Question

Calculate the molar solubility of CaF2 in a solution containing 0.205 M of Ca(NO3)2. The Ksp value for CaF2 is 1.46×10−10.

Answer 1

Ca(NO3)2 --------------> Ca^2+ (aq) + 2NO3^-

0.205M                          0.205M
CaF2 (s) ----------------> Ca^2+ (aq) + 2F^- (aq)

                                        s+0.205         2s

Ksp   = [Ca^2+][F^-]^2

1.46*10^-10   = (s+0.205)(2s)^2

1.46*10^-10   = 0.205*4s^2                     [s+0.205 = 0.205]

s                     = 1.33*10^-5

molar solubility of CaF2   = 1.33*10^-5 M >>>>answer