MCQ
The $pH$ of a $0.02\,M\,\,NH_4Cl$ solution will be [given $K_b\,(NH_4OH) = 10^{-5}$ and $log\,2 = 0.301$ ]
- A$2.65$
- ✓$5.35$
- C$4.35$
- D$4.65$
✓
Answer
Correct option: B.
$5.35$
b
For the salt of strong acid and weak base
For the salt of strong acid and weak base
${H^ + } = \sqrt {\frac{{{K_w} \times C}}{{{K_b}}}} $
$[{H^ + }] = \sqrt {\frac{{{{10}^{ - 14}} \times 2 \times {{10}^{ - 2}}}}{{{{10}^{ - 5}}}}} $
$ - \log [{H^ + }] = 6 - \frac{1}{2}\log \,20$
$\therefore \,pH = 5.35$
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