A weak base, `B`, has basicity constant `K_(b)=2xx10^(-5)`. The `pH` of any solution in which `[B]=[BH^(+)]` is

Options

A. 4.7
B. 7.9
C. 9.3
D. 9.7

Answer

C

Explanation

`B+H_(2)O hArr BH^(+) +OH^(-)`
`K_(b)=([BH^(+)][OH^(-)])/([B])`
`[BH^(+)]=[B]("Given")`
`:. K_(b)=[OH^(-)]`
or `[OH^(-)]=2xx10^(-5)`
`[H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(2xx10^(-5))=(10^(-9))/(2)`
`pH=-log [H^(+)]=-log . (10^(-9))/(2)`
`=9+log 2 =9+0.3010=9.3010`