Calculate the solubility (in g/l) of calcium fluoride in water at 25°c if the ksp for caf2 is 1.5 * 10-10.

When CaF₂ dissolves, it dissociates as follows;

CaF₂ - - > Ca²⁺ + 2F⁻

If molar solubility of CaF₂ is x, then molar solubility of Ca²⁺ is x and F⁻ is 2x.

ksp is solubility product constant and it can be calculated as follows;

ksp = [Ca²⁺][F⁻]²

ksp = (x) (2x) ²

ksp = 4x³

4x³ = 1.5 x 10⁻¹⁰

x³ = 0.375 x 10⁻¹⁰

x = 3.3 x 10⁻⁴ M

To calculate solubility in terms of g/L, we have to know the molar mass of CaF₂.

Solubility of CaF₂ is 3.3 x 10⁻⁴ mol/L x 78 g/mol = 0.0257 g/L