Question

The $pH$ of a $0.02MN{H}_{4}Cl$ solution will be: [given $K_b\left(N{H}_{4}OH\right)={10}^{-5}$ and $\log 2=0.301]$.

• A. $4.65$
• B. $5.35$
• C. $4.35$
• D. $2.65$

Answer 1

Answer: (B)

$5.35$

For the salt of strong acid and weak base
$H^{+}=\sqrt{\frac{K_w\times C}{K_b}}$
$\left[H^{+}\right]=\sqrt{\frac{{10}^{-14}\times 2\times {10}^{-2}}{{10}^{-5}}}$
$-\log \left[H^{+}\right]=6-\frac{1}{2}\log 20$
$\therefore pH=5.35$.