At `25^(@)C`, the dissociation constant of a base. `BOH` is `1.0xx10^(-12)`. The concentration of hydroxyl ions in `0.01`M aqueous solution of the base would be

Options

A. `2.0xx10^(-6)mol L^(-1)`
B. `1.0xx10^(-5)mol L^(-1)`
C. `1.0xx10^(-6)mol L^(-1)`
D. `1.0xx10^(-7)mol L^(-1)`

Answer

D

Explanation

Base, BOH is dissociated as follows
`BOH hArr B^(+)+OH^(-)`
So, the dissociation constant of BOH base
`K_(b)=([B^(+)][OH^(-)])/([BOH])` ….(i)
At equilibrium `[B^(+)]=[OH^(-)]`
`therefore K_(b)=([OH^(-)]^(2))/([BOH])`
Given that `K_(b)=1.0xx10^(-12)`
and `[BOH]=0.01 M`
Thus, `1.0xx10^(-12)=([OH^(-)]^(2))/(0.01)`
`[OH^(-)]^(2)=1xx10^(-14)`
`[OH^(-)]=1.0xx10^(-7)"mol L"^(-1)`