Calculate the molar solubility of CaF2 in a solution containing 0.200 M Ca(NO3)2. [Ksp (CaF2) = 1.46x10^-10]

Ca(NO3)2 here is Strong electrolyte

It will dissociate completely to give [Ca2+] = 0.200 M

At equilibrium:

CaF2 <----> Ca2+ + 2 F-

   0.2 +s 2s

Ksp = [Ca2+][F-]^2

1.46*10^-10=(0.2 + s)*(2s)^2

Since Ksp is small, s can be ignored as compared to 0.2

Above expression thus becomes:

1.46*10^-10=(0.2)*(2s)^2

1.46*10^-10= 0.2 * 4(s)^2

s = 1.35*10^-5 M

Answer: 1.35*10^-5 M