What is the solubility constant when the concentration of Ca 2+is 0.002M and the concentration of F -is 0.004M?

Question

CaF2 (s)== Ca 2+ (aq) + 2F -(aq) What is the solubility constant when the concentration of Ca 2+is 0.002M and the concentration of F -is 0.004M?

Answer 1

CaF₂(s) <==> Ca²⁺(aq) + 2F^-(aq)

Ksp = [Ca²⁺][F-]²

Ksp = [0.002][0.004]²

Ksp = 3.2x10^-8