What is the aq. ammonia concentration of a solution prepared by dissolving 0.15 mole of${\mathrm{NH}}_4^{+}{\mathrm{CH}}_3{\mathrm{COO}}^{-}$ in $1{\mathrm{LH}}_2\mathrm{O}$?

$\left[{\mathrm{K}}_{\mathrm{a}\left({\mathrm{CH}}_3\mathrm{COOH}\right)}=1.8\times {10}^{-5}={\mathrm{K}}_{\mathrm{b}\left({\mathrm{NH}}_3\right)}\right]$

$\underset{\mathrm{c}(1-\mathrm{h})}{{\mathrm{CH}}_3{\mathrm{COONH}}_4}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \underset{\mathrm{ch}}{{\mathrm{CH}}_3\mathrm{COH}}+\underset{\mathrm{ch}}{{\mathrm{NH}}_4\mathrm{OH}}$

$\left.\begin{array}{r}\sqrt{\frac{{\mathrm{k}}_{\mathrm{u}}}{{\mathrm{k}}_{\mathrm{a}}\times {\mathrm{k}}_{\mathrm{b}}}}=\frac{\mathrm{h}}{1-\mathrm{h}}\\ \frac{{10}^{-2}}{1.8}=\frac{\mathrm{h}}{1-\mathrm{h}}\end{array}\right\}\begin{array}{r}\Rightarrow \mathrm{h}~5.55\times {10}^{-3}\\ \left[{\mathrm{NH}}_4\mathrm{OH}\right]=\mathrm{Ch}\to 0.15\end{array}$

${\mathrm{k}}_{\mathrm{h}}={\mathrm{h}}^2$

$\begin{array}{l}\Rightarrow \frac{100}{(18{)}^2}\times {10}^{-4}={\mathrm{h}}^2\\ \Rightarrow \mathrm{h}=\frac{10}{18}\times {10}^{-2}\end{array}$

$=\frac{{10}^{-1}}{18}=\frac{1}{180}$

$\left[{\mathrm{NH}}_3\right]=\mathrm{ch}=\frac{0.15}{100}\times \frac{1}{180}$

$=0.83\times {10}^{-3}=8.3\times {10}^{-4}\mathrm{M}$