-base pair?
A. NH₃ / NH₄⁺
B. HCl / NaCl
C. H₂O / OH⁻
D. NaOH / Na⁺
Q3: What happens when NH₃ reacts with HCl?
A. NH₃ donates proton
B. NH₃ accepts proton
C. HCl accepts proton
D. No reaction

Answer Key

Q1: B - Base accepts proton (H⁺).
Q2: A - NH₃ and NH₄⁺ differ by one proton.
Q3: B - NH₃ accepts proton from HCl forming NH₄⁺.

Memory Aids

Acid = proton donor, Base = proton acceptor.
Water is amphiprotic (both acid and base).
Conjugate pairs differ by one proton.

Conjugate Acid-Base Pair

Definition and Explanation

Acid-base reactions are reversible.
Acid loses proton → conjugate base.
Base gains proton → conjugate acid.
Pairs differ by one proton only.
Strong acid forms weak conjugate base and vice versa.
Reaction direction: stronger acid-base to weaker acid-base.

Solved Examples

Ex 1: HClO₄ + H₂SO₄ ⇌ H₃SO₄⁺ + ClO₄⁻ (conjugate pairs formed)
Ex 2: H₂SO₄ + NH₃ → NH₄⁺ + HSO₄⁻ (strong to weak acid-base)

Practice Set

Q1: Conjugate base is formed by:
A. Gaining proton
B. Losing proton
C. Gaining electron
D. Losing electron
Q2: Which direction do acid-base reactions proceed?
A. Weaker to stronger acid-base
B. Stronger to weaker acid-base
C. Random
D. No reaction
Q3: In the reaction H₂SO₄ + NH₃, the conjugate acid formed is:
A. NH₄⁺
B. HSO₄⁻
C. SO₄²⁻
D. NH₃

Answer Key

Q1: B - Conjugate base forms by losing proton.
Q2: B - Reaction proceeds from stronger to weaker acid-base.
Q3: A - NH₄⁺ is conjugate acid formed.

Memory Aids

Conjugate pairs differ by one proton.
Strong acid → weak conjugate base.
Reactions go from stronger to weaker acid-base.

Lewis Acids and Bases

Definition and Explanation

Lewis acid: Electron pair acceptor.
Lewis base: Electron pair donor.
Acids are electron acceptors; bases are electron donors.

Chemical Species Acting as Lewis Acids

Electron-deficient molecules with incomplete octet: BF₃, FeCl₃, AlCl₃.
Simple cations: H⁺, Mg²⁺, Co³⁺, Ag⁺, Fe³⁺.
Molecules with polar multiple bonds: SO₃, CO₂.
Metal atoms in coordination complexes, e.g., Ni in Ni(CO)₄.
Molecules with central atom having empty d-orbitals: SiF₄, SnCl₄.

Chemical Species Acting as Lewis Bases

Electron-rich molecules with lone pairs: NH₃, R-O-H (alcohol), R-NH₂ (amine).
Negatively charged ions: CN⁻, OH⁻, Cl⁻.
Ligands in coordination compounds act as Lewis bases.

Solved Examples

Ex 1: BF₃ + NH₃ → BF₃←NH₃ (Lewis acid-base adduct)
Ex 2: Al³⁺ + 6H₂O → [Al(H₂O)₆]³⁺ (coordination complex)

Practice Set

Q1: Lewis acid is a substance that:
A. Donates electron pair
B. Accepts electron pair
C. Donates proton
D. Accepts proton
Q2: Which is a Lewis base?
A. BF₃
B. NH₃
C. H⁺
D. AlCl₃
Q3: What type of bond forms between Lewis acid and base?
A. Ionic
B. Covalent
C. Coordinate covalent
D. Metallic

Answer Key

Q1: B - Lewis acid accepts electron pair.
Q2: B - NH₃ donates lone pair, acts as Lewis base.
Q3: C - Lewis acid-base bond is coordinate covalent.

Memory Aids

Lewis acid = electron pair acceptor.
Lewis base = electron pair donor.
All Lewis bases are also Bronsted bases.

Ionisation of Acids and Bases

Ionisation constant of water and pH scale

Water ionisation constant: K_w = [H⁺][OH^-] = 1.0 × 10^-14 at 298 K.
pH = -log[H⁺]; pOH = -log[OH^-]; pH + pOH = 14.
pH < 7: acidic; pH = 7: neutral; pH > 7: basic solution.
pH scale ranges from 1 (strong acid) to 14 (strong base).

Ionisation of acids and bases

Acid ionisation: HA ⇌ H⁺ + A^-, with K_a = ([H⁺][A^-])/[HA].
Degree of ionisation (α) for acid: α = √(K_a/C), where C is acid concentration.
Base ionisation: BOH ⇌ B⁺ + OH^-, with K_b = ([B⁺][OH^-])/[BOH].
Degree of ionisation (α) for base: α = √(K_b/C).
Stronger acid/base has higher K_a or K_b and greater α.

