`BOH` is a weak base, molar concentration of `BOH` that provides a `[OH]^(-)` of `1.5xx10^(-3) M[K_(b)(BOH)=1.5xx10^(-5) M]` is

Options

A. `1.5xx10^(-5)M`
B. `0.015M`
C. `0.0015`
D. `0.15M`

Answer

D

Explanation

`BOH hArr B^(+) + OH^(-) `
`K_(b)=([B^(+)][OH^(-)])/([BOH])`
`[B^(+)]=[OH^(-)]`
`:. K_(b)=([OH^(-)]^(2))/([BOH])`
`[BOH]=([OH^(-)]^(2))/(K_(b))`
`=((1.5 xx 10^(-3))^(2))/(1.5 xx 10^(-5))=1.5 xx 10^(-1)M`
`=0.15 M`