weirdfacts
Year 12 / Ka, Kb and pKa

Small constant. Useful prediction.

For HA + H₂O ⇌ H₃O⁺ + A−A^{-}, Ka = [H₃O⁺][A−A^{-}]/[HA]. At the same temperature, larger Ka means a stronger acid; smaller pKa = −log₁₀Ka also means a stronger acid.

Start with concentration c and let x dissociate. Then Ka = x2x^{2}/(c − x). If x is small relative to c, x ≈ Kac\sqrt{K_{\mathrm a}c}. State that approximation and check x/c afterwards; otherwise solve the quadratic.

For a conjugate pair, KaKb = Kw. At 25 °C this gives pKa + pKb = 14.00. The acid and base must be a conjugate pair, not any two substances.

INTERACTIVE MODEL

Strong is not concentrated

H⁺H⁺H⁺H⁺H⁺H⁺H⁺H⁺H⁺H⁺H⁺H⁺H⁺H⁺H⁺H⁺H⁺H⁺HA: undissociated · H⁺ and A⁻: dissociatedpH 2.00014
c = 0.010 M · pH 2.00 · ionised ≈ 100.0%

At 25 °C. Compare HCl with a model weak acid (Ka = 1.8 × 10⁻⁵). The pH calculation includes water; particle counts are schematic.

Ka = x²/(c − x) pKa = −log₁₀Ka
Assumed knowledge

ICE-table stoichiometry; the small-x approximation is checked at about 5% or less here.

Learning checkpoints & sourceYOUR SYLLABUS CHECKPOINT
  • Write dissociation-constant expressions.
  • Calculate weak-acid equilibrium concentrations.
  • Relate Ka, Kb and Kw.
QCAA Chemistry 2025 v1.3 · p. 34 ↗
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Turn the idea into an answer.

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