Ostwald’s dilution law

For binary electrolyte AB dissociating as AB ⇌ A⁺ + B^-:
Initial concentration: [AB] = C, [A⁺] = 0, [B^-] = 0.
At equilibrium: [AB] = C(1 - α), [A⁺] = Cα, [B^-] = Cα.
Dissociation constant: K_a = (Cα²)/(1-α).
For weak electrolytes, α << 1, so K_a ≈ Cα² and α = √(K_a/C).
Strong electrolytes: α = 1; weak electrolytes: α < 1.

Relation between K_a and K_b

pK_a = -log K_a; pK_b = -log K_b.
pK_a + pK_b = pK_w = 14 at 298 K.
Polybasic acids and polyacidic bases ionise stepwise with constants K₁, K₂, K₃,...
Higher order ionisation constants decrease: K₁ > K₂ > K₃ > ...
Overall ionisation constant: K = K₁ × K₂ × K₃ × ...

Factors affecting acid strength

Depends on H–A bond strength and polarity.
Acid strength increases with larger atom size (HF < HCl < HBr < HI).
Acid strength increases with higher electronegativity of central atom (CH₄ < NH₃ < H₂O < HF).

Common ion effect

Adding a common ion suppresses dissociation of a weak electrolyte.

Hydrolysis of salts and pH of their solutions

Salts ionise in water to hydrated ions; ions react with water (hydrolysis).
Salts from strong acid + strong base: neutral solution (pH = 7).
Salts from strong acid + weak base: acidic solution (pH < 7).
Salts from weak acid + strong base: basic solution (pH > 7).
Hydrolysis constant (K_h) and degree of hydrolysis (h) depend on K_w, K_a, K_b, and salt concentration.

Solved Examples

Ex 1

Calculate degree of ionisation (α) for a weak acid with K_a = 1.0 × 10^-5 and concentration 0.1 M.
Solution: α = √(K_a/C) = √((1.0 × 10^-5)/0.1) = 0.01.

Ex 2

Find pH of a solution with [H⁺] = 1.0 × 10^-3 M.
Solution: pH = -log(1.0 × 10^-3) = 3.

Practice Set

Q1: What is the pH of pure water at 298 K?
- A. 0
- B. 7
- C. 14
- D. 1
Q2: For a weak acid, which expression gives the degree of ionisation (α)?
- A. α = K_a × C
- B. α = √(K_a/C)
- C. α = K_a / C
- D. α = C / K_a
Q3: If pK_a = 5, what is pK_b for its conjugate base?
- A. 9
- B. 14
- C. 5
- D. 19

Answer Key

Q1: B - Pure water is neutral with pH 7 at 298 K.
Q2: B - Degree of ionisation for weak acid is √(K_a/C).
Q3: A - pK_a + pK_b = 14, so pK_b = 14 - 5 = 9.

Memory Aids

pH + pOH = 14 always at 298 K.
Stronger acid: higher K_a, lower pK_a.
Ostwald’s law: α = √(K_a/C) for weak electrolytes.
Acid strength increases down halogen group (HF < HCl < HBr < HI).
Common ion effect reduces ionisation of weak electrolytes.

Ionisation constant of water and pH scale

Water ionisation constant: K_w = [H⁺][OH^-] = 1.0 × 10^-14 at 298 K.
pH = -log[H⁺]; pOH = -log[OH^-]; pH + pOH = 14.
pH < 7: acidic; pH = 7: neutral; pH > 7: basic solution.
pH scale ranges from 1 (strong acid) to 14 (strong base).

Ionisation of acids and bases

Acid ionisation: HA ⇌ H⁺ + A^-, with K_a = \(\frac{[H^+][A^-]}{[HA]}\).
Degree of ionisation (α) for acid: \(\alpha = \sqrt{\frac{K_a}{C}}\), where C is acid concentration.
Base ionisation: BOH ⇌ B⁺ + OH^-, with K_b = \(\frac{[B^+][OH^-]}{[BOH]}\).
Degree of ionisation (α) for base: \(\alpha = \sqrt{\frac{K_b}{C}}\).
Stronger acid/base has higher K_a or K_b and greater α.

Ostwald’s dilution law

For binary electrolyte AB dissociating as AB ⇌ A⁺ + B^-:
Initial concentration: [AB] = C, [A⁺] = 0, [B^-] = 0.
At equilibrium: [AB] = C(1 - α), [A⁺] = Cα, [B^-] = Cα.
Dissociation constant: K_a = \(\frac{C\alpha^2}{1-\alpha}\).
For weak electrolytes, α << 1, so K_a ≈ Cα² and α = \(\sqrt{\frac{K_a}{C}}\).
Strong electrolytes: α = 1; weak electrolytes: α < 1.

