What is the difference between Arrhenius, Brønsted-Lowry, and Lewis theories of acids and bases?

Arrhenius Theory: Acid produces H⁺ ions, base produces OH⁻ ions in water. Brønsted-Lowry Theory: Acid is a proton (H⁺) donor, base is a proton acceptor. Lewis Theory: Acid is an electron pair acceptor, base is an electron pair donor. Lewis theory is the most general and includes reactions without protons.

What is the relationship between Ka and Kb for a conjugate acid-base pair?

For a conjugate acid-base pair, Ka × Kb = Kw = 10⁻¹⁴ at 25°C. Also, pKa + pKb = pKw = 14. A strong acid has a weak conjugate base (high Ka, low Kb) and vice versa.

How do you calculate the pH of a weak acid solution?

For a weak acid HA with concentration C and ionization constant Ka: [H⁺] = √(Ka × C) when α << 1. Therefore, pH = ½(pKa - log C). For more accurate calculation without approximation, solve the quadratic equation from Ka expression.

What is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation for acidic buffer is: pH = pKa + log([Salt]/[Acid]) or pH = pKa + log([A⁻]/[HA]). For basic buffer: pOH = pKb + log([Salt]/[Base]). This equation is used to calculate pH of buffer solutions.

What are the different types of salt hydrolysis?

Four types: (1) Strong acid + Strong base salt: No hydrolysis, pH = 7. (2) Weak acid + Strong base salt: Anion hydrolysis, pH > 7 (basic). (3) Strong acid + Weak base salt: Cation hydrolysis, pH < 7 (acidic). (4) Weak acid + Weak base salt: Both hydrolyze, pH depends on Ka vs Kb.

How do you predict precipitation using ionic product and Ksp?

Compare Ionic Product (IP) with Solubility Product (Ksp): If IP Ksp: Supersaturated, precipitation occurs until IP = Ksp.

What is the common ion effect?

Common ion effect is the suppression of ionization of a weak electrolyte by adding a strong electrolyte containing a common ion. It shifts the equilibrium backward (Le Chatelier's principle). Example: Adding NaCl to AgCl solution reduces AgCl solubility due to common Cl⁻ ion.

How many marks does Ionic Equilibrium carry in JEE?

Ionic Equilibrium carries 8-12% weightage in JEE Main (2-3 questions, 8-12 marks) and 10-15% in JEE Advanced (3-5 questions, 12-20 marks). It's one of the highest-scoring topics in Physical Chemistry with direct formula-based numericals.

Ionic Equilibrium for JEE 2025-26 | Complete Notes on pH, Acids, Bases, Buffers, Ksp & Hydrolysis

Q: What is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation for acidic buffer is: pH = pKa + log([Salt]/[Acid]) or pH = pKa + log([A⁻]/[HA]). For basic buffer: pOH = pKb + log([Salt]/[Base]). This equation is used to calculate pH of buffer solutions.

Q: What are the different types of salt hydrolysis?

Four types: (1) Strong acid + Strong base salt: No hydrolysis, pH = 7. (2) Weak acid + Strong base salt: Anion hydrolysis, pH > 7 (basic). (3) Strong acid + Weak base salt: Cation hydrolysis, pH < 7 (acidic). (4) Weak acid + Weak base salt: Both hydrolyze, pH depends on Ka vs Kb.

Q: How do you predict precipitation using ionic product and Ksp?

Compare Ionic Product (IP) with Solubility Product (Ksp): If IP Ksp: Supersaturated, precipitation occurs until IP = Ksp.

Q: What is the common ion effect?

Common ion effect is the suppression of ionization of a weak electrolyte by adding a strong electrolyte containing a common ion. It shifts the equilibrium backward (Le Chatelier's principle). Example: Adding NaCl to AgCl solution reduces AgCl solubility due to common Cl⁻ ion.

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

Ionic Equilibrium carries 8-12% weightage in JEE Main (2-3 questions, 8-12 marks) and 10-15% in JEE Advanced (3-5 questions, 12-20 marks). It's one of the highest-scoring topics in Physical Chemistry with direct formula-based numericals.

Electrolytes and Ionization

Electrolytes are substances that dissociate into ions when dissolved in water or in molten state and conduct electricity. Understanding the behavior of electrolytes is fundamental to ionic equilibrium.

