Class 12

Tamil Nadu Class 12 Chemistry Ionic Equilibrium — Samacheer Kalvi

Q: What is common ion effect?

Common ion effect is the suppression of ionisation of a weak electrolyte by adding a strong electrolyte containing a common ion.

Plus Two Chemistry notes on acids, bases, pH, buffers, solubility product, common ion effect, and hydrolysis for Tamil Nadu board exams.

Updated 6 June 20263 min read

Quick Summary

Ionic equilibrium explains how weak acids, weak bases, salts, and sparingly soluble compounds behave in solution. The chapter is numerical-heavy: pH, buffer equations, hydrolysis, and solubility product are the central exam areas.

Core Relations

pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14 at 298 K
For a weak acid: Ka = [H+][A-] / [HA]
For a weak base: Kb = [BH+][OH-] / [B]
Kw = Ka x Kb for conjugate acid-base pairs

Buffers

Acidic Buffer

Weak acid + salt of weak acid, for example acetic acid + sodium acetate.

pH = pKa + log([salt]/[acid])

Basic Buffer

Weak base + salt of weak base, for example ammonium hydroxide + ammonium chloride.

pOH = pKb + log([salt]/[base])

Solubility Product

For AB(s) <=> A+ + B-,

Ksp = [A+][B-]

If ionic product < Ksp, no precipitation
If ionic product = Ksp, saturated solution
If ionic product > Ksp, precipitation occurs

Worked Examples

Example 1 — pH of a strong acid

Find the pH of 0.01 M HCl.

HCl fully ionises → [H⁺] = 0.01 = 10⁻²
pH = −log(10⁻²) = 2

Example 2 — pH of a weak acid

0.1 M acetic acid, Ka = 1.8 × 10⁻⁵. Find [H⁺] and pH.

[H⁺] = √(Ka·C) = √(1.8×10⁻⁵ × 0.1) = √(1.8×10⁻⁶) = 1.34×10⁻³
pH = −log(1.34×10⁻³) ≈ 2.87

Example 3 — Buffer (Henderson)

Acetic acid–acetate buffer, [salt] = [acid], pKa = 4.74. Find pH.

pH = pKa + log([salt]/[acid]) = 4.74 + log 1 = 4.74

Example 4 — Solubility product

Solubility of AgCl is 1.3 × 10⁻⁵ mol/L. Find Ksp.

AgCl ⇌ Ag⁺ + Cl⁻ → Ksp = s² = (1.3×10⁻⁵)² = 1.69 × 10⁻¹⁰

Example 5 — Common ion effect (precipitation test)

Will precipitation occur if [Ag⁺][Cl⁻] = 2 × 10⁻⁸ and Ksp(AgCl) = 1.8 × 10⁻¹⁰?

Ionic product (2×10⁻⁸) > Ksp (1.8×10⁻¹⁰) → precipitation occurs.

Salt Hydrolysis (quick guide)

Strong acid + strong base salt (NaCl)        → neutral, pH 7
Strong acid + weak base salt (NH₄Cl)         → acidic, pH < 7
Weak acid + strong base salt (CH₃COONa)      → basic, pH > 7
Weak acid + weak base salt                   → depends on Ka vs Kb

Important Questions

One Mark

1. Define ionic product of water. 2. What is common ion effect? 3. Write the Henderson equation for an acidic buffer. 4. Define solubility product. 5. State the pH of neutral water at 298 K.

Short Answer

1. Distinguish strong and weak electrolytes. 2. Explain buffer action using acetic acid and sodium acetate. 3. Explain how common ion effect helps in selective precipitation. 4. Write the relation between Ka, Kb, and Kw.

Long Answer / Problems

1. Calculate the pH of a weak acid or buffer solution. 2. Derive the expression for solubility product and predict precipitation. 3. Explain hydrolysis of salts formed from strong and weak acids or bases.

Common Mistakes

Using concentration directly as [H+] for weak acids without ionisation calculation
Mixing pH and pOH
Forgetting stoichiometric powers in Ksp
Using buffer formula when one component is absent

Revision Checklist

[ ] pH and pOH numericals
[ ] Buffer equations
[ ] Common ion effect examples
[ ] Ksp and precipitation problems
[ ] Salt hydrolysis cases

Frequently Asked Questions

A buffer solution resists change in pH when small amounts of acid or base are added. An acidic buffer commonly contains a weak acid and its salt.