Question

Calculate the solubility of AgCN in a buffer solution of pH 3 . Given k_sp of AgCN 1.2×10^-16 and k_a for HCN is 4.8×10^-10.

Answer 1

The dissociation of AgCN can be written as:

AgCN (s) → Ag+ (aq) + CN- (aq) ...... Ksp

The pH is acidic so there is H+ ion in solution that will react with CN- as:
H+ (aq) + CN- (aq) → HCN (aq) ..... 1/Ka
Overall reaction can be written as:
AgCN (s) + H+ (aq) → Ag+ (aq) + HCN (aq) ....Ko= Ksp/Ka.
Ksp = 4 x 10-16 given

Ka = 6.2x10-10 (standard dissociation constant for HCN)

Ko = 4 x 10-16/ 6.2x10-10 = 6.45 x10-7,
At equilibrium :
AgCN (s) + H+ (aq) ⇔ Ag+ (aq) + HCN (aq) ....Ko= Ksp/Ka.

AgCN= X moles

[H+] = 0.001 as pH of solution is 3

[Ag+] = X moles

HCN = X moles

Now Ko = [Ag+][HCN]/[H+]
Putting the values in given expression we get:
6.45 x10-7= x² / 0.001;
solving, x = 2.54 x10-5 M.