What is the Law of Mass Action?

The Law of Mass Action states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the active masses (concentrations) of reactants, each raised to their stoichiometric coefficients. For aA + bB ⇌ cC + dD, the equilibrium constant Kc = [C]^c[D]^d / [A]^a[B]^b

What is the relationship between Kp and Kc?

The relationship between Kp and Kc is given by: Kp = Kc(RT)^Δn, where R is the gas constant (0.0821 L·atm/mol·K), T is temperature in Kelvin, and Δn is the difference between moles of gaseous products and reactants. When Δn = 0, Kp = Kc.

What is Le Chatelier's Principle?

Le Chatelier's Principle states that if a system at equilibrium is subjected to a change (in concentration, pressure, or temperature), the equilibrium will shift in a direction that tends to counteract the change. This principle helps predict how equilibrium position changes when conditions are altered.

What is the difference between Kc and Q (Reaction Quotient)?

Kc is the equilibrium constant calculated at equilibrium, while Q (Reaction Quotient) is calculated using instantaneous concentrations at any point. If Q Kc, reaction proceeds backward; if Q = Kc, system is at equilibrium.

How to calculate pH of a buffer solution?

For acidic buffer: pH = pKa + log([Salt]/[Acid]) - Henderson-Hasselbalch equation. For basic buffer: pOH = pKb + log([Salt]/[Base]). Buffer capacity is maximum when [Salt] = [Acid] or [Salt] = [Base], where pH = pKa or pOH = pKb.

What is Solubility Product (Ksp)?

Solubility Product (Ksp) is the equilibrium constant for a sparingly soluble salt dissolving in water. For salt AxBy: Ksp = [A^y+]^x[B^x-]^y. If ionic product (IP) > Ksp, precipitation occurs; if IP < Ksp, solution is unsaturated; if IP = Ksp, solution is saturated.

How many marks does Chemical Equilibrium carry in JEE?

Chemical Equilibrium carries 10-15% weightage in JEE Main (3-4 questions, 12-16 marks) and 12-18% in JEE Advanced (4-5 questions, 16-20 marks). It's one of the highest-scoring chapters combining both conceptual and numerical problems.

Chemical Equilibrium for JEE 2025-26 | Complete Notes with Kc, Kp, Le Chatelier's Principle & Solved Examples

Q: How many marks does Chemical Equilibrium carry in JEE?

Chemical Equilibrium carries 10-15% weightage in JEE Main (3-4 questions, 12-16 marks) and 12-18% in JEE Advanced (4-5 questions, 16-20 marks). It's one of the highest-scoring chapters combining both conceptual and numerical problems.

Equilibrium Basics

Chemical equilibrium is a dynamic state where the rates of forward and backward reactions become equal, and the concentrations of reactants and products remain constant over time. This is the foundation for understanding all equilibrium phenomena.

1.1 Reversible and Irreversible Reactions

Irreversible Reactions

Reactions that proceed only in one direction until reactants are completely converted to products.

Examples:

Combustion: CH₄ + 2O₂ → CO₂ + 2H₂O
Neutralization: HCl + NaOH → NaCl + H₂O
Precipitation: AgNO₃ + NaCl → AgCl↓ + NaNO₃

Reversible Reactions

Reactions that proceed in both directions simultaneously under same conditions.

Examples:

N₂ + 3H₂ ⇌ 2NH₃
PCl₅ ⇌ PCl₃ + Cl₂
H₂ + I₂ ⇌ 2HI
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

1.2 Characteristics of Chemical Equilibrium

Key Characteristics

Dynamic Nature: Both forward and backward reactions continue to occur at equal rates
Constant Concentrations: Concentrations of reactants and products remain unchanged with time
Closed System: Equilibrium can only be achieved in a closed system (no escape of matter)
Same Temperature: Equilibrium constant value depends on temperature
Approach from Either Side: Same equilibrium can be reached from reactants or products
Catalyst Effect: Catalyst speeds up attainment of equilibrium but doesn't change equilibrium position
Constant ΔG: At equilibrium, ΔG = 0

1.3 Types of Chemical Equilibrium

Type	Description	Example	Kc Expression
Homogeneous	All reactants and products in same phase	H₂(g) + I₂(g) ⇌ 2HI(g)	Kc = [HI]²/[H₂][I₂]
Heterogeneous	Reactants and products in different phases	CaCO₃(s) ⇌ CaO(s) + CO₂(g)	Kc = [CO₂]

⚠️ JEE Important Note

In heterogeneous equilibrium, pure solids and pure liquids are not included in the equilibrium expression because their concentrations (activities) are taken as 1.

Example:

For: 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)

Kp = (P_H₂)⁴ / (P_H₂O)⁴

(Fe and Fe₃O₄ excluded as they are solids)

💡 Quick Memory Trick

"DYNAMIC CLOSED CONSTANT"
Dynamic equilibrium + Closed system + Constant concentrations = Chemical Equilibrium

Equilibrium Constants (Kc, Kp, Kx)

The equilibrium constant is a quantitative measure of the extent to which a reaction proceeds. Understanding the various forms of equilibrium constant and their relationships is crucial for JEE.

