10 M 4 Calculate the molar solubility of CaF2 at 25 C in a s

10 M 4) Calculate the molar solubility of CaF2 at 25 °C in a solution containing Ca(NO32 and (b) 0.010 M NaF. Ks 3.9 x 10 (NOTE: Unless you really like to solve messy equations you need to make an assumption that will greatly simplify the equilibrium expression, Hint: look at the solubility of all the compounds)

Solution

Ca(NO3)2 ----------------> Ca^2+ (aq) + 2No3^- (aq)

0.01M                              0.01M

CaF2 ----------------> Ca^2+ (aq) + 2F^- (aq)

                                     s+0.01         2s

Ksp   = [Ca^2+][F^-]^2

    3.9*10^-11 = (s+0.01)(2s)^2               [ s+0.01 = 0.01]

3.9*10^-11 = (0.01)(2s)^2    

  3.9*10^-11 = (0.01)*4s^2

s^2      = 3.9*10^-11/0.01*4   = 9.75*10^-10

s        = 3.12*10^-5 M

molar solubity of CaF2 = 3.12*10^-5 M

b. NaF (aq)---------------> Na^+ (aq) + F^- (aq)

     0.01M                                             0.01M

CaF2 ----------------> Ca^2+ (aq) + 2F^- (aq)

                                     s       2s +0.01

Ksp   = [Ca^2+][F^-]^2

    3.9*10^-11 = (s)(2s +0.01)^2               [ 2s+0.01 = 0.01]

3.9*10^-11 = (s)(0.01)^2    

  3.9*10^-11 = (0.01)^2 *s

s                  = 3.9*10^-11/(0.01)^2   = 3.9*10^-7 M

solubility = 3.9*10^-7 M