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Find the solubility of AgCN in a buffer solution having pH=3.0 (given Lap for AgCN = 1.2*10^-15 and Ka For HCN = 4.8*10^-10

Find the solubility of AgCN in a buffer solution having pH=3.0 (given Lap for AgCN = 1.2*10^-15 and Ka For HCN = 4.8*10^-10

Grade:11

1 Answers

Arun
25750 Points
6 years ago
Dear Piyush
 
AgCN (s) Ag+ (aq) + CN- (aq) ...... Ksp 
H+ (aq) + CN- (aq) HCN (aq) ..... 1/Ka 
Overall: 
AgCN (s) + H+ (aq) Ag+ (aq) + HCN (aq) ....Ko= Ksp/Ka. 
Ko = (6.00x10^-17) / (4.93x10^-10) = 1.22x10^-7, and 
Ko = [Ag+][HCN]/[H+] 
Also, [H+] = -log[H+]; so [H+] = 10^-3.00 = 0.0010; since the solution is buffered to 3.00. 
For every x moles of AgCN that dissolves in a liter of the buffer, x moles of Ag+ and x moles of HCN are in solution (ie., the solubility), so substituting into the Ko expression: 
1.22x10^-7 = x² / 0.001; 
solving, x = 1.10x10^-5 M.
 
 
 
Regards
Arun (askIITians forum expert)

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