2.1 Law of Mass Action

Statement

"At a constant temperature, the rate of a chemical reaction is directly proportional to the product of the active masses (molar concentrations) of the reactants, each raised to a power equal to the stoichiometric coefficient."

For the general reaction: aA + bB ⇌ cC + dD

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

where [A], [B], [C], [D] are molar concentrations at equilibrium

2.2 Types of Equilibrium Constants

Kc (Concentration)

Based on molar concentrations (mol/L)

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

Used for reactions in solution or gases

Kp (Pressure)

Based on partial pressures (atm or bar)

K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}

Used only for gaseous reactions

Kx (Mole Fraction)

Based on mole fractions

K_x = \frac{(x_C)^c(x_D)^d}{(x_A)^a(x_B)^b}

Used for gaseous reactions

2.3 Relationship Between Kp and Kc

The Fundamental Relationship

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

R = Gas constant = 0.0821 L·atm/(mol·K) or 8.314 J/(mol·K)

T = Temperature in Kelvin

Δn = (Sum of stoichiometric coefficients of gaseous products) − (Sum of stoichiometric coefficients of gaseous reactants)

Special Cases:

If Δn = 0: K_p = K_c
If Δn > 0: K_p > K_c
If Δn < 0: K_p < K_c

Relationship Between Kp and Kx

K_p = K_x \cdot (P_{\text{total}})^{\Delta n}

where P_total is the total pressure of the system

📝 Solved Example 1 (JEE Main 2023 Type)

Question: For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), if Kc = 0.5 mol⁻² L² at 400°C, calculate Kp. (R = 0.0821 L·atm/mol·K)

Solution:

Step 1: Calculate Δn

\Delta n = n_{\text{products}} - n_{\text{reactants}}

\Delta n = 2 - (1 + 3) = 2 - 4 = -2

Step 2: Convert temperature to Kelvin

T = 400 + 273 = 673 \text{ K}

Step 3: Apply the relationship

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

K_p = 0.5 \times (0.0821 \times 673)^{-2}

K_p = 0.5 \times (55.25)^{-2}

K_p = 0.5 \times \frac{1}{3052.56}

K_p = 1.64 \times 10^{-4} \text{ atm}^{-2}

2.4 Reaction Quotient (Q)

Definition

The reaction quotient (Q) has the same expression as equilibrium constant but uses instantaneous concentrations (not equilibrium concentrations).

Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \text{ (at any instant)}

Comparing Q with K:

Condition	Direction	Reaction Status
Q < K	Forward →	More products will form
Q > K	Backward ←	More reactants will form
Q = K	No net change	At equilibrium

2.5 Properties of Equilibrium Constant

Important Properties

1. Reverse Reaction

If K is equilibrium constant for A ⇌ B

\text{Then for } B \rightleftharpoons A: K' = \frac{1}{K}

2. Multiplying Equation by n

If equation is multiplied by n:

\[K' = K^n\]

3. Dividing Equation by n

If equation is divided by n:

K' = K^{1/n} = \sqrt[n]{K}

4. Adding Equations

If reaction 1 + reaction 2 = reaction 3:

K_3 = K_1 \times K_2

5. Temperature Dependence

For endothermic: K increases with temperature

For exothermic: K decreases with temperature

📝 Solved Example 2 (JEE Advanced Pattern)

Question: Given:
(i) A ⇌ B, K₁ = 2
(ii) B ⇌ C, K₂ = 3
Find K for: (a) A ⇌ C (b) 2A ⇌ 2C (c) C ⇌ A

Solution:

(a) A ⇌ C:

Adding equations (i) and (ii):

K = K_1 \times K_2 = 2 \times 3 = 6

(b) 2A ⇌ 2C:

This is 2 times the equation in (a):

K = (K_1 \times K_2)^2 = 6^2 = 36

(c) C ⇌ A:

This is reverse of equation in (a):

K = \frac{1}{K_1 \times K_2} = \frac{1}{6}

2.6 Degree of Dissociation (α)

Definition

\alpha = \frac{\text{Number of moles dissociated}}{\text{Initial number of moles}}

α always lies between 0 and 1 (or 0% to 100%)

Important Formulas:

For PCl₅ ⇌ PCl₃ + Cl₂:

K_p = \frac{\alpha^2 P}{1 - \alpha^2}

where P is total pressure

For N₂O₄ ⇌ 2NO₂:

K_p = \frac{4\alpha^2 P}{1 - \alpha^2}

If α << 1 (very small):

\alpha \approx \sqrt{\frac{K_p}{P}} \text{ or } \alpha \approx \sqrt{K_c \cdot V}

Le Chatelier's Principle

Le Chatelier's Principle is one of the most important concepts in equilibrium, allowing us to predict how a system at equilibrium responds to external changes. This is a guaranteed topic in JEE with conceptual and numerical questions.

Statement

"If a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium shifts in a direction that tends to counteract (oppose) the change."

