If 147 g of znf2 (ksp = 0.0300) is added to water to make a 1.00 l solution, what is the [f−] concentration in the solution in molarity?

When the balanced reaction equation is:

and by using the ICE table:

                    ZnF2(s) ↔Zn2+(aq)  +  2 F-(aq)

initial                                0                    0

change                           +X                   +2X

Equ                                   X                    2X

so we have the Ksp expression of this reaction as the following:

Ksp = [Zn2+][F-]^2

when we have Ksp = 0.03

and we assumed [Zn2+] = X 

and [F-] = 2X 

so, we can substitute and solve to get the value of X:

0.03 = X * (2X)^2

0.03 = X* 4X^2

0.03 = 4X^3

∴X = 0.196 M

and when the [F-] = 2X 

∴[F-] = 2 * 0.196 = 0.192 M