Explain why a buffer solution formed by mixing excess ethanoic acid with sodium hydroxide solution resists a large change in pH when a small amount of strong base is added.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
A buffer solution contains significant concentrations of both ethanoic acid (CH₃COOH) and sodium ethanoate (CH₃COO⁻Na⁺) in equilibrium: CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq).
Model answer (4 marks)
A buffer contains significant amounts of the weak acid CH₃COOH and its conjugate base CH₃COO⁻. When a small amount of strong base (OH⁻) is added, the OH⁻ reacts with CH₃COOH:
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.
This reaction consumes most of the added OH⁻, so the concentration of free OH⁻ – and therefore the pH – changes only slightly. The ratio [CH₃COO⁻]/[CH₃COOH] is only slightly altered, so by the Henderson–Hasselbalch equation the pH changes only a little.
CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.
This reaction consumes most of the added OH⁻, so the concentration of free OH⁻ – and therefore the pH – changes only slightly. The ratio [CH₃COO⁻]/[CH₃COOH] is only slightly altered, so by the Henderson–Hasselbalch equation the pH changes only a little.
Examiner tips
- Mention the reservoir of acid and base, the reaction that removes OH⁻, and the small change in the acid/base ratio. Use the Henderson–Hasselbalch equation to justify the small pH change. Keep the answer concise and to the point.
Common mistakes
- Failing to state that the OH⁻ reacts with CH₃COOH, or confusing the direction of the reaction.\nUsing the wrong equation (e.g. pOH = pKb + log([BH⁺]/[B])) instead of Henderson–Hasselbalch.\nOver‑explaining the buffer mechanism beyond the 4 marks.
Mark scheme (4 marks)
- The buffer contains a reservoir of the weak acid (CH₃COOH) and its conjugate base (CH₃COO⁻) in significant concentrations.
- When strong base (OH⁻) is added, it reacts with the ethanoic acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.
- This reaction removes most of the added OH⁻, preventing a large increase in [OH⁻] / preventing a large decrease in [H⁺].
- The ratio [CH₃COO⁻]/[CH₃COOH] changes only slightly, so by the Henderson–Hasselbalch equation (pH = pKa + log([A⁻]/[HA])), the pH changes only slightly.
Key terms in this question
Related
- All IB DP Chemistry Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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