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

Answer

C

Explanation

`CH_3COOH hArr CH_3COO^(-) + H^(+)`
`K_1=([CH_3COO^-][H^+])/([CH_3COOH])`
`=1.5xx10^(-5)`
`HCN hArr H^(+)+CN^-`
`K_2=([CN^-][H^+])/"[HCN]"`
`=4.5xx10^(-10)`
`CN^(-) + CH_3COOH hArr HCN + CH_3COO^-`
`K=([HCN][CH_3COO^-])/([CN^-][CH_3COOH])`
`=K_1/K_2=(1.5xx10^(-5))/(4.5xx10^(-10))=0.3 xx10^5`
`K=3xx10^4`