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. `3.0 xx 10^(4)`
B. `3.0 xx 10^(5)`
C. `3.0 xx 10^(-5)`
D. `3.0 xx 10^(-4)`

Answer

A

Explanation

`K_(c)=K_(a(CH_(3)COOH)) xx (1)/(K_(a(HCN)))`
`=1.5 xx 10^(-5) xx (1)/(4.5 xx 10^(-10))`
`~= 3 xx 10^(4)`