Calculate the concentration of HF, needed to just begin the precipitation of CaF2, from 0.387 M Ca(NO3)2

Question

CaF2 (s)<>Ca 2 (aq) 2F- (aq) Ksp= 3.9 x 10^ -11

HF (aq) <> H (aq) F- (aq) Ka= 7.2 x 10^ -4

Answer 1

[Ca2+][F-]^2 = 3.9 X 10^-11

[Ca2+] = 0.387 mol dm-3

So [F-]^2 = (3.9 X 10^-11)/0.387 = 1 X 10^-10.

Ka = ([H+][F-])/[HF].

Since [H+] must be equal to [F-] this can be rewritten as ...

[HF] = ([F-]^2)/Ka

So at point of just precipitating ...

[HF] = 1 X 10^-10/7.2 X 10^-4 = 1.39 X 10^-7

So you don&#039;t have to add much HF to get a precipitate!!