In a buffer solution containing equal concentration of `B^(-)` and `HB`, the `K_(b)` for `B^(-)` is `10^(-10)`. The `pH` of buffer solution is

Options

A. 10
B. 7
C. 6
D. 4

Answer

D

Explanation

Key Idea - (i) For basic buffer,
`pOH=pK_(b)+log.(["salt"])/(["base"])`
(ii) `pH+pOH = 14`
Given, `K_(b)=1xx10^(-10)`, [salt]=[base]
`pOH=-log K_(b)+log.(["salt"])/(["base"])`
`therefore pOH=-log(1xx10^(-10))+log1=10`
`pH+pOH=14 [because "concentration of "[B^(-)]=[HB]`
`pH=14-10=4`