Solubility / precipitates!?

Question

what must be the F- concentration to form precipitation of SrF2, in a solution that is 1.663x10^5 M Sr(NO3)2 ? Ksp=2.9x10^-9

i know that to form a precipitate the ion product has to be greater that the Ksp, but i don't know how to go about solving this problem / setting it up... help!!

Answer 1

SrF2 (s)  ----->  Sr2+ (aq)  +  2F1- (aq)

Qsp = [Sr2+][F1-]^2

2.9 x 10^-9 = 1.663 x 10^-5 x [F1-]^2

[F1-]^2 = 2.9 x 10^-9 / 1.663 x 10^-5 = 0.000174

[F1-] = 0.0132 M

The [F1-] concentration must be equal to or greater than 0.0132 M to cause precipitation.

I assumed the [Sr2+] concentration is 1.663 x 10^-5, not 1.663 x 10^5.

Answer 2

SrF2 <==> Sr2+ + 2F-  Ksp = 2.9 x 10^-9

I'm sure you meant that the concentration of strontium nitrate is 1.66 x 10^-5 M and not 1.66 x 10^5.

Ksp = [Sr2+] [F-]^2

Ksp = (1.66 x 10^-5)(x)^2 = 2.9 x 10^-9

solve for x

x = 0.0132N

The [F-] must be greater than 0.0132M to get a precipitate of SrF2.