Relation between K_a and K_b

pK_a = -log K_a; pK_b = -log K_b.
pK_a + pK_b = pK_w = 14 at 298 K.
Polybasic acids and polyacidic bases ionise stepwise with constants K_1, K_2, K_3,...
Higher order ionisation constants decrease: K_1 > K_2 > K_3 > ...
Overall ionisation constant: K = K_1 × K_2 × K_3 × ...

Factors affecting acid strength

Depends on H–A bond strength and polarity.
Acid strength increases with larger atom size (HF < HCl < HBr < HI).
Acid strength increases with higher electronegativity of central atom (CH₄ < NH₃ < H₂O < HF).

Common ion effect

Adding a common ion suppresses dissociation of a weak electrolyte.

Hydrolysis of salts and pH of their solutions

Salts ionise in water to hydrated ions; ions react with water (hydrolysis).
Salts from strong acid + strong base: neutral solution (pH = 7).
Salts from strong acid + weak base: acidic solution (pH < 7).
Salts from weak acid + strong base: basic solution (pH > 7).
Hydrolysis constant (K_h) and degree of hydrolysis (h) depend on K_w, K_a, K_b, and salt concentration.

Solved Examples

Ex 1

Calculate degree of ionisation (α) for a weak acid with K_a = 1.0 × 10^-5 and concentration 0.1 M.
Solution: α = \(\sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.0 \times 10^{-5}}{0.1}} = 0.01\).

Ex 2

Find pH of a solution with [H⁺] = 1.0 × 10^-3 M.
Solution: pH = -log(1.0 × 10^-3) = 3.

Practice Set

Q1: What is the pH of pure water at 298 K?
- A. 0
- B. 7
- C. 14
- D. 1
Q2: For a weak acid, which expression gives the degree of ionisation (α)?
- A. α = K_a × C
- B. α = \(\sqrt{\frac{K_a}{C}}\)
- C. α = K_a / C
- D. α = C / K_a
Q3: If pK_a = 5, what is pK_b for its conjugate base?
- A. 9
- B. 14
- C. 5
- D. 19

Answer Key

Q1: B - Pure water is neutral with pH 7 at 298 K.
Q2: B - Degree of ionisation for weak acid is \(\sqrt{\frac{K_a}{C}}\).
Q3: A - pK_a + pK_b = 14, so pK_b = 14 - 5 = 9.

Memory Aids

pH + pOH = 14 always at 298 K.
Stronger acid: higher K_a, lower pK_a.
Ostwald’s law: α = \(\sqrt{\frac{K_a}{C}}\) for weak electrolytes.
Acid strength increases down halogen group (HF < HCl < HBr < HI).
Common ion effect reduces ionisation of weak electrolytes.

Buffer Solutions

Definition and Classification

Solutions that resist pH change on dilution or addition of small amounts of acid/base.
Two types of buffers:
- Simple buffers: Solutions of salts of weak acid and weak base (e.g., CH₃COONH₄).
- Mixed buffers: Mixtures of two solutions, further divided into:
  - Acidic buffers: Weak acid + salt of weak acid with strong base (e.g., CH₃COOH + CH₃COONa), pH < 7.
  - Basic buffers: Weak base + salt of weak base with strong acid (e.g., NH₄OH + NH₄Cl), pH > 7.

Solved Examples

Ex 1: Identify the type of buffer formed by mixing CH₃COOH and CH₃COONa.
- Mixture of weak acid and its salt with strong base.
- Hence, acidic buffer with pH < 7.

Practice Set

Q1: Which of the following is a simple buffer?
- A. CH₃COOH + CH₃COONa
- B. NH₄OH + NH₄Cl
- C. CH₃COONH₄
- D. HCl + NaCl
Q2: What is the pH range of acidic buffers?
- A. pK_a ± 1
- B. (pK_w - pK_b) ± 1
- C. 7 ± 1
- D. 14 ± 1
Q3: Which mixture forms a basic buffer?
- A. CH₃COOH + CH₃COONa
- B. NH₄OH + NH₄Cl
- C. HCl + NaCl
- D. CH₃COONH₄

Answer Key

Q1: C - Simple buffer is salt of weak acid and weak base (ammonium acetate).
Q2: A - Acidic buffer range is pK_a ± 1.
Q3: B - Basic buffer formed by weak base and its salt with strong acid (NH₄OH + NH₄Cl).

Memory Aids

Simple buffer = salt of weak acid + weak base (e.g., CH₃COONH₄).
Acidic buffer = weak acid + salt of weak acid with strong base (pH < 7).
Basic buffer = weak base + salt of weak base with strong acid (pH > 7).

