Question

The $pH$ of a solution of weak base at neutralization with strong acid is $8 . K_b$ for the base is

• A. $1.0 \times 10^{-4}$
• B. $1.0 \times 10^{-6}$
• C. $1.0 \times 10^{-8}$
• D. None of there

Answer 1

Answer: (B)

At half neutralization,
${\left[ B ^{\oplus}\right]=[ BOH ] ;[\text { Salt }]=[\text { base }]}$
$ pOH = p K_b+\log \frac{\left[ B ^{\oplus}\right]}{[ BOH ]}$
$ pOH = p K_b,( pH =8, pOH =6) $
$\therefore p K_b=6, K_b=1 \times 10^{-6}$