Solubility and ion concentrations

Problem Summary

If the molar solubility of CaF2 at 35°C in water is 1.24 x 10–3 mol L–1, what is the [Ca2+] and [F–] in solution?

Solubility and ion concentrations are related through the dissolution equation.

s(CaF2) = change in [Ca2+] = change in [F–]
112
CaF2(s) Ca2+ + 2F–

[Ca2+] = 1.24 x 10–3 mol L–1
[F–] = 2.48 x 10–3 mol L–1

Note that in water, where the only source of Ca2+ and F–, the equilibrium concentrations of BOTH [Ca2+] and [F–] are directly related to s(CaF2) through the dissolving equation.
Dr Sheila Woodgate
© The University of Auckland 2002