3.1 Effect of Concentration Change

Change	Direction of Shift	Effect on K	Effect on Q
Add reactant	Forward →	No change	Q decreases initially
Remove reactant	Backward ←	No change	Q increases initially
Add product	Backward ←	No change	Q increases initially
Remove product	Forward →	No change	Q decreases initially

3.2 Effect of Pressure Change

General Rule

Increasing pressure shifts equilibrium towards the side with fewer moles of gas.
Decreasing pressure shifts equilibrium towards the side with more moles of gas.

Examples:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Δn = 2 - 4 = -2 (products have fewer moles)

↑ Pressure → Forward direction → More NH₃

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Δn = 2 - 1 = +1 (products have more moles)

↑ Pressure → Backward direction → More PCl₅

H₂(g) + I₂(g) ⇌ 2HI(g)

Δn = 2 - 2 = 0 (equal moles on both sides)

No effect of pressure change

⚠️ Important Notes on Pressure

Addition of inert gas at constant volume: No effect on equilibrium (partial pressures don't change)
Addition of inert gas at constant pressure: Volume increases → equilibrium shifts toward more moles side
Pressure changes don't affect Kp or Kc (only position changes)

3.3 Effect of Temperature Change

Reaction Type	↑ Temperature	↓ Temperature	Effect on K
Exothermic (ΔH < 0)	Backward ←	Forward →	K decreases with ↑T
Endothermic (ΔH > 0)	Forward →	Backward ←	K increases with ↑T

Van't Hoff Equation

\ln\frac{K_2}{K_1} = \frac{\Delta H°}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

or: \[\log\frac{K_2}{K_1} = \frac{\Delta H°}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]

3.4 Effect of Catalyst

Key Points

Does NOT change equilibrium position or K value
Increases rate of both forward and backward reactions equally
Helps system reach equilibrium faster
Lowers activation energy for both reactions

📝 Solved Example 3 (JEE Main 2024 Type)

Question: For the reaction: N₂(g) + O₂(g) ⇌ 2NO(g), ΔH = +180 kJ
Predict the effect on equilibrium when:
(a) Temperature is increased
(b) Pressure is increased
(c) N₂ concentration is increased
(d) Catalyst is added

Solution:

(a) Temperature increased:

Reaction is endothermic (ΔH > 0)

Equilibrium shifts FORWARD → More NO formed, K increases

(b) Pressure increased:

Δn = 2 - (1 + 1) = 0 (equal moles on both sides)

NO EFFECT on equilibrium position

(c) N₂ concentration increased:

Reactant is added

Equilibrium shifts FORWARD → More NO formed, K unchanged

(d) Catalyst added:

NO EFFECT on equilibrium position. Only reaches equilibrium faster.

💡 Summary Table for Le Chatelier's Principle

Factor	Effect on K	Effect on Position
Concentration	No change	Yes, shifts
Pressure (if Δn ≠ 0)	No change	Yes, shifts
Temperature	Changes!	Yes, shifts
Catalyst	No change	No change
Inert gas (const. V)	No change	No change

Ionic Equilibrium

Ionic equilibrium deals with equilibria involving ions in solution. This includes the study of acids, bases, salts, and their ionization behavior. This section carries 40-50% weightage within equilibrium.

4.1 Electrolytes

Strong Electrolytes

Completely dissociate in water (α ≈ 1)

Strong Acids: HCl, HNO₃, H₂SO₄, HClO₄, HBr, HI
Strong Bases: NaOH, KOH, Ca(OH)₂, Ba(OH)₂
Most Salts: NaCl, KNO₃, Na₂SO₄

Weak Electrolytes

Partially dissociate in water (α << 1)

Weak Acids: CH₃COOH, H₂CO₃, H₃PO₄, HF, HCN
Weak Bases: NH₃, NH₄OH, amines
Some Salts: PbCl₂, HgCl₂

4.2 Arrhenius Theory of Acids and Bases

Arrhenius Acid

Produces H⁺ ions in aqueous solution

HCl → H⁺ + Cl⁻

CH₃COOH → H⁺ + CH₃COO⁻

Arrhenius Base

Produces OH⁻ ions in aqueous solution

NaOH → Na⁺ + OH⁻

NH₄OH → NH₄⁺ + OH⁻

Limitation: Only applicable to aqueous solutions

4.3 Brønsted-Lowry Theory

Brønsted Acid

Proton (H⁺) donor

HCl + H₂O → H₃O⁺ + Cl⁻

(HCl donates H⁺)

Brønsted Base

Proton (H⁺) acceptor

NH₃ + H₂O → NH₄⁺ + OH⁻

(NH₃ accepts H⁺)

Conjugate Acid-Base Pairs

A conjugate acid-base pair differs by one proton (H⁺)

Acid + Base ⇌ Conjugate Base + Conjugate Acid

HCl + H₂O ⇌ Cl⁻ + H₃O⁺

HCl/Cl⁻ is conjugate acid-base pair
H₂O/H₃O⁺ is conjugate acid-base pair

Important Relationships:

