A saturated aqueous solution of calcium fluoride, CaF₂, is in equilibrium with excess undissolved solid. Explain the effect on the position of equilibrium and on the value of Ksp when a small volume of concentrated hydrofluoric acid, HF(aq), is added to this system at constant temperature.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
The dissolution equilibrium for calcium fluoride is: CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq)
Model answer (4 marks)
Adding HF increases the concentration of F⁻ in solution (HF partially dissociates to give F⁻, or the common ion F⁻ is added). The higher [F⁻] raises the reaction quotient Qsp above Ksp, so the equilibrium shifts to the left – CaF₂(s) is favoured and dissolution is reduced. Consequently the solubility of CaF₂ decreases (common‑ion effect). The numerical value of Ksp does not change, because Ksp depends only on temperature and the temperature is held constant.
Examiner tips
- Show the common‑ion effect by stating that added F⁻ raises Qsp and shifts equilibrium left.
- Explain that Ksp is a constant at fixed temperature – it does not change with concentration changes.
- Use the correct symbol Ksp and mention the solid phase is unchanged in the expression.
- Keep the answer concise – 4 marks only.
Common mistakes
- Confusing the effect on Ksp with a change in solubility – Ksp stays the same.
- Failing to recognise that HF is a weak acid and only partially dissociates, yet still contributes F⁻.
- Not stating that the shift is to the left (towards the solid) rather than to the right.
Mark scheme (4 marks)
- Adding HF(aq) increases the concentration of F⁻ ions in solution (since HF is a weak acid that partially dissociates, contributing F⁻, or equivalently the common ion F⁻ is added).
- The increased [F⁻] causes the reaction quotient Qsp to exceed Ksp, so the position of equilibrium shifts to the left (towards the solid / towards less dissolution).
- As the equilibrium shifts left, Ca²⁺ ions are removed from solution (and more solid CaF₂ is deposited), so the solubility / extent of dissolution of CaF₂ decreases (common ion effect).
- The value of Ksp remains unchanged because temperature is constant; Ksp depends only on temperature.
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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