Question

Determine the molar solubility of MX (Ksp=4.2x10-8) in 0.047 M NaCN. The metal ion M+ forms M(CN)2- (Kf=3.1x1010) in the presence of CN- ion

Answer 1

..................MX ==> M^+ + X^- and Ksp = (M^+)(X^-) = 4.2E-8
I................solid.........0.........0
C...............solid.........S.........S where S = solubility

But M^+ + 2CN^- ==> [M(CN)2]^- and Kf = 3.1E10
3,1E10 = [M(CN)2]^-/(M^+)(CN^-)^2

You know total solubility, S = (M^+)[M(CN)2]-
Use Kf to solve for ratio of [M(CN)2]^-/(M^+),plug in CN^- and solve for [M(CN)2]- in terms of (M^+), plug that in to the total S equation above and finally solve for (M^+) in terms of S. Plug that into Ksp expxression and solve for S. Post your work if you get stuck.