K_a \times K_b = K_w

pK_a + pK_b = pK_w = 14 \text{ (at 25°C)}

4.4 Lewis Theory

Lewis Acid

Electron pair acceptor

Examples: H⁺, BF₃, AlCl₃, FeCl₃, CO₂, SO₃, metal ions (Ag⁺, Cu²⁺, Zn²⁺)

Lewis Base

Electron pair donor

Examples: OH⁻, NH₃, H₂O, F⁻, Cl⁻, molecules with lone pairs

4.5 Ionization of Water

Ionic Product of Water (Kw)

H₂O + H₂O ⇌ H₃O⁺ + OH⁻

or simply: H₂O ⇌ H⁺ + OH⁻

K_w = [H^+][OH^-] = 10^{-14} \text{ at 25°C}

In pure water at 25°C:

[H⁺] = [OH⁻] = 10⁻⁷ M (neutral)

Note: Kw increases with temperature (ionization is endothermic)

4.6 Ionization Constants (Ka and Kb)

Ka (Acid Dissociation Constant)

For weak acid HA:

HA ⇌ H⁺ + A⁻

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

Higher Ka = Stronger acid
Lower pKa = Stronger acid

Kb (Base Dissociation Constant)

For weak base B:

B + H₂O ⇌ BH⁺ + OH⁻

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

Higher Kb = Stronger base
Lower pKb = Stronger base

Ostwald's Dilution Law

For a weak electrolyte with degree of dissociation α and concentration C:

K = \frac{C\alpha^2}{1-\alpha}

If α << 1 (very small):

K \approx C\alpha^2

\alpha = \sqrt{\frac{K}{C}}

pH and pOH Calculations

pH scale provides a convenient way to express the acidity or basicity of a solution. Understanding pH calculations is essential for JEE as it forms the basis for numerous numerical problems.

5.1 Definitions

pH Scale

pH (Power of Hydrogen)

pH = -\log[H^+]

[H^+] = 10^{-pH}

pOH

pOH = -\log[OH^-]

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

Fundamental Relationships at 25°C:

\[pH + pOH = 14\]

\[pK_w = pH + pOH = 14\]

5.2 pH Scale Interpretation

pH Range	Nature	[H⁺] vs [OH⁻]	Examples
pH < 7	Acidic	[H⁺] > [OH⁻]	Lemon (pH 2), Vinegar (pH 3), Coffee (pH 5)
pH = 7	Neutral	[H⁺] = [OH⁻]	Pure water at 25°C
pH > 7	Basic	[H⁺] < [OH⁻]	Soap (pH 9), Milk of Magnesia (pH 10)

5.3 pH Calculation Formulas

Strong Acid (HA)

Complete dissociation: HA → H⁺ + A⁻

[H^+] = C_{\text{acid}}

pH = -\log C_{\text{acid}}

For diprotic acid H₂A: [H⁺] = 2C

Strong Base (BOH)

Complete dissociation: BOH → B⁺ + OH⁻

[OH^-] = C_{\text{base}}

pOH = -\log C_{\text{base}}

For Ca(OH)₂: [OH⁻] = 2C

Weak Acid (HA)

Partial dissociation: HA ⇌ H⁺ + A⁻

[H^+] = \sqrt{K_a \cdot C}

pH = \frac{1}{2}(pK_a - \log C)

Valid when α << 1

Weak Base (B)

B + H₂O ⇌ BH⁺ + OH⁻

[OH^-] = \sqrt{K_b \cdot C}

pOH = \frac{1}{2}(pK_b - \log C)

Valid when α << 1

📝 Solved Example 4 (JEE Main 2023 Type)

Question: Calculate the pH of:
(a) 0.01 M HCl
(b) 0.1 M CH₃COOH (Ka = 1.8 × 10⁻⁵)
(c) 0.001 M NaOH

Solution:

(a) 0.01 M HCl (strong acid):

[H^+] = 0.01 = 10^{-2} M

pH = -\log(10^{-2}) = 2

(b) 0.1 M CH₃COOH (weak acid):

[H^+] = \sqrt{K_a \cdot C} = \sqrt{1.8 \times 10^{-5} \times 0.1}

[H^+] = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} M

pH = -\log(1.34 \times 10^{-3}) = 2.87

(c) 0.001 M NaOH (strong base):

[OH^-] = 0.001 = 10^{-3} M

pOH = -\log(10^{-3}) = 3

\[pH = 14 - pOH = 14 - 3 = 11\]

⚠️ Very Dilute Solutions

For very dilute strong acids (C < 10⁻⁶ M), contribution of H⁺ from water must be considered.

Example: 10⁻⁸ M HCl

If we use [H⁺] = 10⁻⁸, then pH = 8 (basic!) - This is wrong!

Correct approach:

[H^+]_{\text{total}} = [H^+]_{\text{acid}} + [H^+]_{\text{water}}

Solving: [H⁺] ≈ 1.05 × 10⁻⁷ M, so pH ≈ 6.98 (slightly acidic)

Salt Hydrolysis

Salt hydrolysis is the reaction of salt with water to produce an acidic or basic solution. Understanding hydrolysis helps predict the pH of salt solutions.

