Question

Ignoring activities, determine the molar solubility (S) of Zn(CN)2 in a solution with a pH = 1.02. Ksp (Zn(CN)2) = 3.0 × 10-16; Ka (HCN) = 6.2 × 10-10.

Answer 1

Zn(CN)2 <---> Zn2+ (aq) + 2CN-(aq)      , K = Ksp = 3 x 10^ -16

2H+ (aq) + 2CN-(aq) <---> 2HCN (aq)          K = 1/ Ka^2 = ( 1/(6.2x10^-10)^2 ) = 2.6 x 10^18

combining bith equations we get

Zn(CN)2(s) + 2H+ (aq) <---> Zn2+(aq) + 2HCN (aq) , K combined = ( 3x10^-16) x ( 2.6x10^18) =780.44

pH = 1.02 which means [H+] = 10^ -1.02 = 0.0955 M

at equilibrium [H+] = 0.0955-2X , [Zn2+] = X , [HCN] = 2X ,

K = [Zn2+] [HCN]^2 /[H+]^2

780.44 = ( X) ( 2X)^2 / ( 0.0955-2X)^2

4X^3 = 780.44( 4X^2 -0.382X + 0.00912)

4X^3 = 3121.76X^2 -298.13X + 7.1176

X = 0.048 M= solubility of Zn(CN)2