- Home
- Standard 11
- Chemistry
6-2.Equilibrium-II (Ionic Equilibrium)
medium
$0.1 \,\,M$ $N{H_3}$ के जलीय विलयन की $pH$ है $({K_b} = 1.8 \times {10^{ - 5}})$
A
$11.13$
B
$12.5$
C
$13.42$
D
$11.55$
Solution
(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$
$pOH = \log 10\frac{1}{{1.34 \times {{10}^{ – 4}}}}$; $pOH = 2.87$
$pH + pOH = 14$; $pH + 2.87 = 14$
$pH = 14 – 2.87$; $pH = 11.13$
Standard 11
Chemistry