6.1 Types of Salts and Their Hydrolysis

Salt Type	Formed From	Hydrolysis	Solution pH	Example
Strong acid + Strong base	HCl + NaOH	No hydrolysis	pH = 7 (Neutral)	NaCl, KNO₃
Strong acid + Weak base	HCl + NH₄OH	Cation hydrolysis	pH < 7 (Acidic)	NH₄Cl, FeCl₃
Weak acid + Strong base	CH₃COOH + NaOH	Anion hydrolysis	pH > 7 (Basic)	CH₃COONa, Na₂CO₃
Weak acid + Weak base	CH₃COOH + NH₄OH	Both hydrolyze	Depends on Ka vs Kb	CH₃COONH₄

6.2 pH Formulas for Hydrolysis

Salt of Weak Acid + Strong Base

A⁻ + H₂O ⇌ HA + OH⁻ (basic)

K_h = \frac{K_w}{K_a}

pH = 7 + \frac{1}{2}(pK_a + \log C)

Example: CH₃COONa, Na₂CO₃

Salt of Strong Acid + Weak Base

BH⁺ + H₂O ⇌ B + H₃O⁺ (acidic)

K_h = \frac{K_w}{K_b}

pH = 7 - \frac{1}{2}(pK_b + \log C)

Example: NH₄Cl, FeCl₃

Salt of Weak Acid + Weak Base

Both cation and anion hydrolyze

pH = 7 + \frac{1}{2}(pK_a - pK_b)

Nature of Solution:

If Ka > Kb → Acidic (pH < 7)
If Ka < Kb → Basic (pH > 7)
If Ka = Kb → Neutral (pH = 7)

Note: pH is independent of concentration for this type of salt!

📝 Solved Example 5 (JEE Main Pattern)

Question: Calculate the pH of 0.1 M CH₃COONa solution. (Ka of CH₃COOH = 1.8 × 10⁻⁵)

Solution:

This is a salt of weak acid (CH₃COOH) + strong base (NaOH)

Hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

Method 1: Using formula

pH = 7 + \frac{1}{2}(pK_a + \log C)

pK_a = -\log(1.8 \times 10^{-5}) = 4.74

pH = 7 + \frac{1}{2}(4.74 + \log 0.1)

pH = 7 + \frac{1}{2}(4.74 - 1)

pH = 7 + \frac{1}{2}(3.74) = 7 + 1.87

\[pH = 8.87\]

Buffer Solutions

Buffer solutions resist change in pH upon addition of small amounts of acid or base. This is one of the most important topics in ionic equilibrium with direct applications in biology and chemistry.

7.1 Types of Buffer Solutions

Acidic Buffer

Weak acid + Salt with strong base

Example: CH₃COOH + CH₃COONa

pH < 7

Common Acidic Buffers:

Acetic acid + Sodium acetate
H₂CO₃ + NaHCO₃
H₂PO₄⁻ + HPO₄²⁻

Basic Buffer

Weak base + Salt with strong acid

Example: NH₄OH + NH₄Cl

pH > 7

Common Basic Buffers:

Ammonia + Ammonium chloride
Glycine + Glycinium ion

7.2 Henderson-Hasselbalch Equation

The Most Important Buffer Formula

For Acidic Buffer:

pH = pK_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}

or: \[pH = pK_a + \log\frac{[A^-]}{[HA]}\]

For Basic Buffer:

pOH = pK_b + \log\frac{[\text{Salt}]}{[\text{Base}]}

Then: pH = 14 - pOH

7.3 Buffer Capacity and Range

Buffer Capacity

Ability of buffer to resist pH change

Maximum when [Salt] = [Acid]
At this point: pH = pKa
Increases with concentration

Buffer Range

Effective pH range of buffer

\text{Buffer Range} = pK_a \pm 1

Buffer works best in this range

📝 Solved Example 6 (JEE Main 2024 Type)

Question: Calculate the pH of a buffer containing 0.2 M CH₃COOH and 0.1 M CH₃COONa. (pKa = 4.74)

Solution:

Using Henderson-Hasselbalch equation:

pH = pK_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}

pH = 4.74 + \log\frac{0.1}{0.2}

pH = 4.74 + \log(0.5)

\[pH = 4.74 + (-0.301)\]

\[pH = 4.44\]

💡 Buffer Action Mechanism

When acid (H⁺) is added:

CH₃COO⁻ + H⁺ → CH₃COOH

Salt reacts with added H⁺, minimizing pH change

When base (OH⁻) is added:

CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

Acid reacts with added OH⁻, minimizing pH change

Solubility Product (Ksp)

Solubility product is the equilibrium constant for a sparingly soluble salt dissolving in water. This concept is crucial for predicting precipitation and understanding qualitative analysis.

