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

Answer

D

Explanation

Given `CH_3COOH rArrCH_3COO^(-)+H^(+)K=1..5xx10^(-5)`.....(i)
`HCNhArrH^(+)+CN^(),K_(a)._1=4.5xx10^(-10)` ....(ii)
For `CN^(-)+CH_3COOHhArr HCN+CH_3COO^(-)`
on subtracting Eq. (ii) from Eq. (i). We get `CH_3COOH+CN^(-) hArrHCN+CH_3COO^(-)`
`K=(K_3)/(K_(a._1))=(1.5xx10^(-5))/(4.5xx10^(-10))`
`10^(5)/(3)=3.33xx10^4=3xx10^4`