MCQ
$pH$ of $0.1\,\, M$ $N{H_3}$ aqueous solution is $({K_b} = 1.8 \times {10^{ - 5}})$
- ✓$11.13$
- B$12.5$
- C$13.42$
- D$11.55$
✓
Answer
Correct option: A.
$11.13$
(a) $N{H_4}OH$ $ \rightleftharpoons $ $NH_4^ + + O{H^ - }$
${K_b} = C{\alpha ^2}$ ; $\frac{{1.8 \times {{10}^{ - 5}}}}{{.1}} = {\alpha ^2}$;$\alpha = 1.34 \times {10^{ - 3}}$
$[O{H^ - }] = \alpha \;.\;C$$ = 1.34 \times {10^{ - 3}} \times .1$
$pOH = \log 10\frac{1}{{1.34 \times {{10}^{ - 4}}}}$; $pOH = 2.87$
$pH + pOH = 14$; $pH + 2.87 = 14$
$pH = 14 - 2.87$; $pH = 11.13$
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