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

`CH_(3)COOH hArr CH_(3)COO ^(-) +H^(+)`
`K_(a)(CH_(3)COOH)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])=1.5xx10^(-5)`
`HCN hArr H^(+)+CN^(-)`
`K_(a)(HCN)=([H^(+)][CN^(-)])/([HCN])=4.5 xx 10^(-10)`
`CN^(-)+CH_(3)COOH hArr HCN+CH_(3)COO^(-)`
`K=([HCN][CH_(3)COO^(-)])/([CN^(-)][CH_(3)COOH])`
`K=(K_(a)(CH_(3)COOH))/(K_(a)(HCN))=(1.5 xx 10^(-5))/(4.5 xx 10^(-10))`
`=(15)/(45)xx10^(5)=(150)/(45)xx10^(4)`
`=3.33 xx 10^(4)`