Weak Acids and Ka

Core Concept

A weak acid ionizes only partially in water, so it establishes a real equilibrium described by an acid-dissociation constant Ka = [H₃O⁺][A⁻]/[HA]. The smaller the Ka (or the larger the pKa), the weaker the acid. Solving for pH means setting up an ICE table and solving for the equilibrium concentration of H₃O⁺, usually with a justified approximation (x ≪ initial concentration) or, when necessary, the quadratic formula.

Key Ideas

• Partial ionization: HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq), with a double arrow.
• Ka expression: products over reactants, water omitted, concentrations at equilibrium.
• pKa = −log Ka: smaller pKa = stronger acid; each pKa unit is a ten-fold change in Ka.
• ICE table: track Initial, Change, Equilibrium concentrations; the change column uses −x for reactants and +x for products.
• Percent ionization = ([H₃O⁺] at equilibrium / initial [HA]) × 100%; it increases as the acid is diluted.

Equations and Variables

• Ka = [H₃O⁺][A⁻] / [HA]
• pKa = −log Ka
• ICE: [HA]₀ − x, [H₃O⁺] = x, [A⁻] = x → Ka = x² / ([HA]₀ − x)
• Approximation (valid when x is small): Ka ≈ x² / [HA]₀, so x ≈ √(Ka × [HA]₀)
• Rule of thumb for the approximation: valid if x / [HA]₀ ≤ 5% (or [HA]₀ / Ka ≥ ~400)
• % ionization = (x / [HA]₀) × 100%

How It Works

1. Write the equilibrium and the Ka expression.
2. Build the ICE table with the given initial concentration [HA]₀.
3. Substitute the equilibrium row into Ka, giving Ka = x²/([HA]₀ − x), where x = [H₃O⁺] = [A⁻].
4. Try the approximation: if [HA]₀ is much larger than Ka, x is negligible compared to [HA]₀, so Ka ≈ x²/[HA]₀ and x ≈ √(Ka·[HA]₀).
5. Check the 5% rule (x/[HA]₀ × 100%). If it fails, solve the full quadratic instead.
6. Convert x to pH = −log x, and optionally compute % ionization.

Worked Example

Find the pH of 0.10 M acetic acid (CH₃COOH), Ka = 1.8 × 10⁻⁵.

ICE table: [CH₃COOH]₀ = 0.10, change −x; [H₃O⁺] = [CH₃COO⁻] = x.

Ka = x² / (0.10 − x) = 1.8 × 10⁻⁵

Approximate (0.10 − x ≈ 0.10):

x ≈ √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M

Check: (1.34 × 10⁻³ / 0.10) × 100% = 1.3% < 5% ✓ (approximation valid).

pH = −log(1.34 × 10⁻³) = 2.87

How it works

1. Write the equilibrium and the Ka expression.
2. Build the ICE table with the given initial concentration [HA]₀.
3. Substitute the equilibrium row into Ka, giving Ka = x²/([HA]₀ − x), where x = [H₃O⁺] = [A⁻].
4. Try the approximation: if [HA]₀ is much larger than Ka, x is negligible compared to [HA]₀, so Ka ≈ x²/[HA]₀ and x ≈ √(Ka·[HA]₀).
5. Check the 5% rule (x/[HA]₀ × 100%). If it fails, solve the full quadratic instead.
6. Convert x to pH = −log x, and optionally compute % ionization.

Common confusions

• "Weak acid pH = −log[HA]." — Wrong; that ignores partial ionization. You must solve Ka = x²/([HA]₀ − x).
• "Small Ka means the acid is dilute." — Ka is a constant for the acid; concentration is a separate quantity you set.
• "The approximation is always fine." — It fails for high Ka relative to [HA]₀ (e.g., 0.010 M HF, Ka = 6.8 × 10⁻⁴); then you must use the quadratic.
• "Percent ionization is constant." — It depends on concentration and increases as the solution is diluted.
• "A stronger weak acid ionizes 100%." — No; once it ionizes completely it is no longer weak — it is strong.

Quick review

• Ka = [H₃O⁺][A⁻]/[HA]; pKa = −log Ka.
• ICE table → Ka = x²/([HA]₀ − x).
• Approximate x ≈ √(Ka·[HA]₀) and check the 5% rule; else quadratic.
• % ionization = x/[HA]₀ × 100%; increases on dilution.
• pH = −log x.