The correct option is C 5.9×10−5 M
In precipitation reaction we first assume complete precipitation.
Ag++HCN→AgCN+H+
Since equal volume of AgNO3 and HCN are mixed.
Let , v be the volume of AgNO3 and HCN
So , [H+]=0.02×v2v=0.01 M
Now for the equilibria
AgCN⇌Ag++CN−;Ksp=[Ag+][CN−]=2.2×10−16HCN⇌H++CN−;Ka=[H+][CN−][HCN]=6.2×10−10
Let "s" be the solubility of AgCN.
The equilibrium reaction of AgCN is given below:
AgCN(s)⇌Ag+(aq)+CN−(aq)..(1)
c−s s s−x
CN−(aq)+H+(aq)1(Ka)HCN⇌HCN(aq)−−−−(2)
s c 0
s−x c−x x
[Ag+]=s[Ag+]=(s−x)+x=[CN−]+[HCN]
From the ratio of [HCN]/[CN−] we see , [CN−]<<[HCN]. So, s=[Ag+]=[HCN]
Thus from the equation (1) we have ,
[Ag+]=[CN−]+[HCN]∴2.2×10−16[CN−]=[CN−]+[H+][CN−]6.2×10−10Or,2.2×10−16[CN−]=[CN−]+(0.01)[CN−]6.2×10−10∴[CN−<<<(0.01)[CN−]6.2×10−10Or,2.2×10−16[CN−]=(0.01)[CN−]6.2×10−10[CN−]2=2.2×10−16×6.2×10−100.01[CN−]2=13.64×10−24[CN−]=3.7×10−12 M[Ag+]=Ksp[CN−][Ag+]=2.2×10−163.7×10−12[Ag+]=5.9×10−5 M