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

Options

A. `3.0xx10^(5)`
B. `3.0xx10^(-5)`
C. `3.0xx10^(-4)`
D. `3.0xx10^(4)`

Answer

D

Explanation

Given, `CH_(3)COOHhArrCH_(3)COO^(-) + H^(+)` ,
`K_(a1) = 1.5xx10^(-5)` ….(i)
`HCN hArrH^(+) + CN^(-)` ,
`K_(a2) = 4.5xx10^(-10)` …(ii)
`CN^(-) + CH_(3)COOHhArrHCN + CH_(3)COO^(-)` K = ?
On subtracting Eq. (ii) form Eq. (i), we get
`CH_(3)COOH + CN^(-)hArrHCN + CH_(3)COO^(-)`
`K = (K_(a1))/(K_(a2)) = (1.5xx10^(-5))/(4.5xx10^(-10)) = (10^(5))/(3) = 3.33xx10^(4)`