Weak Bases and Kb

Core Concept

A weak base accepts only a fraction of the protons available, establishing an equilibrium described by the base-dissociation constant Kb = [BH⁺][OH⁻]/[B]. The smaller the Kb (or larger the pKb), the weaker the base. Solving for pH mirrors the weak-acid procedure: build an ICE table, find [OH⁻] = x, compute pOH, then pH = 14.00 − pOH. Two structural families matter — neutral molecules with a lone pair (NH₃, amines) and anions that are conjugate bases of weak acids (CH₃COO⁻, F⁻, CN⁻).

Key Ideas

• Partial protonation: B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq), double arrow.
• Kb expression: Kb = [BH⁺][OH⁻] / [B]; water (the solvent) is omitted.
• pKb = −log Kb: smaller pKb = stronger base.
• Two families: neutral bases (NH₃, CH₃NH₂) and anionic bases (the conjugate bases of weak acids).
• Relationship to the conjugate acid: for any conjugate pair, Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C.

Equations and Variables

• Kb = [BH⁺][OH⁻] / [B]
• pKb = −log Kb
• ICE: [B]₀ − x, [BH⁺] = x, [OH⁻] = x → Kb = x² / ([B]₀ − x)
• Approximation: x ≈ √(Kb × [B]₀), valid when x/[B]₀ ≤ 5%
• pH = 14.00 − pOH = 14.00 + log[OH⁻]
• Ka × Kb = Kw (conjugate pair, 25 °C)

How It Works

1. Write the base ionization with water and the Kb expression.
2. Build the ICE table; the equilibrium row gives [BH⁺] = [OH⁻] = x and [B] = [B]₀ − x.
3. Substitute into Kb → Kb = x²/([B]₀ − x).
4. Approximate x ≈ √(Kb·[B]₀) and verify the 5% rule; if it fails, solve the quadratic.
5. Compute pOH = −log x, then pH = 14.00 − pOH.
6. For an anionic base like CH₃COO⁻, get Kb from the conjugate acid's Ka: Kb = Kw / Ka.

Worked Example

Find the pH of 0.20 M ammonia (NH₃), Kb = 1.8 × 10⁻⁵.

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻; Kb = [NH₄⁺][OH⁻]/[NH₃].

ICE: [NH₃] = 0.20 − x; [NH₄⁺] = [OH⁻] = x.

Kb = x² / (0.20 − x) ≈ x² / 0.20 = 1.8 × 10⁻⁵

x = √(1.8 × 10⁻⁵ × 0.20) = √(3.6 × 10⁻⁶) = 1.90 × 10⁻³ M = [OH⁻]

Check: (1.90 × 10⁻³ / 0.20) × 100% = 0.95% < 5% ✓

pOH = −log(1.90 × 10⁻³) = 2.72 → pH = 14.00 − 2.72 = 11.28

How it works

1. Write the base ionization with water and the Kb expression.
2. Build the ICE table; the equilibrium row gives [BH⁺] = [OH⁻] = x and [B] = [B]₀ − x.
3. Substitute into Kb → Kb = x²/([B]₀ − x).
4. Approximate x ≈ √(Kb·[B]₀) and verify the 5% rule; if it fails, solve the quadratic.
5. Compute pOH = −log x, then pH = 14.00 − pOH.
6. For an anionic base like CH₃COO⁻, get Kb from the conjugate acid's Ka: Kb = Kw / Ka.

Common confusions

• "A weak acid has a strong conjugate base." — No; only the conjugate of a strong acid is negligible. For a weak acid, its conjugate base is also weak, with Kb = Kw/Ka.
• "All bases contain OH⁻." — NH₃ and amines have no hydroxide in their formula; they produce OH⁻ by pulling a proton from water.
• "pOH is the same as pH for bases." — A basic solution has high pH and low pOH; always convert pOH → pH with 14.00 − pOH.
• "The 5% rule is optional." — It is the test that decides whether the approximation is legitimate; ignoring it risks wrong answers for dilute or relatively strong bases.

Quick review

• Kb = [BH⁺][OH⁻]/[B]; pKb = −log Kb.
• ICE → Kb = x²/([B]₀ − x); approximate and check 5%.
• [OH⁻] = x; pOH = −log x; pH = 14.00 − pOH.
• Ka × Kb = Kw links a base to its conjugate acid.
• Neutral bases (NH₃, amines) and anionic bases (conjugate bases of weak acids).