# 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.

> IB DP Chemistry Higher Level (2023 syllabus) — R2.3 How far? The extent of chemical change · Explain · 4 marks

> The dissolution equilibrium for calcium fluoride is: CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq)

## Mark scheme (4 marks)

1. 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).
2. 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).
3. 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).
4. The value of Ksp remains unchanged because temperature is constant; Ksp depends only on temperature.

## Key terms

- [position of equilibrium](https://www.gradenine.co.uk/glossary/position-of-equilibrium)

## Related

- [Revision notes for IB DP Chemistry Higher Level (2023 syllabus)](https://www.gradenine.co.uk/learn)
- [How to answer "Explain" questions](https://www.gradenine.co.uk/tools/command-word-cheatsheet)
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Source: [GradeNine](https://www.gradenine.co.uk/q/a-saturated-aqueous-solution-of-calcium-13cd604c) · Published by Druglandscape Ltd.