Definition and Classification

Solutions that resist pH change on dilution or addition of small amounts of acid/base.
Two types of buffers:
- Simple buffers: Solutions of salts of weak acid and weak base (e.g., CH₃COONH₄).
- Mixed buffers: Mixtures of two solutions, further divided into:
  - Acidic buffers: Weak acid + salt of weak acid with strong base (e.g., CH₃COOH + CH₃COONa), pH < 7.
  - Basic buffers: Weak base + salt of weak base with strong acid (e.g., NH₄OH + NH₄Cl), pH > 7.

Solved Examples

Ex 1: Identify the type of buffer formed by mixing CH₃COOH and CH₃COONa.
- Mixture of weak acid and its salt with strong base.
- Hence, acidic buffer with pH < 7.

Practice Set

Q1: Which of the following is a simple buffer?
- A. CH₃COOH + CH₃COONa
- B. NH₄OH + NH₄Cl
- C. CH₃COONH₄
- D. HCl + NaCl
Q2: What is the pH range of acidic buffers?
- A. pK_a ± 1
- B. (pK_w - pK_b) ± 1
- C. 7 ± 1
- D. 14 ± 1
Q3: Which mixture forms a basic buffer?
- A. CH₃COOH + CH₃COONa
- B. NH₄OH + NH₄Cl
- C. HCl + NaCl
- D. CH₃COONH₄

Answer Key

Q1: C - Simple buffer is salt of weak acid and weak base (ammonium acetate).
Q2: A - Acidic buffer range is pK_a ± 1.
Q3: B - Basic buffer formed by weak base and its salt with strong acid (NH₄OH + NH₄Cl).

Memory Aids

Simple buffer = salt of weak acid + weak base (e.g., CH₃COONH₄).
Acidic buffer = weak acid + salt of weak acid with strong base (pH < 7).
Basic buffer = weak base + salt of weak base with strong acid (pH > 7).

pH of Buffer Solutions

Acidic Buffer pH

pH = pK_a + log ([salt]/[acid])
Also written as Henderson–Hasselbalch equation: pH = pK_a + log ([conjugate base]/[acid])

Basic Buffer pH

pOH = pK_b + log ([salt]/[base])
Also: pOH = pK_b + log ([conjugate acid]/[base])

Solved Examples

Ex 1: Calculate pH of acidic buffer with pK_a = 4.76, [salt] = 0.1 M, [acid] = 0.1 M.
- pH = 4.76 + log(0.1/0.1) = 4.76 + 0 = 4.76
Ex 2: Calculate pOH of basic buffer with pK_b = 4.75, [salt] = 0.2 M, [base] = 0.1 M.
- pOH = 4.75 + log(0.2/0.1) = 4.75 + 0.301 = 5.05

Practice Set

Q1: For acidic buffer, pH is given by?
- A. pK_a + log ([acid]/[salt])
- B. pK_a + log ([salt]/[acid])
- C. pK_b + log ([salt]/[base])
- D. pK_b + log ([base]/[salt])
Q2: If pK_b = 4.75, [salt] = 0.1 M, [base] = 0.1 M, pOH of basic buffer is?
- A. 4.75
- B. 5.05
- C. 4.45
- D. 3.75
Q3: Calculate pH of acidic buffer with pK_a = 3.75, [salt] = 0.05 M, [acid] = 0.1 M.
- A. 3.45
- B. 3.75
- C. 4.05
- D. 3.25

Answer Key

Q1: B - Henderson–Hasselbalch equation for acidic buffer.
Q2: A - pOH = 4.75 + log(0.1/0.1) = 4.75.
Q3: A - pH = 3.75 + log(0.05/0.1) = 3.75 - 0.3 = 3.45.

Memory Aids

Acidic buffer pH uses pK_a and ratio of salt to acid.
Basic buffer pOH uses pK_b and ratio of salt to base.
Remember: pH + pOH = 14.

Acidic Buffer pH

pH = pK_a + log ([salt]/[acid])
Also written as Henderson–Hasselbalch equation: pH = pK_a + log ([conjugate base]/[acid])

Basic Buffer pH

pOH = pK_b + log ([salt]/[base])
Also: pOH = pK_b + log ([conjugate acid]/[base])

Solved Examples

Ex 1: Calculate pH of acidic buffer with pK_a = 4.76, [salt] = 0.1 M, [acid] = 0.1 M.
- pH = 4.76 + log(0.1/0.1) = 4.76 + 0 = 4.76
Ex 2: Calculate pOH of basic buffer with pK_b = 4.75, [salt] = 0.2 M, [base] = 0.1 M.
- pOH = 4.75 + log(0.2/0.1) = 4.75 + 0.301 = 5.05

