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

Answer

C

Explanation

We are given
`CH_(3)COOH hArr CH_(3)COO^(-)+H^(+), K_(1)=1.5xx10^(-5)`
`HCN hArr H^(+)+CN^(-), K_(2)= 4.5xx10^(-10)`
If we reverse the second equilibrium expression and add it to the first one, we get
`CN^(-)+CH_(3)COOH hArr CH_(3)COO^(-)+HCN`
`:. K_(eq)=K_(1)xx(1)/(K_(2))=(1.5xx10^(-5))/(4.5xx10^(-10))`
`=0.33xx10^(5)`
`=3.3xx10^(4)`
Note that equilibrium constant expressions are multiplied when we add the expressions. When w reverse the expression, the new equilibrium constant is always the reciprocal of the previous one.