Explain why a mixture of ammonia solution and ammonium chloride solution acts as a buffer, maintaining a nearly constant pH when a small amount of strong acid is added.

IB DP Chemistry Standard 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)

A buffer contains a weak base (NH₃) and its conjugate acid (NH₄⁺) in significant concentrations.
When a small amount of strong acid is added, the NH₃ reacts with H⁺ to form NH₄⁺:
NH₃ + H⁺ → NH₄⁺.
This removes the added protons, so the concentration of H⁺ (and therefore pH) changes only slightly.
The equilibrium NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ shifts to keep the ratio [NH₄⁺]/[NH₃] almost constant.
Because the buffer components are present in large excess, the ratio changes minimally, maintaining a nearly constant pH.

Examiner tips

  • State that the buffer contains NH₃ and NH₄⁺. Explain the reaction of NH₃ with added H⁺. Mention the equilibrium shift that keeps the ratio constant. Show why the ratio changes only slightly due to large concentrations.

Common mistakes

  • Confusing NH₃ with NH₄⁺ as the base. Forgetting to mention the equilibrium shift. Claiming the pH stays exactly constant rather than nearly constant.

Mark scheme (4 marks)

  1. A buffer contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in significant concentrations.
  2. When a small amount of strong acid (H⁺/H₃O⁺) is added, the base component (NH₃) reacts with and removes the added protons.
  3. The equilibrium NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ (or NH₄⁺ ⇌ NH₃ + H⁺) shifts to maintain the ratio of weak acid to conjugate base, so [H⁺] (and hence pH) remains nearly unchanged.
  4. The system works because the concentrations of both NH₃ and NH₄⁺ are large relative to the amount of added acid, so the ratio [NH₄⁺]/[NH₃] (and therefore pH) changes only minimally.

Related

More Proton transfer reactions questions

▶ Try answering this question with AI marking (free) →