Explain why the addition of a small amount of hydrochloric acid to a buffer solution containing ethanoic acid and sodium ethanoate produces only a negligible change in pH, with reference to the relevant equilibrium and the species present.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
A buffer contains a large reservoir of the conjugate base CH₃COO⁻. When a small amount of H⁺ from HCl is added, the H⁺ reacts rapidly with CH₃COO⁻:
CH₃COO⁻ + H⁺ → CH₃COOH.
This consumes the added H⁺ and shifts the equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ to the left. Because the concentrations of CH₃COOH and CH₃COO⁻ are both large compared with the added acid, the ratio [CH₃COO⁻]/[CH₃COOH] changes only slightly. According to the Henderson–Hasselbalch equation, pH = pKₐ + log([CH₃COO⁻]/[CH₃COOH]), so the pH changes negligibly.
CH₃COO⁻ + H⁺ → CH₃COOH.
This consumes the added H⁺ and shifts the equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ to the left. Because the concentrations of CH₃COOH and CH₃COO⁻ are both large compared with the added acid, the ratio [CH₃COO⁻]/[CH₃COOH] changes only slightly. According to the Henderson–Hasselbalch equation, pH = pKₐ + log([CH₃COO⁻]/[CH₃COOH]), so the pH changes negligibly.
Examiner tips
- Mention the buffer’s conjugate base reservoir first. Show the reaction of H⁺ with CH₃COO⁻. Explain the shift of the acid–base equilibrium. Relate the small change in the ratio to the Henderson–Hasselbalch equation.
Common mistakes
- Failing to state that the buffer’s base reacts with the added H⁺. Using the wrong equilibrium (e.g. CH₃COOH ⇌ CH₃COO⁻ + H⁺ instead of the reverse). Ignoring the Henderson–Hasselbalch relationship or the relative concentrations.
Mark scheme (4 marks)
- The buffer contains a reservoir of ethanoate ions (CH₃COO⁻) as the conjugate base.
- The added H⁺ ions are consumed by reaction with ethanoate ions: CH₃COO⁻ + H⁺ → CH₃COOH.
- This shifts the equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ to the left, so [H⁺] / pH changes only slightly.
- Because the concentrations of both CH₃COOH and CH₃COO⁻ remain large relative to the small amount of acid added, the ratio [CH₃COO⁻]/[CH₃COOH] and hence pH (via the Henderson–Hasselbalch relationship) changes only negligibly.
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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