A solution is prepared by adding some solid PbF2 to water and allowing the solid to come to equilibrium with the solution. The equilibrium constant for the reaction below is Kc = 3.6x10−8.

Question

PbF2(s) ⇌ Pb2+(aq) + 2 F−(aq)

What is the equilibrium concentration of Pb2+ in the solution? Assume that some of the solid remains undissolved.

Answer 1

......................PbF2(s) ⇌ Pb2+(aq) + 2 F−(aq)
I.....................solid.............0......................0
C....................solid.............x.....................2x
E.....................solid.............x.....................2x
K = (Pb^2+)(F^-)^2
Plug the E line into the Kc expression and solve for x = (Pb^2+) = ?

Answer 2

what do you mean plug it into the kc expression? do i just use x and 2x to plug into K = (Pb^2+)(F^-)^2?

Answer 3

3.6x10−8 = x(2x)^2
x^3=9.0x10^-9
x=0.002