Question

pH of 0.1 M aqueous $N{H}_{4}OH$ will be:
[Given : $K_{a\left[N{H}_4^{+}\right]}={10}^{-9}$]

• A. 3
• B. 11
• C. 3.5
• D. 10.5

Answer 1

The correct option is B 11

$\because N{H}_{4}OH$ is a weak base hence, ionisation follows as
$N{H}_{4}OH\to N{H}_4^{+}+O{H}^{-}N{H}_4^{+}\to N{H}_3+H^{+}\because K_a.K_b=K_w={10}^{-14}{K}_{b}ofN{H}_{4}OH\to \frac{K_w}{K_a}={10}^{-5}p{K}_b=-\log \left(K_b\right)=-\log \left({10}^{-5}\right)=5pOH=\frac{1}{2}(p{K}_b-\log C)=\frac{1}{2}[5-\log (0.1\left)\right]=\frac{1}{2}[5+1]pOH=3\because pH+pOH=14\therefore pH=14-3=11$