Question

The pH of a 0.02M $N{H}_{4}Cl$ solution will be [given $K_b(N{H}_{4}OH)=10^{-5}$ and log2=0.301]

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

Answer 1

Answer: (B)

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