Question

The pH of a 0.02M $NH_4Cl$ solution will be [given $K_b(NH_4OH) = 10^{-5}$ and log2=0.301]

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

Answer 1

Answer: (B)

5.35

Answer 2

For the salt of strong acid and weak base
$H^{+} = \sqrt{\frac{K_{W} \times C}{K_{b}}}$
$\left[H^{+}\right] = \sqrt{\frac{10^{-14} \times2 \times10^{-2}}{10^{-5}} }$
$- \log \left[H^{+}\right] = 6 - \frac{1}{2} \log 20$
$\therefore pH = 5.35$