Acid Dissociation Constant
Ka, pKa, and Weak Acid/Base Equilibria
Acid Dissociation Constant (Ka)
Definition
Equilibrium constant for the dissociation of a weak acid in water
HA + H₂O ⇌ H₃O⁺ + A⁻
Ka = [H₃O⁺][A⁻] / [HA]
HA = weak acid
A⁻ = conjugate base
H₃O⁺ = hydronium ion (or H⁺)
pKa Definition
pKa = -log(Ka)
Ka = 10-pKa
Low pKa (0-5): Strong acid
Medium pKa (5-10): Moderate acid
High pKa (>10): Weak acid
Base Dissociation Constant (Kb)
Weak Base Equilibrium
B + H₂O ⇌ BH⁺ + OH⁻
Kb = [BH⁺][OH⁻] / [B]
Relationship: Ka and Kb
Ka × Kb = Kw
Ka × Kb = 1.0 × 10-14 (at 25°C)
Also:
pKa + pKb = 14.00
pH Calculations for Weak Acids
ICE Table Method
For HA ⇌ H⁺ + A⁻:
| [HA] | [H⁺] | [A⁻] | |
|---|---|---|---|
| Initial | C | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium | C - x | x | x |
Ka = x² / (C - x)
Simplification (5% Rule)
If C/Ka > 100, then x << C, so (C - x) ≈ C
Ka ≈ x² / C
x = √(Ka × C)
Then: pH = -log(x)
Percent Dissociation
% dissociation = (x / C) × 100%
Typical weak acids: 1-10% dissociation
Common pKa Values
| Acid | Formula | pKa | Ka |
|---|---|---|---|
| Hydrochloric acid | HCl | -7 | ~10⁷ (strong) |
| Sulfuric acid (1st) | H₂SO₄ | -3 | ~10³ (strong) |
| Phosphoric acid (1st) | H₃PO₄ | 2.15 | 7.1 × 10⁻³ |
| Acetic acid | CH₃COOH | 4.75 | 1.8 × 10⁻⁵ |
| Carbonic acid (1st) | H₂CO₃ | 6.35 | 4.5 × 10⁻⁷ |
| Hydrocyanic acid | HCN | 9.21 | 6.2 × 10⁻¹⁰ |
| Ammonium ion | NH₄⁺ | 9.25 | 5.6 × 10⁻¹⁰ |
| Water | H₂O | 15.7 | 2.0 × 10⁻¹⁶ |
Worked Examples
Example 1: Calculate pH of Weak Acid
Problem: Calculate pH of 0.10 M acetic acid (Ka = 1.8 × 10⁻⁵).
Solution:
Check: C/Ka = 0.10/(1.8×10⁻⁵) = 5,556 > 100 ✓ (can use simplification)
x = √(Ka × C)
x = √(1.8×10⁻⁵ × 0.10)
x = √(1.8×10⁻⁶)
x = 1.34×10⁻³ M = [H⁺]
pH = -log(1.34×10⁻³)
pH = 2.87
Check: % dissociation = (1.34×10⁻³/0.10) × 100% = 1.34% < 5% ✓
Example 2: Calculate Ka from pH
Problem: A 0.50 M weak acid has pH = 2.70. Calculate Ka.
Solution:
pH = 2.70 → [H⁺] = 10⁻²·⁷⁰ = 2.0 × 10⁻³ M
From ICE table: [H⁺] = [A⁻] = x = 2.0 × 10⁻³
[HA] = 0.50 - 0.002 ≈ 0.50 M
Ka = [H⁺][A⁻] / [HA]
Ka = (2.0×10⁻³)² / 0.50
Ka = (4.0×10⁻⁶) / 0.50
Ka = 8.0 × 10⁻⁶
pKa = -log(8.0×10⁻⁶) = 5.10
Example 3: Ka and Kb Relationship
Problem: NH₃ has Kb = 1.8 × 10⁻⁵. What is Ka for NH₄⁺?
Solution:
Ka × Kb = Kw
Ka = Kw / Kb
Ka = (1.0×10⁻¹⁴) / (1.8×10⁻⁵)
Ka(NH₄⁺) = 5.6 × 10⁻¹⁰
pKa + pKb = 9.25 + 4.75 = 14.00 ✓
Common Mistakes
Using Simplification When Invalid
Always check C/Ka > 100 before simplifying!
Confusing Ka and pKa
Lower pKa = stronger acid (higher Ka)!
Forgetting Equilibrium
For weak acids, [H⁺] ≠ initial concentration!