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–] |
| 1 | 2 |
CaF2(s) 
Ca2+ + 2F–
Ca2+ + 2F–[Ca2+] = 1.24 × 10–3 mol L–1
[F–] = 2.48 × 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.