1.1 Classification of Electrolytes

Strong Electrolytes

Completely dissociate into ions in solution (α ≈ 1 or 100%)

Examples:

Strong Acids: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄
Strong Bases: NaOH, KOH, Ca(OH)₂, Ba(OH)₂
Most Salts: NaCl, KNO₃, Na₂SO₄, CuSO₄

HCl → H⁺ + Cl⁻ (100% dissociation)

Weak Electrolytes

Partially dissociate into ions in solution (α << 1)

Examples:

Weak Acids: CH₃COOH, H₂CO₃, HCN, HF, H₂S, H₃PO₄
Weak Bases: NH₃, NH₄OH, amines (CH₃NH₂)
Few Salts: HgCl₂, PbCl₂, CdBr₂

CH₃COOH ⇌ H⁺ + CH₃COO⁻ (partial)

1.2 Degree of Ionization (α)

Definition

\alpha = \frac{\text{Number of moles ionized}}{\text{Total number of moles dissolved}}

• α = 1 for strong electrolytes (complete ionization)

• α << 1 for weak electrolytes (partial ionization)

• α is usually expressed as fraction (0 to 1) or percentage (0% to 100%)

1.3 Ostwald's Dilution Law

For Weak Electrolytes

For weak acid HA ⇌ H⁺ + A⁻ with initial concentration C and degree of ionization α:

Species	Initial	At Equilibrium
HA	C	C(1-α)
H⁺	0	Cα
A⁻	0	Cα

K_a = \frac{[H^+][A^-]}{[HA]} = \frac{C\alpha \cdot C\alpha}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}

If α << 1 (weak electrolyte approximation):

K_a \approx C\alpha^2

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

Key insight: As dilution increases (C decreases), α increases. "Dilution increases ionization"

📝 Solved Example 1 (JEE Main Pattern)

Question: Calculate the degree of ionization of 0.1 M acetic acid solution. Ka = 1.8 × 10⁻⁵

Solution:

Using Ostwald's dilution law (assuming α << 1):

\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.8 \times 10^{-5}}{0.1}}

\alpha = \sqrt{1.8 \times 10^{-4}} = 1.34 \times 10^{-2}

\alpha = 0.0134 \text{ or } 1.34\%

Verification: α = 0.0134 << 1, so our approximation is valid!

💡 When to NOT Use Approximation

If α > 0.1 (10%), the approximation α << 1 fails. In such cases, solve the quadratic equation:

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

Rearranging: Cα² + Kaα - Ka = 0

Theories of Acids and Bases

Understanding the different theories of acids and bases is crucial for JEE. Each theory has its own scope and applications. Let's study them in detail.

2.1 Arrhenius Theory (1887)

Definition

Arrhenius Acid

A substance that produces H⁺ (or H₃O⁺) ions in aqueous solution

HCl(aq) → H⁺(aq) + Cl⁻(aq)

H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq)

CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq)

Arrhenius Base

A substance that produces OH⁻ ions in aqueous solution

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

Ca(OH)₂(aq) → Ca²⁺(aq) + 2OH⁻(aq)

NH₄OH(aq) ⇌ NH₄⁺(aq) + OH⁻(aq)

Neutralization Reaction:

H⁺ + OH⁻ → H₂O

ΔH = -57.1 kJ/mol (for strong acid + strong base)

Limitations:

Only applicable to aqueous solutions
Cannot explain acidic nature of CO₂, SO₂, etc.
Cannot explain basic nature of NH₃ (no OH⁻ in formula)
Cannot explain acid-base reactions in non-aqueous solvents

2.2 Brønsted-Lowry Theory (1923)

Definition

Brønsted Acid

Proton (H⁺) donor

Must have at least one hydrogen that can be donated

Brønsted Base

Proton (H⁺) acceptor

Must have a lone pair to accept proton

Example Reactions:

HCl + H₂O → H₃O⁺ + Cl⁻

HCl (acid - donates H⁺) + H₂O (base - accepts H⁺)

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

NH₃ (base - accepts H⁺) + H₂O (acid - donates H⁺)

HCl + NH₃ → NH₄⁺ + Cl⁻

Reaction in gas phase (no water needed!)

2.3 Conjugate Acid-Base Pairs

Definition

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

General Equation:

Acid₁ + Base₂ ⇌ Conjugate Base₁ + Conjugate Acid₂

Acid	Conjugate Base	Relationship
HCl	Cl⁻	Strong acid → Weak conjugate base
CH₃COOH	CH₃COO⁻	Weak acid → Stronger conjugate base
H₂O	OH⁻	Very weak acid → Strong conjugate base
NH₄⁺	NH₃	Weak acid → Weak conjugate base
H₃O⁺	H₂O	Strong acid → Very weak base

Important Relationships:

K_a \times K_b = K_w = 10^{-14} \text{ (at 25°C)}

\[pK_a + pK_b = pK_w = 14\]

Rule: Stronger the acid, weaker its conjugate base (and vice versa)

2.4 Lewis Theory (1923)

Definition (Most General)

Lewis Acid

Electron pair acceptor

Must have vacant orbital or ability to expand octet

Examples:

All cations: H⁺, Ag⁺, Cu²⁺, Fe³⁺, Zn²⁺
Electron deficient: BF₃, AlCl₃, BCl₃
Molecules with vacant d-orbitals: SiF₄, SnCl₄, SF₄
Central atom with multiple bonds: CO₂, SO₂, SO₃

Lewis Base

Electron pair donor

Must have lone pair of electrons

Examples:

