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6-2.Equilibrium-II (Ionic Equilibrium)
medium
$0.1\,M\,N{H_3}$ विलयन की $pH$ होगी
A
$11.27$
B
$11.13$
C
$12$
D
$9.13$
Solution
$NH _3+ H _2 O \rightleftharpoons NH _4+ OH ^{-}$
$Kb =\frac{\left[ NH _4^{+}\right]\left[ OH ^{-}\right]}{\left[ NH _3\right]}$
$1.8 \times 10^{-5}=\frac{ x ^2}{0.1}$
$\Rightarrow x =\sqrt{1.8 \times 10^{-6}}$
$=1.34 \times 10^{-3}$
$\therefore\left[ OH ^{-}\right]\left[ H ^{+}\right]=1.34 \times 10^{-3}\left[ H ^{+}\right]=10^{-14}$
$\Rightarrow pH =11.13$
Standard 11
Chemistry
