$\left[\right.$ આપેલ છે: ${K}_{m}=1 \times 10^{-14}$ અને $\left.{K}_{{b}}=1.8 \times 10^{-5}\right]$
So ${K}_{b}=\frac{\left[{NH}_{4}^{+}\right]\left[{HO}^{-}\right]}{\left[{NH}_{3}\right]}$
$\left[{HO}^{-}\right]=\frac{{K}_{{b}} \times\left[{NH}_{3}\right]}{\left[{NH}_{4}^{+}\right]} =1.8 \times 10^{-5} \times \frac{2}{5} \times \frac{210}{504}$
$=3 \times 10^{-6}$
$\begin{array}{*{20}{c}}
{C{H_3}C{H_2}OH{\text{ }} + {\text{ }}{H_3}{O^ + }\, \to \,C{H_3}C{H_2} - {O^ + } - H\, + {H_2}O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,H\,\,\,\,\,\,}
\end{array}$