Question

Determine the percentage degree of dissociation of $0.05M$ $N{H}_3$ at ${25}^{o}C$ in a solution of $pH=11$.

• A. 2%
• B. 4%
• C. 5%
• D. 10%

Answer 1

The correct option is A 2%

$\begin{array}{cccccc}N{H}_3+H_{2}O\to & N{H}_{4}OH& \rightleftharpoons & N{H}_4^{+}& +& O{H}^{-}\\ & C& & 0& & 0\\ & C-C\alpha & & C\alpha & & C\alpha \end{array}$
$\alpha =\sqrt{\frac{K_b}{C}}$

$pH=-log\left[H^{+}\right]$
$\left[H^{+}\right]={10}^{-11}$
$\left[H^{+}\right]\left[O{H}^{-}\right]=k_w={10}^{-14}$
$\left[O{H}^{-}\right]=\frac{{10}^{-14}}{{10}^{-11}}={10}^{-3}$
${10}^{-3}=C\alpha$
$\alpha =\frac{{10}^{-3}}{0.05}$
$\alpha =2\times {10}^{-2}$
%$\alpha =2\times {10}^{-2}\times 100$
%$\alpha =2$%