Calculate `[OH^(-)]` in `0.20M` solution of `NH_(3)`, if `K_(b)` for `NH_(3)` is `1.8xx10^(-5)`.

Explanation

`underset(1-x)(NH_(3))H_(2)OhArr underset(x)(NH_(4)^(+))+underset(x)(OH^(-))`
`K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])= (X^(2))/((1-X))=X^(2)` , `(.: 1-X ~~1)`
`:. X= [OH^(-)]= sqrt(K_(b))=sqrt(1.8xx10^(-5))`
`= 4.24xx10^(-3)M`