All anions: OH⁻, Cl⁻, F⁻, CN⁻, S²⁻
Molecules with lone pairs: NH₃, H₂O, ROH, ROR
Amines: CH₃NH₂, (CH₃)₃N
Carbanions, alkenes, alkynes (π electrons)

Lewis Acid-Base Reactions (Coordinate Bond Formation):

BF₃ + :NH₃ → F₃B←NH₃ (Adduct formation)

AlCl₃ + :Cl⁻ → [AlCl₄]⁻

Ag⁺ + 2:NH₃ → [Ag(NH₃)₂]⁺

H⁺ + :OH⁻ → H₂O

2.5 Comparison of Theories

Aspect	Arrhenius	Brønsted-Lowry	Lewis
Acid Definition	H⁺ producer	H⁺ donor	e⁻ pair acceptor
Base Definition	OH⁻ producer	H⁺ acceptor	e⁻ pair donor
Scope	Most limited (aqueous only)	Broader (any solvent)	Most general
Water Role	Solvent only	Can be acid or base	Lewis base
BF₃	Not acid	Not acid	Acid

💡 Amphoteric/Amphiprotic Substances

Substances that can act as both acid and base are called amphoteric (Lewis) or amphiprotic (Brønsted).

Examples:

H₂O: Can donate H⁺ (acid) or accept H⁺ (base)
HSO₄⁻: Conjugate base of H₂SO₄, conjugate acid of SO₄²⁻
HCO₃⁻: Conjugate base of H₂CO₃, conjugate acid of CO₃²⁻
HPO₄²⁻, H₂PO₄⁻: Intermediate species of polyprotic acid
Al(OH)₃, Zn(OH)₂: Amphoteric hydroxides

Ionization of Water, pH and pOH

The self-ionization of water and the pH scale are fundamental concepts in ionic equilibrium. Understanding these concepts is essential for all pH calculations.

3.1 Self-Ionization of Water

Auto-protolysis of Water

Water undergoes self-ionization to a very small extent:

H₂O + H₂O ⇌ H₃O⁺ + OH⁻

or simply: H₂O ⇌ H⁺ + OH⁻

Ionic Product of Water (Kw):

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

In pure water at 25°C:

[H⁺] = [OH⁻] = 10⁻⁷ M (neutral)

Temperature dependence: Kw increases with temperature because ionization is endothermic (ΔH > 0)

Temperature (°C)	Kw	[H⁺] in pure water	pH of pure water
0	0.114 × 10⁻¹⁴	0.34 × 10⁻⁷ M	7.47
25	1.0 × 10⁻¹⁴	1.0 × 10⁻⁷ M	7.00
60	9.6 × 10⁻¹⁴	3.1 × 10⁻⁷ M	6.51
100	51 × 10⁻¹⁴	7.1 × 10⁻⁷ M	6.14

⚠️ Important Note for JEE

At temperatures other than 25°C, neutral solution has pH ≠ 7! A solution is neutral when [H⁺] = [OH⁻], regardless of the actual pH value. At 60°C, neutral pH = 6.51, not 7!

3.2 pH and pOH Scale

Definitions

pH (Potential of Hydrogen)

pH = -\log[H^+] = -\log[H_3O^+]

Inversely:

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

pOH

pOH = -\log[OH^-]

Inversely:

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

Fundamental Relationships at 25°C:

\[pH + pOH = pK_w = 14\]

where pKw = -log Kw

3.3 pH Scale Interpretation

pH Range	Nature	[H⁺] Range	Examples
0 - 3	Strongly Acidic	1 - 10⁻³ M	Gastric juice (1.5), Lemon (2.4)
3 - 6	Weakly Acidic	10⁻³ - 10⁻⁶ M	Vinegar (3), Coffee (5), Milk (6.5)
7	Neutral	10⁻⁷ M	Pure water at 25°C
8 - 11	Weakly Basic	10⁻⁸ - 10⁻¹¹ M	Blood (7.4), Sea water (8.5), Soap (9-10)
11 - 14	Strongly Basic	10⁻¹¹ - 10⁻¹⁴ M	Ammonia (11), NaOH 0.1M (13)

💡 Quick pH Calculation Tricks

If [H⁺] = a × 10⁻ⁿ M:

pH = n - log a

Example: [H⁺] = 2 × 10⁻⁵ → pH = 5 - log 2 = 5 - 0.3 = 4.7

Useful log values to memorize:

log 2 = 0.3, log 3 = 0.48, log 5 = 0.7, log 7 = 0.85

📝 Solved Example 2

Question: Calculate [H⁺], [OH⁻], and pOH for a solution with pH = 4.5

Solution:

Step 1: Find [H⁺]

[H^+] = 10^{-pH} = 10^{-4.5} = 10^{-5} \times 10^{0.5}

[H^+] = 10^{-5} \times 3.16 = 3.16 \times 10^{-5} M

Step 2: Find pOH

\[pOH = 14 - pH = 14 - 4.5 = 9.5\]

Step 3: Find [OH⁻]

[OH^-] = 10^{-pOH} = 10^{-9.5} = 3.16 \times 10^{-10} M

Answers:

[H⁺] = 3.16 × 10⁻⁵ M

[OH⁻] = 3.16 × 10⁻¹⁰ M

pOH = 9.5

Ionization Constants Ka and Kb

The ionization constants Ka and Kb quantify the strength of weak acids and bases respectively. These constants are fundamental for pH calculations.