8.1 Definition and Expression

Solubility Product (Ksp)

For sparingly soluble salt A_xB_y:

A_xB_y(s) ⇌ xA^y+(aq) + yB^x-(aq)

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

Common Examples:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) → Ksp = [Ag⁺][Cl⁻]

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq) → Ksp = [Pb²⁺][I⁻]²

Ca₃(PO₄)₂(s) ⇌ 3Ca²⁺(aq) + 2PO₄³⁻(aq) → Ksp = [Ca²⁺]³[PO₄³⁻]²

8.2 Relationship between Ksp and Solubility

Solubility (S) and Ksp Relationship

If solubility of A_xB_y is S mol/L, then:

[A^y+] = xS and [B^x-] = yS

K_{sp} = (xS)^x \cdot (yS)^y = x^x \cdot y^y \cdot S^{(x+y)}

Salt Type	Example	Ksp Expression	S in terms of Ksp
AB (1:1)	AgCl, BaSO₄	Ksp = S²	S = √Ksp
AB₂ (1:2)	PbI₂, CaF₂	Ksp = 4S³	S = ∛(Ksp/4)
A₂B (2:1)	Ag₂CrO₄	Ksp = 4S³	S = ∛(Ksp/4)
AB₃ (1:3)	AlF₃, Fe(OH)₃	Ksp = 27S⁴	S = ⁴√(Ksp/27)
A₃B₂ (3:2)	Ca₃(PO₄)₂	Ksp = 108S⁵	S = ⁵√(Ksp/108)

⚠️ Important JEE Note

Comparing Solubility of Different Salts:

For salts with same stoichiometry (e.g., AgCl, AgBr): Compare Ksp directly. Higher Ksp = Higher solubility
For salts with different stoichiometry (e.g., AgCl vs Ag₂CrO₄): Calculate S for each salt and then compare

8.3 Ionic Product (IP) vs Ksp

Predicting Precipitation

Ionic Product (IP) has same expression as Ksp but uses instantaneous concentrations:

IP = [A^{y+}]^x[B^{x-}]^y \text{ (at any instant)}

Condition	Result	Solution Status
IP < Ksp	No precipitation	Unsaturated solution (more salt can dissolve)
IP = Ksp	At equilibrium	Saturated solution
IP > Ksp	Precipitation occurs	Supersaturated → salt precipitates until IP = Ksp

📝 Solved Example 7 (JEE Main 2024 Type)

Question: Calculate the solubility of AgCl in water at 25°C. Ksp of AgCl = 1.8 × 10⁻¹⁰

Solution:

AgCl is an AB type salt:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Let solubility = S mol/L

At equilibrium: [Ag⁺] = S, [Cl⁻] = S

K_{sp} = [Ag^+][Cl^-] = S \times S = S^2

S^2 = 1.8 \times 10^{-10}

S = \sqrt{1.8 \times 10^{-10}}

S = 1.34 \times 10^{-5} \text{ mol/L}

📝 Solved Example 8 (JEE Advanced Pattern)

Question: Equal volumes of 2 × 10⁻⁴ M AgNO₃ and 2 × 10⁻⁴ M NaCl are mixed. Will AgCl precipitate? (Ksp of AgCl = 1.8 × 10⁻¹⁰)

Solution:

Step 1: Calculate concentrations after mixing

When equal volumes are mixed, concentrations are halved:

[Ag^+] = \frac{2 \times 10^{-4}}{2} = 10^{-4} M

[Cl^-] = \frac{2 \times 10^{-4}}{2} = 10^{-4} M

Step 2: Calculate Ionic Product (IP)

IP = [Ag^+][Cl^-] = 10^{-4} \times 10^{-4} = 10^{-8}

Step 3: Compare IP with Ksp

IP = 10^{-8}

K_{sp} = 1.8 \times 10^{-10}

IP (10^{-8}) > K_{sp} (1.8 \times 10^{-10})

Since IP > Ksp, AgCl WILL precipitate!

Common Ion Effect

The common ion effect describes the shift in equilibrium when a common ion is added to a solution. This is a direct application of Le Chatelier's principle and is important for understanding solubility and buffer behavior.

9.1 Definition

Common Ion Effect

The suppression of ionization of a weak electrolyte by the addition of a strong electrolyte containing a common ion.

Example:

CH₃COOH ⇌ CH₃COO⁻ + H⁺

When CH₃COONa (which gives CH₃COO⁻) is added:

→ [CH₃COO⁻] increases

→ Equilibrium shifts BACKWARD (Le Chatelier)

→ Ionization of CH₃COOH decreases

→ [H⁺] decreases, pH increases

9.2 Effect on Solubility

Solubility in Presence of Common Ion

The solubility of a sparingly soluble salt decreases when a common ion is added.

Example: Solubility of AgCl in NaCl solution

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

When NaCl is added → [Cl⁻] increases

By Le Chatelier → Equilibrium shifts backward

→ Solubility of AgCl decreases

Mathematical Treatment:

If AgCl is dissolved in x M NaCl solution:

K_{sp} = [Ag^+][Cl^-] = S(S + x) \approx S \cdot x

(assuming x >> S)

S = \frac{K_{sp}}{x}

📝 Solved Example 9 (JEE Main 2023 Type)

Question: Calculate the solubility of AgCl in 0.1 M NaCl solution. (Ksp of AgCl = 1.8 × 10⁻¹⁰)

Solution:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Let solubility of AgCl = S mol/L

At equilibrium:

[Ag⁺] = S

[Cl⁻] = S + 0.1 ≈ 0.1 M (since S << 0.1)

K_{sp} = [Ag^+][Cl^-]

1.8 \times 10^{-10} = S \times 0.1

S = 1.8 \times 10^{-9} \text{ mol/L}

Comparison:

Solubility in pure water: 1.34 × 10⁻⁵ M

Solubility in 0.1 M NaCl: 1.8 × 10⁻⁹ M

Solubility decreased by ~7500 times due to common ion effect!

