Question

Determine the [ZN] [CN] and [HCN] in a saturated solution of Zn(CN)2 with a fixed pH of 1.5

The Ksp for Zn(CN)2 is 3.0x10^16. The Ka for HCN is 6.2x10^-10

Answer 1

pH=-log [H+]=1.5

[H+]=10^-1.5=0.0316M=[H3O+]

HCN+H2O <---> CN- + H3O+ ,ka=6.2*10^-10

using henderson -hasselbach eqn,

pH=pka+log [base]/[acid]=pka+log [CN-]/[HCN]

pka=-log ka=-log (6.2*10^-10)=9.21

1.5=9.21+ log [CN-]/[HCN]

[CN-]/[HCN]=2.0*10^-8      [HCN]=[H+]=0.0316M

or,[CN-]=[HCN]*(2.0*10^-8 )=(2.0*10^-8 )*0.0316M=0.0632*10^-8 M

Zn(CN)2 <--->Zn2+ 2CN-

ksp=solubility product=[Zn2+][CN-]^2

3.0*10^16=[Zn2+](0.0632*10^-8 M)^2

[Zn2+]=751.081M