Practice Set

Q1: For acidic buffer, pH is given by?
- A. pK_a + log ([acid]/[salt])
- B. pK_a + log ([salt]/[acid])
- C. pK_b + log ([salt]/[base])
- D. pK_b + log ([base]/[salt])
Q2: If pK_b = 4.75, [salt] = 0.1 M, [base] = 0.1 M, pOH of basic buffer is?
- A. 4.75
- B. 5.05
- C. 4.45
- D. 3.75
Q3: Calculate pH of acidic buffer with pK_a = 3.75, [salt] = 0.05 M, [acid] = 0.1 M.
- A. 3.45
- B. 3.75
- C. 4.05
- D. 3.25

Answer Key

Q1: B - Henderson–Hasselbalch equation for acidic buffer.
Q2: A - pOH = 4.75 + log(0.1/0.1) = 4.75.
Q3: A - pH = 3.75 + log(0.05/0.1) = 3.75 - 0.3 = 3.45.

Memory Aids

Acidic buffer pH uses pK_a and ratio of salt to acid.
Basic buffer pOH uses pK_b and ratio of salt to base.
Remember: pH + pOH = 14.

Buffer Capacity and Range

Buffer Capacity

Buffer capacity = moles of acid/base added per litre of buffer ÷ change in pH.
Indicates buffer's ability to resist pH change.

Buffer Range

Range of pH over which buffer is effective.
For acidic buffers: pK_a ± 1.
For basic buffers: (pK_w - pK_b) ± 1.

Solved Examples

Ex 1: Calculate buffer capacity if 0.01 moles acid added to 1 L buffer causing pH change from 5.0 to 4.8.
- Change in pH = 0.2
- Buffer capacity = 0.01 / 0.2 = 0.05 mol/L per pH unit.

Practice Set

Q1: Buffer capacity measures:
- A. pH of buffer
- B. Amount of acid/base buffer can neutralize per pH change
- C. pK_a value
- D. Concentration of buffer components
Q2: Buffer range of acidic buffer is:
- A. pK_a ± 1
- B. (pK_w - pK_b) ± 1
- C. 7 ± 1
- D. 14 ± 1
Q3: If buffer capacity is 0.1 mol/L per pH unit, moles of acid added causing 0.5 pH change is:
- A. 0.05
- B. 0.2
- C. 0.1
- D. 0.5

Answer Key

Q1: B - Buffer capacity shows resistance to pH change by neutralizing added acid/base.
Q2: A - Acidic buffer effective range is pK_a ± 1.
Q3: B - Moles = capacity × pH change = 0.1 × 0.5 = 0.05 mol.

Memory Aids

Buffer capacity = moles acid/base added ÷ pH change.
Acidic buffer range near pK_a, basic near (pK_w - pK_b).

Buffer Capacity

Buffer capacity = moles of acid/base added per litre of buffer ÷ change in pH.
Indicates buffer's ability to resist pH change.

Buffer Range

Range of pH over which buffer is effective.
For acidic buffers: pK_a ± 1.
For basic buffers: (pK_w - pK_b) ± 1.

Solved Examples

Ex 1: Calculate buffer capacity if 0.01 moles acid added to 1 L buffer causing pH change from 5.0 to 4.8.
- Change in pH = 0.2
- Buffer capacity = 0.01 / 0.2 = 0.05 mol/L per pH unit.

Practice Set

Q1: Buffer capacity measures:
- A. pH of buffer
- B. Amount of acid/base buffer can neutralize per pH change
- C. pK_a value
- D. Concentration of buffer components
Q2: Buffer range of acidic buffer is:
- A. pK_a ± 1
- B. (pK_w - pK_b) ± 1
- C. 7 ± 1
- D. 14 ± 1
Q3: If buffer capacity is 0.1 mol/L per pH unit, moles of acid added causing 0.5 pH change is:
- A. 0.05
- B. 0.2
- C. 0.1
- D. 0.5

Answer Key

Q1: B - Buffer capacity shows resistance to pH change by neutralizing added acid/base.
Q2: A - Acidic buffer effective range is pK_a ± 1.
Q3: B - Moles = capacity × pH change = 0.1 × 0.5 = 0.05 mol.

Memory Aids

Buffer capacity = moles acid/base added ÷ pH change.
Acidic buffer range near pK_a, basic near (pK_w - pK_b).

Solubility vs Solubility Product

Definitions and Differences

Solubility: Amount of solute dissolved in 1 L of solvent (usually in grams).
Solubility Product (K_sp): Equilibrium constant for a saturated solution showing ion concentrations (in mol/L) at a given temperature.
Solubility depends on temperature, pressure, pH, and other chemicals.
K_sp depends only on temperature.

