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