Question

The Ksp of CaF2 is 5.3 mc003-1.jpg 10-9. What is the molar solubility of CaF2 in 0.050 M Ca(NO3)2?

Answer 1

Calcium-fluoride and calcium-nitrite dissolve and dissociate as follows:
CaF₂(s) ⇄ Ca²⁺(aq) + 2 F⁻(aq)  Ksp = [Ca²⁺] x ([F⁻])^2 = 5.3 x 10^-9
Ca(NO₃)₂(s) ⇄ Ca²⁺(aq) + 2 NO₃⁻(aq)  [Ca²⁺]₀ = [Ca(NO₃)₂] = 0.050 M

If s is the molar solubility of calcium-fluoride, then, at equilibrium, we can write:
[Ca²⁺] = s +[Ca²⁺]₀ = s + 0.050 M
[F⁻]= 2s

5.3 x 10^-9 = (s + 0.050 M) x (2s)^2

S is very small compared to 0.050 M so the s inside the first brackets can be ignored:

5.3 x 10^-9 = (0.050 M) x (2s)^2
5.3 x 10^-9 / 0.050 M = 4 x s^2
1.06 x 10^-7 = 4 x s^2
1.06 x 10^-7 / 4 = s^2
2.65 x 10^-8 = s^2
s = √2.65 x 10^-8
s = 1.63 x 10^-4

The molar solubility of calcium-fluoride in 0.050 M of calcium-nitrite is 1.63 x 10 ^-4 moles / liter.