Barium fluoride, BaF2, has a measured solubility of 0.02 mol/L at 25°C. What is the solubility product constant?

Question

Barium fluoride, BaF2, has a measured solubility of 0.02 mol/L at 25°C. What is the solubility product constant?

Answer 1

BaF₂(s) <==> Ba²⁺(aq) + 2F^-(aq)

Solubility BaF₂ = 0.02 M

[Ba²⁺] = 0.02 M

[F^-] = 0.02 M x 2 = 0.04 M

Ksp = [Ba²⁺][F^-]²

Ksp = (0.02)(0.04)²

Ksp = 6.4x10^-5