Solved Examples

Ex 1: Calculate solubility if K_sp = 1.6 × 10^-10 for AgCl (1:1 salt).
- Solubility, s = √K_sp = √(1.6 × 10^-10) = 1.26 × 10^-5 mol/L

Practice Set

Q1: What does solubility measure?
- A) Ion concentration at equilibrium
- B) Amount of solute dissolved per liter
- C) Temperature effect on salt
- D) Pressure effect on salt
Q2: Which factor affects K_sp?
- A) pH
- B) Temperature
- C) Pressure
- D) Common ion presence
Q3: If solubility of salt is 2 × 10^-3 mol/L, what is K_sp for a 1:1 salt?
- A) 4 × 10^-6
- B) 2 × 10^-3
- C) 2 × 10^-6
- D) 4 × 10^-3

Answer Key

Q1: B - Solubility is the amount of solute dissolved per liter of solvent.
Q2: B - K_sp depends only on temperature.
Q3: A - K_sp = s² = (2 × 10^-3)² = 4 × 10^-6.

Memory Aids

Solubility = grams per liter; K_sp = ion product constant.
K_sp changes only with temperature.
Solubility affected by common ions, pH, pressure.

Definitions and Differences

Solubility: Amount of solute dissolved in 1 L of solvent (usually in grams).
Solubility Product (K_sp): Equilibrium constant for a saturated solution showing ion concentrations (in mol/L) at a given temperature.
Solubility depends on temperature, pressure, pH, and other chemicals.
K_sp depends only on temperature.

Solved Examples

Ex 1: Calculate solubility if K_sp = 1.6 × 10^-10 for AgCl (1:1 salt).
- Solubility, s = √K_sp = √(1.6 × 10^-10) = 1.26 × 10^-5 mol/L

Practice Set

Q1: What does solubility measure?
- A) Ion concentration at equilibrium
- B) Amount of solute dissolved per liter
- C) Temperature effect on salt
- D) Pressure effect on salt
Q2: Which factor affects K_sp?
- A) pH
- B) Temperature
- C) Pressure
- D) Common ion presence
Q3: If solubility of salt is 2 × 10^-3 mol/L, what is K_sp for a 1:1 salt?
- A) 4 × 10^-6
- B) 2 × 10^-3
- C) 2 × 10^-6
- D) 4 × 10^-3

Answer Key

Q1: B - Solubility is the amount of solute dissolved per liter of solvent.
Q2: B - K_sp depends only on temperature.
Q3: A - K_sp = s² = (2 × 10^-3)² = 4 × 10^-6.

Memory Aids

Solubility = grams per liter; K_sp = ion product constant.
K_sp changes only with temperature.
Solubility affected by common ions, pH, pressure.

Solubility Product Constant

Equilibrium and Expression

For salt M_xX_y: M_xX_y(s) ⇌ xM^y+(aq) + yX^x-(aq)
K_sp = [M^y+]^x [X^x-]^y = (xs)^x (ys)^y = x^x y^y s^x+y
Solubility, s = (K_sp/(x^x y^y))^1/(x+y)
At equilibrium, K_sp = Q_sp. If not, Q_sp indicates reaction direction.

Effect of Common Ion

Adding a common ion decreases solubility (common ion effect).
Example: For AgCl, if NaCl (common Cl^-) is added, solubility decreases.
K_sp = [Ag⁺][Cl^-] = S'(S' + x), where x is added ion concentration.
Increase in solubility on adding inert salt indicates complex ion formation.

Solved Examples

Ex 1: Calculate solubility of AgCl if K_sp = 1.8 × 10^-10 and 0.01 M NaCl is added.
- K_sp = S'(S' + 0.01) ≈ S' × 0.01 (if S' << 0.01)
- S' = K_sp / 0.01 = 1.8 × 10^-8 mol/L (much less than pure water solubility)

Practice Set

Q1: What happens to solubility when a common ion is added?
- A) Increases
- B) Decreases
- C) No change
- D) Depends on temperature
Q2: For salt M_xX_y, K_sp expression is?
- A) [M^y+][X^x-]
- B) [M^y+]^x[X^x-]^y
- C) (xs)(ys)
- D) (x+y)s
Q3: If Q_sp > K_sp, what occurs?
- A) No precipitation
- B) Precipitation
- C) Equilibrium
- D) Complex formation

Answer Key

Q1: B - Common ion decreases solubility by shifting equilibrium.
Q2: B - K_sp = [M^y+]^x[X^x-]^y.
Q3: B - Ionic product greater than K_sp causes precipitation.

Memory Aids

Common ion effect lowers solubility.
K_sp expression includes ion concentrations raised to their stoichiometric powers.
Q_sp > K_sp means precipitation.

Equilibrium and Expression

For salt M_xX_y: M_xX_y(s) ⇌ xM^y+(aq) + yX^x-(aq)
K_sp = [M^y+]^x [X^x-]^y = (xs)^x (ys)^y = x^x y^y s^x+y
Solubility, s = \( \left( \frac{K_{sp}}{x^x y^y} \right)^{\frac{1}{x+y}} \)
At equilibrium, K_sp = Q_sp. If not, Q_sp indicates reaction direction.

