what are the equilibrium concentrations of Pb2+ and F- in a saturated solution of lead (II)flouride if the Ksp for PbF2 is 3.2*10^-8?

Answer 1

..........PbF2 ==> Pb^2+ + 2F^-
I.........solid.....0.......0
C.........solid.....x.......2x
E.........solid.....x.......2x

Ksp = (Pb^2+)(F^-)^2
Substitute the E line into Ksp expression and solve for x and 2x.

Answer 2

.000000006