4.1 Acid Dissociation Constant (Ka)

For Weak Acid HA

HA + H₂O ⇌ H₃O⁺ + A⁻

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

Higher Ka = Stronger acid (more dissociation)

Lower pKa = Stronger acid (pKa = -log Ka)

Acid	Ka	pKa	Strength
HF	6.6 × 10⁻⁴	3.18	Weak (strongest among weak)
HCOOH	1.8 × 10⁻⁴	3.75	Weak
CH₃COOH	1.8 × 10⁻⁵	4.74	Weak
H₂CO₃	4.2 × 10⁻⁷	6.38	Very weak
HCN	6.2 × 10⁻¹⁰	9.21	Very weak

4.2 Base Dissociation Constant (Kb)

For Weak Base B

B + H₂O ⇌ BH⁺ + OH⁻

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

Higher Kb = Stronger base

Lower pKb = Stronger base (pKb = -log Kb)

4.3 Relationship: Ka × Kb = Kw

For Conjugate Acid-Base Pair

K_a \times K_b = K_w = 10^{-14}

\[pK_a + pK_b = pK_w = 14\]

Example:

For CH₃COOH/CH₃COO⁻ pair:

Ka(CH₃COOH) = 1.8 × 10⁻⁵, so pKa = 4.74

Kb(CH₃COO⁻) = Kw/Ka = 10⁻¹⁴/1.8×10⁻⁵ = 5.56 × 10⁻¹⁰

pKb = 14 - 4.74 = 9.26

4.4 Polyprotic Acids

Acids with Multiple Ionizable Protons

Example: H₃PO₄ (Phosphoric Acid)

H₃PO₄ ⇌ H⁺ + H₂PO₄⁻; Ka₁ = 7.5 × 10⁻³

H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻; Ka₂ = 6.2 × 10⁻⁸

HPO₄²⁻ ⇌ H⁺ + PO₄³⁻; Ka₃ = 4.2 × 10⁻¹³

Key Points:

Ka₁ > Ka₂ > Ka₃ (always decreasing)
First proton is easiest to remove
For pH calculation, usually only Ka₁ is considered
Overall Ka = Ka₁ × Ka₂ × Ka₃

📝 Solved Example 3 (JEE Advanced Pattern)

Question: Calculate the Kb of acetate ion (CH₃COO⁻) if Ka of acetic acid is 1.8 × 10⁻⁵.

Solution:

Using the relationship Ka × Kb = Kw:

K_b = \frac{K_w}{K_a} = \frac{10^{-14}}{1.8 \times 10^{-5}}

K_b = 5.56 \times 10^{-10}

This shows acetate ion is a weak base (Kb << 1)

pH Calculations - All Types

This section covers pH calculations for all types of solutions - the most frequently asked numerical problems in JEE.

5.1 Master Formula Chart

Solution Type	[H⁺] Formula	pH Formula	Example
Strong Acid (C M)	[H⁺] = C	pH = -log C	0.01 M HCl → pH = 2
Strong Base (C M)	[OH⁻] = C	pOH = -log C, pH = 14 - pOH	0.001 M NaOH → pH = 11
Weak Acid	[H⁺] = √(Ka·C)	pH = ½(pKa - log C)	0.1 M CH₃COOH
Weak Base	[OH⁻] = √(Kb·C)	pOH = ½(pKb - log C)	0.1 M NH₃
Acidic Buffer	Henderson eq.	pH = pKa + log([Salt]/[Acid])	CH₃COOH + CH₃COONa
Basic Buffer	Henderson eq.	pOH = pKb + log([Salt]/[Base])	NH₃ + NH₄Cl
Salt (WA + SB)	Hydrolysis	pH = 7 + ½(pKa + log C)	CH₃COONa (basic)
Salt (SA + WB)	Hydrolysis	pH = 7 - ½(pKb + log C)	NH₄Cl (acidic)
Salt (WA + WB)	Hydrolysis	pH = 7 + ½(pKa - pKb)	CH₃COONH₄

⚠️ Very Dilute Strong Acid (C < 10⁻⁶ M)

For very dilute strong acids, consider water's contribution to [H⁺]:

[H^+]_{total} = \frac{C + \sqrt{C^2 + 4K_w}}{2}

Example: For 10⁻⁸ M HCl, [H⁺] ≈ 1.05 × 10⁻⁷ M, pH ≈ 6.98 (not 8!)

