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

Answer

C

Explanation

` CH_(3) COOH hArr CH_(3)COO^(-) +H^(+)`
` K_(1) = 1.5 xx 10^(-5) = ([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH]) `
` HCN hArr H^(+) + CN^(-)`
`K_(2) = 4.5 xx 10^(-10) = ([H^(+)] [CN^(-)])/([HCN])`
Aim: `CN^(-) + CH_(3)COOH hArr HCN + CH_(3)COO^(-)`
`K= ([HCN] [ CH_(3)COO^(-)])/([CN^(-)] [CH_(3)COOH])=1/K_(2) xx K_(1)`
` 1/(4.5 xx 10^(-10)) xx 1.5 xx 10^(-5) = 3 xx 10^(4) `