Determine the molar solubility (S) of Zn(CN)2 in a solution with a pH = 4.91. Ksp (Zn(CN)2) = 3.0 × 10-16; Ka (HCN) = 6.2 × 10-10.

Answer 1

pH 4.91; therefore, (H^+) = 1.23E-5M
This is worked like your other solubility problem.
Zn(CN)2 ==> Zn^+ + 2CN^-
2H^+ + 2CN^- ==> 2HCN add the two equns
------------------------------------
Zn(CN)2 + 2H^+ ==> Zn^2+ + 2HCN
.........1.23E-5....x.......2x
Keq = Ksp/Ka = 3.0E-16/(6.2E-10)^2 = about 780 but you need to confirm that.
780 = (x)(x)/(1.23E-5)^2 and solve for x.

Answer 2

3.35E-4