📝 Solved Example 4 (JEE Main 2024 Type)

Question: Calculate pH of:
(a) 0.1 M HNO₃
(b) 0.1 M CH₃COOH (Ka = 1.8 × 10⁻⁵)
(c) 0.1 M NH₃ (Kb = 1.8 × 10⁻⁵)
(d) 0.01 M Ca(OH)₂

(a) 0.1 M HNO₃ (Strong acid):

[H⁺] = 0.1 M, pH = 1

(b) 0.1 M CH₃COOH (Weak acid):

pH = \frac{1}{2}(pK_a - \log C) = \frac{1}{2}(4.74 - (-1)) = \frac{5.74}{2} = 2.87

(c) 0.1 M NH₃ (Weak base):

pOH = \frac{1}{2}(pK_b - \log C) = \frac{1}{2}(4.74 - (-1)) = 2.87

pH = 14 - 2.87 = 11.13

(d) 0.01 M Ca(OH)₂ (Strong base, dibasic):

[OH⁻] = 2 × 0.01 = 0.02 M

pOH = -log(0.02) = -log(2 × 10⁻²) = 2 - 0.3 = 1.7

pH = 14 - 1.7 = 12.3

Buffer Solutions

Buffer solutions resist change in pH upon addition of small amounts of acid or base. They're crucial for biological systems and laboratory work.

6.1 Types of Buffers

Acidic Buffer (pH < 7)

Weak acid + Its salt with strong base

Examples:

CH₃COOH + CH₃COONa (Acetate buffer)
H₂CO₃ + NaHCO₃ (Blood buffer)
H₂PO₄⁻ + HPO₄²⁻ (Phosphate buffer)

Basic Buffer (pH > 7)

Weak base + Its salt with strong acid

Examples:

NH₃ + NH₄Cl (Ammonia buffer)
Glycine + Glycinium chloride

6.2 Henderson-Hasselbalch Equation

The Master Buffer Equations

Acidic Buffer:

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

or: pH = pKa + log([A⁻]/[HA])

Basic Buffer:

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

Then: pH = 14 - pOH

6.3 Buffer Capacity and Range

Buffer Capacity

Maximum when [Salt] = [Acid] (ratio = 1)
At this point: pH = pKa
Increases with total concentration
Decreases when ratio moves away from 1

Buffer Range

\text{Effective Range} = pK_a \pm 1

Buffer works best when:

0.1 < [Salt]/[Acid] < 10

📝 Solved Example 5 (JEE Main Pattern)

Question: Calculate pH of buffer containing 0.3 M CH₃COOH and 0.6 M CH₃COONa. (pKa = 4.74)

pH = pK_a + \log\frac{[Salt]}{[Acid]} = 4.74 + \log\frac{0.6}{0.3}

pH = 4.74 + \log 2 = 4.74 + 0.301

\[pH = 5.04\]

Salt Hydrolysis

Salt hydrolysis is the reaction of salt ions with water to produce acidic or basic solutions. This is a high-scoring topic with direct formula-based questions.

7.1 Complete Hydrolysis Summary

Salt Type	Example	Hydrolysis	pH Formula	Nature
Strong Acid + Strong Base	NaCl, KNO₃	No hydrolysis	pH = 7	Neutral
Weak Acid + Strong Base	CH₃COONa, Na₂CO₃	Anion hydrolysis	pH = 7 + ½(pKa + log C)	Basic
Strong Acid + Weak Base	NH₄Cl, FeCl₃	Cation hydrolysis	pH = 7 - ½(pKb + log C)	Acidic
Weak Acid + Weak Base	CH₃COONH₄	Both hydrolyze	pH = 7 + ½(pKa - pKb)	Depends on Ka vs Kb

⚠️ Special Note: WA + WB Salt

For salt of weak acid + weak base, pH is independent of concentration!

If Ka > Kb → pH < 7 (Acidic)

If Ka < Kb → pH > 7 (Basic)

If Ka = Kb → pH = 7 (Neutral)

📝 Solved Example 6

Question: Calculate pH of 0.1 M NH₄Cl solution. (Kb of NH₃ = 1.8 × 10⁻⁵)

This is salt of SA (HCl) + WB (NH₃), so solution is acidic.

pKb = -log(1.8 × 10⁻⁵) = 4.74

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

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

pH = 7 - \frac{1}{2}(4.74 - 1) = 7 - 1.87

\[pH = 5.13\]

Solubility Product (Ksp)

8.1 Ksp and Solubility Relationship

For Salt A_xB_y

A_xB_y(s) ⇌ xA^y+(aq) + yB^x-(aq)

K_{sp} = [A^{y+}]^x[B^{x-}]^y = x^x \cdot y^y \cdot S^{(x+y)}

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

8.2 Ionic Product vs Ksp

Precipitation Criteria

IP < Ksp:

Unsaturated solution → No precipitation

IP = Ksp:

Saturated solution → At equilibrium

IP > Ksp:

Supersaturated → Precipitation occurs

Common Ion Effect

Definition

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

Effects:

Decreases degree of ionization (α) of weak electrolyte
Decreases solubility of sparingly soluble salt
Ksp and Ka values remain unchanged

📝 Solved Example 7

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

Let solubility = S mol/L

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

K_{sp} = [Ag^+][Cl^-] = S \times 0.1 = 1.8 \times 10^{-10}

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

Compare: Solubility in pure water = 1.34 × 10⁻⁵ M
Solubility decreased ~7500 times due to common ion!

