The \(\mathrm{pH}\) of monoacidic weak base is 10.9. Calculate the percent dissociation in \(0.02 \mathrm{M}\) solution.

Monoacidic weak base \(\mathrm{BOH}\)
\(
\begin{aligned}
& \mathrm{BOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{B}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} \\
& \mathrm{C} \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha \\
& \mathrm{C} \alpha \quad- \\
& \mathrm{pH}=10.9 \\
& \mathrm{pH}+\mathrm{pOH}=14 \\
& \mathrm{pOH}=14-10.9=3.1 \\
& \mathrm{pOH}=-\log _{10}\left[\mathrm{OH}^{-}\right] \\
& {\left[\mathrm{OH}^{-}\right]=10^{-\mathrm{pOH}}} \\
& =10^{-3.1}=10^{-4}+0.9 \\
& =10^{-4} \times 10^{0.9} \\
& {\left[\mathrm{OH}^{-}\right]=7.9 \times 10^{-4} \mathrm{M}} \\
& {\left[\mathrm{OH}^{-}\right]=\mathrm{C}_\alpha} \\
& \alpha=\frac{7.9 \times 10^{-4}}{0.02} \\
& \alpha=3.95 \times 10^{-2} \\
& \% \alpha=3.95 \times 10^{-2} \times 100 \\
& =3.95 \%
\end{aligned}
\)