9.3 Applications of Common Ion Effect

1. Salting Out of Soap

Adding NaCl to soap solution precipitates soap (sodium salt of fatty acid) due to common Na⁺ ion.

2. Qualitative Analysis

Selective precipitation of metal ions by controlling [S²⁻] in presence of HCl.

3. Buffer Solutions

Common ion effect suppresses ionization of weak acid/base, maintaining pH.

4. Hardness Removal

Adding Na₂CO₃ precipitates CaCO₃ from hard water due to CO₃²⁻ common ion.

💡 Summary: Common Ion Effect

Adding common ion → Equilibrium shifts backward (Le Chatelier)
For weak electrolyte → Degree of ionization (α) decreases
For sparingly soluble salt → Solubility decreases
Ksp and Ka values remain unchanged (only equilibrium position changes)

📋 Formula Quick Reference Sheet

Equilibrium Constants

Kp = Kc(RT)^Δn

Kp = Kx(P)^Δn

K_reverse = 1/K

K_new = K^n (if eqn × n)

K_total = K₁ × K₂ (if eqn added)

Ionic Equilibrium

Ka × Kb = Kw = 10⁻¹⁴

pKa + pKb = pKw = 14

pH + pOH = 14

α = √(K/C) (if α << 1)

pH Calculations

Strong acid: pH = -log C

Weak acid: pH = ½(pKa - log C)

Salt (WA+SB): pH = 7 + ½(pKa + log C)

Salt (SA+WB): pH = 7 - ½(pKb + log C)

Salt (WA+WB): pH = 7 + ½(pKa - pKb)

Buffer & Solubility

pH = pKa + log([Salt]/[Acid])

Buffer range = pKa ± 1

AB: Ksp = S²

AB₂: Ksp = 4S³

A₂B₃: Ksp = 108S⁵

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Kc, Kp Calculations: 25%
✓ pH & Ionic Equilibrium: 30%
✓ Buffer Solutions: 15%
✓ Solubility Product: 15%
✓ Le Chatelier's Principle: 10%
✓ Common Ion Effect: 5%

JEE Advanced (Last 5 Years)

✓ Multi-concept equilibrium problems: 35%
✓ Complex ionic equilibrium: 25%
✓ Simultaneous equilibria: 15%
✓ Titration & pH curves: 15%
✓ Precipitation & Ksp: 10%

Weightage Analysis

JEE Main: 12-16 marks (3-4 questions)

JEE Advanced: 16-24 marks (4-6 questions)

Difficulty Level: Medium to Hard

Time Required: 4-5 hours practice

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Write equilibrium constant expressions for: (a) N₂ + 3H₂ ⇌ 2NH₃ (b) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
For the reaction H₂ + I₂ ⇌ 2HI, Kc = 50 at 700 K. Find Kp.
Calculate pH of: (a) 0.01 M HNO₃ (b) 0.1 M NaOH (c) 0.001 M Ca(OH)₂
Define: (a) Lewis acid (b) Buffer capacity (c) Solubility product
Predict the effect of increasing pressure on: PCl₅ ⇌ PCl₃ + Cl₂
Calculate pH of 0.1 M HCN solution (Ka = 6 × 10⁻¹⁰)
What is the pH of a buffer containing equal concentrations of CH₃COOH and CH₃COONa? (pKa = 4.74)
Calculate solubility of AgBr in water (Ksp = 5 × 10⁻¹³)

Level 2: Intermediate (JEE Main/Advanced)

For N₂O₄ ⇌ 2NO₂, Kp = 0.14 at 25°C. Calculate degree of dissociation at 1 atm pressure.
Calculate pH of 0.1 M NH₄Cl solution. (Kb of NH₃ = 1.8 × 10⁻⁵)
A buffer is prepared by mixing 100 mL of 0.2 M CH₃COOH and 100 mL of 0.1 M NaOH. Calculate pH. (pKa = 4.74)
Calculate solubility of PbI₂ in (a) pure water (b) 0.1 M KI solution. (Ksp = 1.4 × 10⁻⁸)
If Q = 10⁻⁸ and Ksp = 10⁻¹⁰ for AgCl, will precipitation occur?
At 500 K, Kc for H₂ + I₂ ⇌ 2HI is 45. Starting with 1 mol each of H₂ and I₂ in 1 L, find equilibrium concentrations.
Calculate the pH when 10 mL of 0.1 M HCl is added to 100 mL of buffer (0.1 M CH₃COOH + 0.1 M CH₃COONa)
The pH of blood is 7.4. Calculate ratio of [HCO₃⁻]/[H₂CO₃] (pKa = 6.1)

