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Question

Equal volume of 0.02 M AgNO3 and 0.02 M HCN were mixed. Calculate [Ag+] at equilibrium assuming no cynao-complex formation.
Given : Ksp(AgCN)=2.2×1016, Ka (HCN)=6.2×1010.

A
2.64×104 M
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B
3.69×108 M
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C
5.9×105 M
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D
4.64×102 M
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Solution

The correct option is C 5.9×105 M
In precipitation reaction we first assume complete precipitation.
Ag++HCNAgCN+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
AgCNAg++CN;Ksp=[Ag+][CN]=2.2×1016HCNH++CN;Ka=[H+][CN][HCN]=6.2×1010


Let "s" be the solubility of AgCN.
The equilibrium reaction of AgCN is given below:
AgCN(s)Ag+(aq)+CN(aq)..(1)
cs s sx
CN(aq)+H+(aq)1(Ka)HCNHCN(aq)(2)
s c 0
sx cx x
[Ag+]=s[Ag+]=(sx)+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]2.2×1016[CN]=[CN]+[H+][CN]6.2×1010Or,2.2×1016[CN]=[CN]+(0.01)[CN]6.2×1010[CN<<<(0.01)[CN]6.2×1010Or,2.2×1016[CN]=(0.01)[CN]6.2×1010[CN]2=2.2×1016×6.2×10100.01[CN]2=13.64×1024[CN]=3.7×1012 M[Ag+]=Ksp[CN][Ag+]=2.2×10163.7×1012[Ag+]=5.9×105 M

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