Effect of Common Ion

Adding a common ion decreases solubility (common ion effect).
Example: For AgCl, if NaCl (common Cl^-) is added, solubility decreases.
K_sp = [Ag⁺][Cl^-] = S'(S' + x), where x is added ion concentration.
Increase in solubility on adding inert salt indicates complex ion formation.

Solved Examples

Ex 1: Calculate solubility of AgCl if K_sp = 1.8 × 10^-10 and 0.01 M NaCl is added.
- K_sp = S'(S' + 0.01) ≈ S' × 0.01 (if S' << 0.01)
- S' = K_sp / 0.01 = 1.8 × 10^-8 mol/L (much less than pure water solubility)

Practice Set

Q1: What happens to solubility when a common ion is added?
- A) Increases
- B) Decreases
- C) No change
- D) Depends on temperature
Q2: For salt M_xX_y, K_sp expression is?
- A) [M^y+][X^x-]
- B) [M^y+]^x[X^x-]^y
- C) (xs)(ys)
- D) (x+y)s
Q3: If Q_sp > K_sp, what occurs?
- A) No precipitation
- B) Precipitation
- C) Equilibrium
- D) Complex formation

Answer Key

Q1: B - Common ion decreases solubility by shifting equilibrium.
Q2: B - K_sp = [M^y+]^x[X^x-]^y.
Q3: B - Ionic product greater than K_sp causes precipitation.

Memory Aids

Common ion effect lowers solubility.
K_sp expression includes ion concentrations raised to their stoichiometric powers.
Q_sp > K_sp means precipitation.

Applications of Solubility Product

Key Uses

Calculate solubility of sparingly soluble salts from K_sp.
Predict precipitation by comparing ionic product (Q) with K_sp.
Explain precipitation of common salt by adding HCl gas (increases Cl^- concentration).
Use in qualitative analysis to separate ions based on solubility and common ion effect.

Solved Examples

Ex 1: Predict if precipitation occurs when [Ag⁺] = 1 × 10^-4 M and [Cl^-] = 1 × 10^-3 M, K_sp of AgCl = 1.8 × 10^-10.
- Ionic product = (1 × 10^-4)(1 × 10^-3) = 1 × 10^-7 > K_sp
- Precipitation occurs.

Practice Set

Q1: What indicates precipitation?
- A) Ionic product < K_sp
- B) Ionic product = K_sp
- C) Ionic product > K_sp
- D) No ions present
Q2: What happens when HCl gas is passed through impure NaCl solution?
- A) NaCl dissolves more
- B) Pure NaCl precipitates
- C) No change
- D) NaCl decomposes
Q3: Which application uses solubility product in qualitative analysis?
- A) Identifying gases
- B) Separating basic radicals
- C) Measuring pH
- D) Calculating molarity

Answer Key

Q1: C - Precipitation occurs if ionic product > K_sp.
Q2: B - Increased Cl^- causes pure NaCl to precipitate.
Q3: B - Solubility product helps separate basic radicals.

Memory Aids

Precipitation if Q > K_sp.
HCl increases Cl^-, causing NaCl precipitation.
Qualitative analysis uses solubility and common ion effect.

Key Uses

Calculate solubility of sparingly soluble salts from K_sp.
Predict precipitation by comparing ionic product (Q) with K_sp.
Explain precipitation of common salt by adding HCl gas (increases Cl^- concentration).
Use in qualitative analysis to separate ions based on solubility and common ion effect.

Solved Examples

Ex 1: Predict if precipitation occurs when [Ag⁺] = 1 × 10^-4 M and [Cl^-] = 1 × 10^-3 M, K_sp of AgCl = 1.8 × 10^-10.
- Ionic product = (1 × 10^-4)(1 × 10^-3) = 1 × 10^-7 > K_sp
- Precipitation occurs.

Practice Set

Q1: What indicates precipitation?
- A) Ionic product < K_sp
- B) Ionic product = K_sp
- C) Ionic product > K_sp
- D) No ions present
Q2: What happens when HCl gas is passed through impure NaCl solution?
- A) NaCl dissolves more
- B) Pure NaCl precipitates
- C) No change
- D) NaCl decomposes
Q3: Which application uses solubility product in qualitative analysis?
- A) Identifying gases
- B) Separating basic radicals
- C) Measuring pH
- D) Calculating molarity

Answer Key

Q1: C - Precipitation occurs if ionic product > K_sp.
Q2: B - Increased Cl^- causes pure NaCl to precipitate.
Q3: B - Solubility product helps separate basic radicals.

Memory Aids

Precipitation if Q > K_sp.
HCl increases Cl^-, causing NaCl precipitation.
Qualitative analysis uses solubility and common ion effect.