Indicators and Titrations

10.1 Acid-Base Indicators

Indicator Theory

Indicators are weak acids/bases whose ionized and unionized forms have different colors.

HIn (color 1) ⇌ H⁺ + In⁻ (color 2)

K_{In} = \frac{[H^+][In^-]}{[HIn]}

Color Change Range:

\text{pH range} = pK_{In} \pm 1

When [HIn]/[In⁻] ≥ 10 → Color 1 predominates
When [In⁻]/[HIn] ≥ 10 → Color 2 predominates
When [HIn] = [In⁻] → pH = pKIn (transition point)

10.2 Common Indicators

Indicator	pH Range	Acid Color	Base Color	pKIn
Methyl Orange	3.1 - 4.4	Red	Yellow	3.7
Methyl Red	4.4 - 6.2	Red	Yellow	5.1
Litmus	5.0 - 8.0	Red	Blue	6.5
Bromothymol Blue	6.0 - 7.6	Yellow	Blue	7.0
Phenolphthalein	8.0 - 9.8	Colorless	Pink	9.4
Thymolphthalein	9.3 - 10.5	Colorless	Blue	10.0

10.3 Types of Acid-Base Titrations

Strong Acid + Strong Base

Example: HCl + NaOH

Equivalence point: pH = 7
Sharp pH change (3 to 11)
Indicator: Methyl orange, Phenolphthalein, or any

pH jump: ~3 to ~11

Weak Acid + Strong Base

Example: CH₃COOH + NaOH

Equivalence point: pH > 7 (basic)
Salt undergoes hydrolysis
Indicator: Phenolphthalein (pH 8-10)

pH at eq. point: ~8.7

Strong Acid + Weak Base

Example: HCl + NH₃

Equivalence point: pH < 7 (acidic)
Salt (NH₄Cl) undergoes hydrolysis
Indicator: Methyl orange (pH 3-4)

pH at eq. point: ~5.3

Weak Acid + Weak Base

Example: CH₃COOH + NH₃

Equivalence point: pH depends on Ka, Kb
Very gradual pH change
No suitable indicator!

Not titrated with indicators

10.4 Titration Curve Analysis

Key Points on Titration Curve

Weak Acid Titration with Strong Base:

Initial point: pH from weak acid formula
Buffer region: pH = pKa + log([Salt]/[Acid])
Half-equivalence point: pH = pKa (when 50% neutralized)
Equivalence point: pH from hydrolysis formula
Beyond equivalence: pH from excess base

Important Formulas for Titration:

At half-equivalence:

pH = pKa (for weak acid)

pOH = pKb (for weak base)

At equivalence point:

meq of acid = meq of base

N₁V₁ = N₂V₂

⚠️ Indicator Selection Rule

Choose indicator whose pKIn is close to pH at equivalence point.

SA + SB: Any indicator (pH = 7, wide range)
WA + SB: Phenolphthalein (pH > 7)
SA + WB: Methyl orange (pH < 7)
WA + WB: Not suitable for indicator titration

📝 Solved Example 8 (JEE Main Pattern)

Question: 25 mL of 0.1 M CH₃COOH is titrated with 0.1 M NaOH. Calculate pH: (a) Initially (b) After adding 12.5 mL NaOH (c) At equivalence point (Ka = 1.8 × 10⁻⁵)

(a) Initial pH (only weak acid):

pH = \frac{1}{2}(pK_a - \log C) = \frac{1}{2}(4.74 - (-1)) = 2.87

(b) After 12.5 mL NaOH (half-equivalence):

This is exactly half the required volume (25 mL × 0.1 M needs 25 mL × 0.1 M NaOH)

At half-equivalence: [Salt] = [Acid]

\[pH = pK_a = 4.74\]

(c) At equivalence point:

Total volume = 25 + 25 = 50 mL

[Salt] = (25 × 0.1)/50 = 0.05 M

This is salt of WA + SB, so solution is basic

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

pH = 7 + \frac{1}{2}(4.74 + \log 0.05)

pH = 7 + \frac{1}{2}(4.74 - 1.30) = 7 + 1.72

\[pH = 8.72\]

10.5 Polyprotic Acid Titrations

Titration of H₂SO₄, H₃PO₄, Na₂CO₃

Example: Na₂CO₃ + HCl Titration

First equivalence point (pH ≈ 8.3):

Na₂CO₃ + HCl → NaHCO₃ + NaCl

Indicator: Phenolphthalein (colorless endpoint)

