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⁻]
InitialC00
Change-x+x+x
EquilibriumC - xxx

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

AcidFormulapKaKa
Hydrochloric acidHCl-7~10⁷ (strong)
Sulfuric acid (1st)H₂SO₄-3~10³ (strong)
Phosphoric acid (1st)H₃PO₄2.157.1 × 10⁻³
Acetic acidCH₃COOH4.751.8 × 10⁻⁵
Carbonic acid (1st)H₂CO₃6.354.5 × 10⁻⁷
Hydrocyanic acidHCN9.216.2 × 10⁻¹⁰
Ammonium ionNH₄⁺9.255.6 × 10⁻¹⁰
WaterH₂O15.72.0 × 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!

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