Level 3: Advanced (JEE Advanced/Olympiad)

For the equilibrium: 2SO₂ + O₂ ⇌ 2SO₃, starting with 2 mol SO₂ and 1 mol O₂ in 1L vessel, the equilibrium mixture contains 0.6 mol O₂. Calculate Kc.
Calculate pH of 10⁻⁸ M HCl solution (consider water ionization).
A solution contains Mg²⁺ and Ba²⁺ ions (0.01 M each). Calculate the pH at which Mg(OH)₂ starts precipitating but Ba(OH)₂ does not. (Ksp Mg(OH)₂ = 1.8 × 10⁻¹¹, Ksp Ba(OH)₂ = 5 × 10⁻³)
Calculate the pH of 0.1 M H₂S solution. (Ka1 = 10⁻⁷, Ka2 = 10⁻¹⁴)
For CH₃COONH₄ solution, prove that pH = ½(pKa + pKw - pKb) and calculate pH.
Equal volumes of 0.02 M AgNO₃ and 0.01 M K₂CrO₄ are mixed. What is [Ag⁺] in solution? (Ksp Ag₂CrO₄ = 1.1 × 10⁻¹²)
Calculate concentration of S²⁻ required to precipitate CuS but not ZnS from a solution containing 0.01 M each of Cu²⁺ and Zn²⁺. (Ksp CuS = 6 × 10⁻³⁷, Ksp ZnS = 2 × 10⁻²⁵)
The solubility of AgCl in 0.2 M NH₃ is 1.6 × 10⁻³ M. Calculate Kf of [Ag(NH₃)₂]⁺. (Ksp AgCl = 1.8 × 10⁻¹⁰)

Get Solutions PDF with Step-by-Step Working

Year	JEE Main	JEE Advanced	Most Asked Topics
2024	4 Questions (16 marks)	5 Questions (18 marks)	pH, Buffer, Ksp, Le Chatelier
2023	3 Questions (12 marks)	4 Questions (16 marks)	Ionic equilibrium, Hydrolysis, Kp-Kc
2022	4 Questions (16 marks)	5 Questions (20 marks)	Buffer capacity, Common ion, α calculation
2021	3 Questions (12 marks)	4 Questions (14 marks)	Degree of dissociation, Precipitation

📑 Quick Navigation - Chemical Equilibrium

Chemical Equilibrium JEE Main & Advanced 2025-26

Equilibrium Basics

1.1 Reversible and Irreversible Reactions

Irreversible Reactions

Reversible Reactions

1.2 Characteristics of Chemical Equilibrium

Key Characteristics

1.3 Types of Chemical Equilibrium

⚠️ JEE Important Note

💡 Quick Memory Trick

Equilibrium Constants (Kc, Kp, Kx)

2.1 Law of Mass Action

Statement

2.2 Types of Equilibrium Constants

Kc (Concentration)

Kp (Pressure)

Kx (Mole Fraction)

2.3 Relationship Between Kp and Kc

The Fundamental Relationship

Relationship Between Kp and Kx

📝 Solved Example 1 (JEE Main 2023 Type)

2.4 Reaction Quotient (Q)

Definition

2.5 Properties of Equilibrium Constant

Important Properties

📝 Solved Example 2 (JEE Advanced Pattern)

2.6 Degree of Dissociation (α)

Definition

Le Chatelier's Principle

Statement

3.1 Effect of Concentration Change

3.2 Effect of Pressure Change

General Rule

⚠️ Important Notes on Pressure

3.3 Effect of Temperature Change

Van't Hoff Equation

3.4 Effect of Catalyst

Key Points

📝 Solved Example 3 (JEE Main 2024 Type)

💡 Summary Table for Le Chatelier's Principle

Ionic Equilibrium

4.1 Electrolytes

Strong Electrolytes

Weak Electrolytes

4.2 Arrhenius Theory of Acids and Bases

Arrhenius Acid

Arrhenius Base

4.3 Brønsted-Lowry Theory

Brønsted Acid

Brønsted Base

Conjugate Acid-Base Pairs

4.4 Lewis Theory

Lewis Acid

Lewis Base

4.5 Ionization of Water

Ionic Product of Water (Kw)

4.6 Ionization Constants (Ka and Kb)

Ka (Acid Dissociation Constant)

Kb (Base Dissociation Constant)

Ostwald's Dilution Law

pH and pOH Calculations

5.1 Definitions

pH Scale

pH (Power of Hydrogen)

pOH

5.2 pH Scale Interpretation

5.3 pH Calculation Formulas

Strong Acid (HA)

Strong Base (BOH)

Weak Acid (HA)

Weak Base (B)

📝 Solved Example 4 (JEE Main 2023 Type)

⚠️ Very Dilute Solutions

Salt Hydrolysis

6.1 Types of Salts and Their Hydrolysis

6.2 pH Formulas for Hydrolysis

Salt of Weak Acid + Strong Base

Salt of Strong Acid + Weak Base

Salt of Weak Acid + Weak Base