Question

In one litre of 3M solution of NaOH, `H_(2)SO_(4)` is added until `27.4 Kcal` energy us released. Then 2 moles of `CH_(3)COOH` is added to the resulting solution and the final volume becomes 2 litre. If x moles of AgCN can be dissolved in the final solution, what is `x xx 10^(4)`?
`H^(+)+OH^(-)rarrH_(2)O " "DeltaH=-13.7Kcal//mol`
`K_(sp)AgCN=10^(-12)K_(a)(CH_(3)COOH)=10^(-5) K_(a)(HCN)=10^(-9)`

Answer 1

Correct Answer - 2
`pH` of the `2` litre solution `= pK_(a) "of" CH_(3)COOH=5`
`AgCNhArr Aunderset(s)(g^(+))+Cunderset(s)(N)^(-)`
`CN^(-)+underset(10^(-5))H^(+)rarrHunderset(s)CN`
`(S^(2))/(10^(-5)xx10^(-12))=10^(9)`
`s=10^(-4)m/L`