Weak Acids and Weak Bases: Complete Guide to Ka and Kb
Unlike strong acids and bases that completely dissociate, weak acids and bases only partially ionize in solution. Understanding their behavior requires knowledge of dissociation constants Ka and Kb. Let’s explore these important concepts.
Weak Acids
Definition: A weak acid is an acid that partially dissociates in aqueous solution, establishing an equilibrium between the molecular form and its ions.
Acid Dissociation Constant (Ka)
For a weak acid HA:
HA(aq) ⇌ H⁺(aq) + A⁻(aq)
The acid dissociation constant is:
Ka = [H⁺][A⁻] / [HA]
The relationship between Ka and pH:
pH = pKa + log([A⁻] / [HA])
Or equivalently:
pH = pKa + log([salt] / [acid])
Where pKa = -log(Ka)
Important: A smaller Ka value means a weaker acid (less dissociation), while a larger Ka value means a stronger acid (more dissociation).
Worked Example – Weak Acid pH Calculation
Problem: A 0.1M solution of acetic acid has a pH of 2.88 at 25°C. Calculate the value of its dissociation constant.
Solution:
For acetic acid (CH₃COOH):
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq)
pH = -log[H⁺]
[H⁺] = 10⁻²·⁸⁸ = 1.32 × 10⁻³ mol/dm³
At equilibrium:
– [CH₃COOH] = 0.1 – 0.00132 ≈ 0.099 M
– [H⁺] = 0.00132 M
– [CH₃COO⁻] = 0.00132 M
Ka = (0.00132)² / 0.099 = 1.77 × 10⁻⁵ mol/dm³
Calculating pH of a Weak Acid Solution
Problem: Calculate the pH of a 0.05M acetic acid solution given that Ka = 1.8 × 10⁻⁵
Solution:
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) Initial conc: 0.05 0 0 Change: -x +x +x Equilibrium: 0.05-x x x
Ka = x² / (0.05 – x) = 1.8 × 10⁻⁵
Assumption: For weak acids, x << 0.05, so:
x² / 0.05 = 1.8 × 10⁻⁵ x² = 0.05 × 1.8 × 10⁻⁵ = 9.0 × 10⁻⁷ x = 9.4 × 10⁻⁴ mol/dm³Therefore: [H⁺] = 9.4 × 10⁻⁴ pH = -log(9.4 × 10⁻⁴) = 3.0
OR using pH formula:
pH = pKa + log([A⁻] / [HA]) = -log(1.8 × 10⁻⁵) + log(0.05 / (0.05 - 0.00094)) = 4.74 + log(0.05 / 0.05) = 4.74 - 1.71 = 3.1
Weak Bases
Definition: A weak base is a base that partially accepts protons in aqueous solution.
Base Dissociation Constant (Kb)
For a weak base B:
B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)
The base dissociation constant is:
Kb = [BH⁺][OH⁻] / [B]
Relationship between Ka and Kb:
For a conjugate acid-base pair:
Ka × Kb = Kw = 1 × 10⁻¹⁴ (at 25°C)
Therefore:
pKa + pKb = 14
Worked Example – Weak Base pH Calculation
Problem: Calculate the pH of a 0.05M ammonia (NH₃) solution. Given Kb = 1.8 × 10⁻⁵
Solution:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) Initial conc: 0.05 0 0 Change: -x +x +x Equilibrium: 0.05-x x x
Kb = x² / (0.05 – x) = 1.8 × 10⁻⁵
Assuming x << 0.05:
x² = 0.05 × 1.8 × 10⁻⁵ = 9.0 × 10⁻⁷ x = 9.4 × 10⁻⁴ mol/dm³pOH = -log(9.4 × 10⁻⁴) = 3.0 pH = 14 - 3.0 = 11.0
Common Weak Acids and Bases
| Weak Acid/Base | Formula | Ka/Kb Value | Type |
|---|---|---|---|
| Acetic acid | CH₃COOH | Ka = 1.8 × 10⁻⁵ | Weak acid |
| Formic acid | HCOOH | Ka = 1.8 × 10⁻⁴ | Weak acid |
| Carbonic acid | H₂CO₃ | Ka = 4.3 × 10⁻⁷ | Weak diprotic acid |
| Ammonia | NH₃ | Kb = 1.8 × 10⁻⁵ | Weak base |
| Methylamine | CH₃NH₂ | Kb = 4.4 × 10⁻⁴ | Weak base |
Key Differences: Strong vs Weak Acids/Bases
| Property | Strong Acid/Base | Weak Acid/Base |
|---|---|---|
| Dissociation | Complete (100%) | Partial |
| Equilibrium | One-way reaction | Two-way equilibrium |
| Equilibrium Constant | Very large (Ka/Kb >> 1) | Small (Ka/Kb << 1) |
| pH Calculation | Direct from concentration | Requires Ka/Kb and equilibrium |
| pH of 0.1M solution | ≈ 1 (acid) or ≈ 13 (base) | Between 2-6 (acid) or 8-12 (base) |
Summary of Formulas
- For weak acids: pH = pKa + log([A⁻]/[HA])
- For weak bases: pOH = pKb + log([B]/[BH⁺])
- Relationship: Ka × Kb = Kw = 1 × 10⁻¹⁴
- At equilibrium: pH + pOH = 14
