A `0.1molar` solution of weak base `BOH` is `1%` dissociated. If `0.2mol` of `BCI` is added in `1L` solution of `BOH`. The degree of dissociation of `BOH` will become

Options

A. `0.02`
B. `0.005`
C. `5xx10^(-5)`
D. `2xx10^(-3)`

Answer

C

Explanation

`alpha = (1)/(100) = 10^(-2)`
`BOH hArr B^(o+) + overset(Theta)OH`
`K_(b) = (Calpha^(2))/(1-alpha) ~~ Calpha^(2) =0.1 xx (10^(-2))^(2) = 10^(-5)`
When `0.2 mol` of `BCI`is added, due to common ion effect of `B^(o+) (BCI hArr B^(o+) +CI^(Theta))`, supression of ionisation of `BOH` occurs, and `alpha` become `alpha'`.
`:. alpha' = (K_(b))/(M) (M =` molarity of common ion added)
`= (10^(-5))/(0.2) = 5 xx 10^(-5)M`