Question

Calculate solubility of AgCN(Ksp = 4 x 10-16) in a buffer solution of pH = 3.

Answer 1

AgCN at equilibrium Ag+  + CN-  Ksp  

H+  + CN-  at equilibrium HCN  1/Ka  

Therefore

AgCN  + H+  at equilibrium Ag+  + HCN  = Ksp/Ka.  

Then (6.0 x 10^-17) / (4.93 x 10^-10) = 1.22 x 10^-7 then

= [Ag+][HCN]/[H+]  

But [H+] = -log[H+]

[H+] = 10^-3.0 = 0.0010

Thus we can say that every x moles of AgCN that dissolves in a liter of the buffer, x moles of Ag+ and x moles of HCN are in solutionand this is the solubility therefore substituting into the in the above expression

1.22 x 10^-7

= x² / 0.001;  

Then the solubility is  = 1.10 x 10^-5 M