Question

A solution contains 1.72×10-2 M lead nitrate and 1.72×10-2 M calcium acetate. Solid ammonium fluoride is added slowly to this mixture. What is the concentration of calcium ion when lead ion begins to precipitate?

Ksp PbF₂ = (3.7e-8)

Ksp CaF₂ = (3.9e-11)

[Ca²⁺] = ______M

Answer 1

Lets find the concentration of F- when lead ion begin to precipitate.

we have [Pb2+] = 1.72*10^-2 M

PbF2 (s)   <—> Pb2+ (aq)   +   2F- (aq)

           1.72*10^-2 + s       2s

Ksp = [Pb2+][F-]^2

3.7*10^-8 = (1.72*10^-2 + s)*(2s)^2

since ksp is small, s will be small and can be ignored as compared to 1.72*10^-2

Above expression thus becomes,

3.7*10^-8 = (1.72*10^-2 )*(2s)^2

3.7*10^-8 = (1.72*10^-2 )*4*s^2

s = 5.38*10^-7 M

[F-] = 2s = 2*5.38*10^-7 M = 1.08*10^-6 M

This is the concentration of F- when PbF2 starts precipitating

Now calculate the concentration of Ca2+ at this point

CaF2 (s)   <—> Ca2+ (aq)   +   2F- (aq)

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

3.9*10^-11 = [Ca2+]*(1.08*10^-6)^2

[Ca2+] = 33.4 M

Answer: 33.4 M