Second equivalence point (pH ≈ 3.8):

NaHCO₃ + HCl → H₂CO₃ + NaCl

Indicator: Methyl orange (red endpoint)

Key Relationships:

Volume for phenolphthalein endpoint = V₁
Volume for methyl orange endpoint = V₂
For pure Na₂CO₃: V₂ = 2V₁
For pure NaHCO₃: V₁ = 0
For mixture: Calculate each component

📋 Master Formula Sheet - Ionic Equilibrium

Basic Relationships

Kw = [H⁺][OH⁻] = 10⁻¹⁴

pH + pOH = 14

Ka × Kb = Kw

pKa + pKb = 14

α = √(K/C)

pH Calculations

Strong acid: pH = -log C

Weak acid: pH = ½(pKa - log C)

Strong base: pOH = -log C

Weak base: pOH = ½(pKb - log C)

Buffer Solutions

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

pOH = pKb + log([Salt]/[Base])

Buffer range = pKa ± 1

Max capacity: [Salt] = [Acid]

Salt Hydrolysis

SA+SB: pH = 7

WA+SB: pH = 7+½(pKa+log C)

SA+WB: pH = 7-½(pKb+log C)

WA+WB: pH = 7+½(pKa-pKb)

Solubility Product

AB: Ksp = S²

AB₂: Ksp = 4S³

A₂B: Ksp = 4S³

AB₃: Ksp = 27S⁴

A₃B₂: Ksp = 108S⁵

Titration

N₁V₁ = N₂V₂

Half-eq: pH = pKa

WA+SB: Use phenolphthalein

SA+WB: Use methyl orange

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ pH Calculations: 30%
✓ Buffer Solutions: 20%
✓ Salt Hydrolysis: 18%
✓ Solubility Product (Ksp): 15%
✓ Ka, Kb Problems: 10%
✓ Indicators & Titration: 7%

JEE Advanced (Last 5 Years)

✓ Multi-step pH calculations: 25%
✓ Buffer & Hydrolysis combined: 22%
✓ Ksp & Precipitation: 20%
✓ Common ion effect: 15%
✓ Titration curve analysis: 10%
✓ Polyprotic acid problems: 8%

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)

JEE Advanced: 12-20 marks (3-5 questions)

Difficulty Level: Medium to Hard

Estimated Study Time: 5-6 hours

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Calculate pH of: (a) 0.01 M HCl (b) 0.001 M NaOH (c) 0.05 M H₂SO₄
What is the degree of ionization of 0.2 M acetic acid? (Ka = 1.8 × 10⁻⁵)
Calculate pH of 0.1 M NH₄OH solution. (Kb = 1.8 × 10⁻⁵)
Define: (a) Brønsted acid (b) Lewis base (c) Amphiprotic substance
Calculate Kb of CH₃COO⁻ if Ka of CH₃COOH = 1.8 × 10⁻⁵
What is the pH of a buffer containing 0.1 M acetic acid and 0.1 M sodium acetate? (pKa = 4.74)
Calculate solubility of AgCl in water. (Ksp = 1.8 × 10⁻¹⁰)
Identify the conjugate acid-base pairs in: HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻
Calculate pH of 0.1 M CH₃COONa solution. (Ka = 1.8 × 10⁻⁵)
Which indicator is suitable for HCl + NaOH titration?

Level 2: Intermediate (JEE Main/Advanced)

Calculate pH of 10⁻⁸ M HCl solution. (Consider water ionization)
A buffer is prepared by mixing 50 mL of 0.2 M acetic acid and 50 mL of 0.1 M NaOH. Calculate pH. (pKa = 4.74)
Calculate solubility of PbI₂ in 0.1 M KI solution. (Ksp of PbI₂ = 1.4 × 10⁻⁸)
Calculate pH of 0.1 M NH₄Cl solution. (Kb of NH₃ = 1.8 × 10⁻⁵)
What volume of 0.1 M NaOH is required to neutralize 25 mL of 0.1 M H₂SO₄?
Calculate pH of a solution containing 0.1 M HCl and 0.2 M acetic acid. (Ka = 1.8 × 10⁻⁵)
Equal volumes of 0.1 M AgNO₃ and 0.1 M NaCl are mixed. Will AgCl precipitate? (Ksp = 1.8 × 10⁻¹⁰)
Calculate [H⁺] at half-equivalence point when 0.1 M CH₃COOH is titrated with 0.1 M NaOH. (Ka = 1.8 × 10⁻⁵)
A solution contains 0.01 M NH₄Cl and 0.01 M NH₃. Calculate pH. (Kb = 1.8 × 10⁻⁵)
Calculate pH of 0.1 M CH₃COONH₄ solution. (Ka = Kb = 1.8 × 10⁻⁵)

Level 3: Advanced (JEE Advanced/Olympiad)

