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)molL^(-1)`
B. `=1.0xx10^(-5)molL^(-1)`
C. `=1.0xx10^(-6)molL^(-1)`
D. `=1.0xx10^(-7)molL^(-1)`

Answer

D

Explanation

`BOH hArr B^(+)OH^(-)`
so, the dissociation constant of BOH base
`K_(b)=([B^(+)][OH^(-)])/([BOH)]`
At equilibrium `[B^(+)]=[OH^(-)]`
`:. ([OH^(-)]^(2))/([BOH])`
Given that `K_(b)=1.0x10^(-12) and [BOH]=0.01 m`
Thus, `1.0x10^(-12)([OH^(-)]^(2))/(0.01)`
`[OH^(-)]^(2)=1x10^(-14)`
`[OH^(-)]=1.0x10^(-7) "mol L "^(-1)`