The dissociation constants for acetic acid and HCN at `25^(@)C` are `1.5xx10^(-5)` and `4.5xx10^(-10)` , respectively. The equilibrium constant for the equilibirum `CN^(-) + CH_(3)COOHhArr HCN + CH_(3)COO^(-)` would be

Options

A. `3xx10^(4)`
B. `3xx10^(5)`
C. `3xx10^(-5)`
D. `3xx10^(-4)`

Explanation

Given
(i) `CH_(3)COOH Leftrightarrow CH_(3)COO^(-)+H^(+)" "K_(1)=1.5 xx 10^(-5)`
(ii) `HCN Leftrightarrow H^(+)+CN^(-) K_(2)=4.5 xx 10^(-10)`
(iii) `H^(+)+CN^(-) Leftrightarrow HCN" "K_(3)=1/K_(2) =(1)/(4.5 xx 10^(-16))`
Adding (i) and (iii) we get
`CH_(3)COOH+CN^(-)Leftrightarrow CH_(3)COO^(-)+HCN`
`K=1.5 xx 10^(-5) xx (1)/(4.5 xx 10^(-10)) =3 xx 10^(4)`