Calculate the pH of 0.1 M H₂S solution. (Ka1 = 10⁻⁷, Ka2 = 10⁻¹⁴)
A solution contains both Mg²⁺ and Zn²⁺ ions (0.01 M each). At what pH will Mg(OH)₂ start precipitating? (Ksp Mg(OH)₂ = 1.8 × 10⁻¹¹)
Derive the expression for degree of hydrolysis of a salt of weak acid and strong base.
Calculate [S²⁻] in a saturated H₂S solution (0.1 M) at pH = 3. (Ka1 = 10⁻⁷, Ka2 = 10⁻¹⁴)
A buffer contains 0.2 M weak acid (pKa = 5) and 0.3 M salt. What volume of 0.1 M HCl can be added to 100 mL of this buffer before pH changes by 1 unit?
Calculate the solubility of AgCl in 0.1 M NH₃ solution. (Ksp AgCl = 1.8 × 10⁻¹⁰, Kf[Ag(NH₃)₂]⁺ = 1.7 × 10⁷)
Plot the titration curve when 25 mL of 0.1 M H₃PO₄ is titrated with 0.1 M NaOH. Calculate pH at key points.
A mixture contains NaOH and Na₂CO₃. 20 mL of mixture requires 25 mL of 0.1 M HCl using phenolphthalein and 35 mL using methyl orange. Calculate moles of each component.
Calculate the pH at which Fe(OH)₃ starts precipitating from 0.001 M Fe³⁺ solution. (Ksp = 1.1 × 10⁻³⁶)
Derive Henderson-Hasselbalch equation starting from the Ka expression.

Get Solutions PDF with Step-by-Step Working

Year	JEE Main	JEE Advanced	Most Asked Topics
2024	3 Questions (12 marks)	4 Questions (16 marks)	pH calculations, Buffer, Ksp
2023	2 Questions (8 marks)	5 Questions (18 marks)	Hydrolysis, Common ion, Titration
2022	3 Questions (12 marks)	4 Questions (14 marks)	Buffer capacity, Precipitation
2021	2 Questions (8 marks)	3 Questions (12 marks)	Ka-Kb relationship, Indicators
2020	3 Questions (12 marks)	4 Questions (16 marks)	Polyprotic acids, Selective precipitation

📑 Quick Navigation - Ionic Equilibrium

Ionic Equilibrium JEE Main & Advanced 2025-26

Electrolytes and Ionization

1.1 Classification of Electrolytes

Strong Electrolytes

Weak Electrolytes

1.2 Degree of Ionization (α)

Definition

1.3 Ostwald's Dilution Law

For Weak Electrolytes

📝 Solved Example 1 (JEE Main Pattern)

💡 When to NOT Use Approximation

Theories of Acids and Bases

2.1 Arrhenius Theory (1887)

Definition

Arrhenius Acid

Arrhenius Base

2.2 Brønsted-Lowry Theory (1923)

Definition

Brønsted Acid

Brønsted Base

2.3 Conjugate Acid-Base Pairs

Definition

2.4 Lewis Theory (1923)

Definition (Most General)

Lewis Acid

Lewis Base

2.5 Comparison of Theories

💡 Amphoteric/Amphiprotic Substances

Ionization of Water, pH and pOH

3.1 Self-Ionization of Water

Auto-protolysis of Water

⚠️ Important Note for JEE

3.2 pH and pOH Scale

Definitions

pH (Potential of Hydrogen)

pOH

3.3 pH Scale Interpretation

💡 Quick pH Calculation Tricks

📝 Solved Example 2

Ionization Constants Ka and Kb

4.1 Acid Dissociation Constant (Ka)

For Weak Acid HA

4.2 Base Dissociation Constant (Kb)

For Weak Base B

4.3 Relationship: Ka × Kb = Kw

For Conjugate Acid-Base Pair

4.4 Polyprotic Acids

Acids with Multiple Ionizable Protons

📝 Solved Example 3 (JEE Advanced Pattern)

pH Calculations - All Types

5.1 Master Formula Chart

⚠️ Very Dilute Strong Acid (C < 10⁻⁶ M)

📝 Solved Example 4 (JEE Main 2024 Type)

Buffer Solutions

6.1 Types of Buffers

Acidic Buffer (pH < 7)

Basic Buffer (pH > 7)

6.2 Henderson-Hasselbalch Equation

The Master Buffer Equations

Acidic Buffer:

Basic Buffer:

6.3 Buffer Capacity and Range

Buffer Capacity

Buffer Range

📝 Solved Example 5 (JEE Main Pattern)

Salt Hydrolysis

7.1 Complete Hydrolysis Summary

⚠️ Special Note: WA + WB Salt

📝 Solved Example 6

Solubility Product (Ksp)

8.1 Ksp and Solubility Relationship

For Salt AxBy

8.2 Ionic Product vs Ksp

Precipitation Criteria

Common Ion Effect

Definition

📝 Solved Example 7

Indicators and Titrations

10.1 Acid-Base Indicators

For Salt A_xB_y