At `25^(@)C K_(b)` for `BOH=1.0xx10^(-12).0.01M` solution of `BOH` has `[OH^(-)]`:

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

`BOHhArrB^(+)+OH^(-)`
So, the dissociation constant of `BOH` base
`K_(b)=([B^(+)][OH^(-)])/([BOH])` (`i`)
At equilibrium `[B^(+)]=[OH^(-)]`
`:. ([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)=1xx0^(-14)`
`[OH^(-)]=1.0xx10^(-7) mol L^(-1)`