Explain why a solution of sodium ethanoate (CH₃COONa) is basic, making reference to the relevant equilibrium and the relative strengths of the acid–base pairs involved.

IB DP Chemistry Higher Level (2023 syllabus) — R3.1 Proton transfer reactions · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (4 marks)

Sodium ethanoate dissociates completely in water:
CH₃COONa → CH₃COO⁻ + Na⁺
The ethanoate ion is a Brønsted–Lowry base and accepts a proton from water:
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
Because ethanoic acid is a weak acid, this equilibrium lies far to the left; only a small amount of OH⁻ is produced. The base strength of CH₃COO⁻ is greater than that of the water/OH⁻ system, so the net result is a slight excess of OH⁻ ions. Consequently the solution is basic (pH > 7).

Examiner tips

  • Show the dissociation step first, then the base reaction with water, and explain the equilibrium position using acid strength.
  • Use the terms ‘Brønsted–Lowry base’, ‘weak acid’, and ‘net excess of OH⁻’ to match the mark scheme.

Mark scheme (4 marks)

  1. Sodium ethanoate fully dissociates / ionises in water to give ethanoate ions (CH₃COO⁻) and Na⁺ ions.
  2. Ethanoate ion acts as a Brønsted–Lowry base and accepts a proton from water: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
  3. The equilibrium lies to the left / is only slightly displaced to the right because ethanoic acid is a weak acid.
  4. The stronger base (CH₃COO⁻) dominates over the weaker base (OH⁻ / water acting as acid), so a small but net excess of OH⁻ is produced, making the solution basic (pH > 7).

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