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

Answer

D

Explanation

Given `K_(b)=1.0xx10^(12)`
`[BOH]=0.01M" "[OH]=?`
`underset(t_(eq))underset(t=o)BOHunderset(c(1-alpha))underset(c)hArrunderset(calpha)underset(0)(B^(+))+underset(calpha)underset(0)(OH^(-))`
`K_(b)=(c^(2)alpha^(2))/(c(1-alpha))=(calpha^(2))/((1-alpha))implies1.0xx10^(-12)=(0.01alpha^(2))/((1-alpha))`
On calculation, we get, `x=1.0xx10^(-5)`
Now, `[OH^(-)]=calpha=0.01xx10^(-5)=1xx10^(-7)molL^(-1)`