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