Solubility Products of Electrolytes

Classification and Formulas

Salts classified by electrolyte formula: binary (1:1), ternary (1:2 or 2:1), quaternary (1:3), 3:2 types.
Examples:
- Binary (1:1): AgCl
- Ternary (1:2): PbCl₂
- Quaternary (1:3): Fe(OH)₃
Solubility (s) related to K_sp differently for each type:
- 1:1 salts: s = √K_sp
- 1:2 salts: s = ∛(K_sp/4)
- 2:1 salts: s = ∛(K_sp/4)
- 1:3 salts: s = ⁴√(K_sp/27)
- 3:2 salts: s = ⁵√(K_sp/108)

Solved Examples

Ex 1: Calculate solubility of PbCl₂ if K_sp = 1.7 × 10^-5.
- For 1:2 salt, s = ∛(K_sp/4) = ∛(1.7 × 10^-5/4) = 0.016 mol/L

Practice Set

Q1: Which salt is binary 1:1?
- A) PbCl₂
- B) AgCl
- C) Fe(OH)₃
- D) Al(OH)₃
Q2: Solubility formula for 1:2 salt is?
- A) s = √K_sp
- B) s = ∛(K_sp/4)
- C) s = K_sp²
- D) s = K_sp/4
Q3: For Fe(OH)₃ (1:3 salt), solubility is?
- A) ⁴√(K_sp/27)
- B) ³√K_sp
- C) √K_sp
- D) ⁵√(K_sp/108)

Answer Key

Q1: B - AgCl is binary 1:1 salt.
Q2: B - For 1:2 salts, s = ∛(K_sp/4).
Q3: A - For 1:3 salts, s = ⁴√(K_sp/27).

Memory Aids

Binary (1:1): s = √K_sp
Ternary (1:2 or 2:1): s = ∛(K_sp/4)
Quaternary (1:3): s = 4th root of (K_sp/27)
3:2 salts: 5th root of (K_sp/108)

Classification and Formulas

Salts classified by electrolyte formula: binary (1:1), ternary (1:2 or 2:1), quaternary (1:3), 3:2 types.
Examples:
- Binary (1:1): AgCl
- Ternary (1:2): PbCl₂
- Quaternary (1:3): Fe(OH)₃
Solubility (s) related to K_sp differently for each type:
- 1:1 salts: s = √K_sp
- 1:2 salts: s = ∛(K_sp/4)
- 2:1 salts: s = ∛(K_sp/4)
- 1:3 salts: s = \( \sqrt[4]{\frac{K_{sp}}{27}} \)
- 3:2 salts: s = \( \sqrt[5]{\frac{K_{sp}}{108}} \)

Solved Examples

Ex 1: Calculate solubility of PbCl₂ if K_sp = 1.7 × 10^-5.
- For 1:2 salt, s = ∛(K_sp/4) = ∛(1.7 × 10^-5/4) = 0.016 mol/L

Practice Set

Q1: Which salt is binary 1:1?
- A) PbCl₂
- B) AgCl
- C) Fe(OH)₃
- D) Al(OH)₃
Q2: Solubility formula for 1:2 salt is?
- A) s = √K_sp
- B) s = ∛(K_sp/4)
- C) s = K_sp²
- D) s = K_sp/4
Q3: For Fe(OH)₃ (1:3 salt), solubility is?
- A) \( \sqrt[4]{\frac{K_{sp}}{27}} \)
- B) \( \sqrt[3]{K_{sp}} \)
- C) \( \sqrt{K_{sp}} \)
- D) \( \sqrt[5]{\frac{K_{sp}}{108}} \)

Answer Key

Q1: B - AgCl is binary 1:1 salt.
Q2: B - For 1:2 salts, s = ∛(K_sp/4).
Q3: A - For 1:3 salts, s = \( \sqrt[4]{\frac{K_{sp}}{27}} \).

Memory Aids

Binary (1:1): s = √K_sp
Ternary (1:2 or 2:1): s = ∛(K_sp/4)
Quaternary (1:3): s = 4th root of (K_sp/27)
3:2 salts: 5th root of (K_sp/108)

NEET Chemistry — ALL CHAPTERS

Some Basic Concepts of Chemistry

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Structure of Atom

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Classification of Elements and Periodicity in Properties

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Chemical Bonding and Molecular Structure

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Thermodynamics

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Equilibrium

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Redox Reactions

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Organic Chemistry - Some Basic Principles and Techniques

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Hydrocarbons

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Solutions

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Electrochemistry

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Chemical Kinetics

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The d- and f- Block Elements

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Coordination Compounds

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Haloalkanes and Haloarenes

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Alcohols, Phenols and Ethers

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Aldehydes, Ketones and Carboxylic Acids

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Amines

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Biomolecules

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The p-Block Elements (Group 13 and 14)

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The p-Block Elements (Group 15 to 18)

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Principles Related to Practical Chemistry

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