Question

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

• A. 10
• B. 7
• C. 6
• D. 4

Answer 1

Answer: (D)

(i) For basic buffer,
$pOH=p{K}_b+\log \frac{[\text{ salt }]}{[\text{ base }]}$
(ii) $pH+pOH=14$
Given, $K_b=1\times 10^{-10},[$ salt $]=[$ base $]$
$pOH=-\log K_b+\log \frac{[\text{ salt }]}{[\text{ base }]}$
$\therefore pOH=-\log \left(1\times 10^{-10}\right)+\log 1=10$
$pH+pOH=14[\because \text{ concentration of }\left[B^{-}\right]=[HB]$
$pH=14-10=4$