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

Answer

A

Explanation

`CH_(3)COOH hArr CH_(4)COO^(-)+H^(+)`
`K_(a1)1.5xx10^(-5)= ([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])`
`HCN hArr H^(+)+CN^(-)`
`K_(a2)=4.5xx10^(-10)=([H^(+)][CH_(3)CN^(-)])/([HCN])`
By (i)(ii)
`K=(K_(aq1))/(k_(a2))=([HCN])/([CN^(-)])([CH_(3)COO^(-)])/([CH_(3)COOH])`
`=(1.5xx10^(-5))/(4.5xx10^(-10))=3xx10^(4)`