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

The $\mathrm{pH}$ of monoacidic weak base is 10.9. Calculate the percent dissociation in $0.02 \mathrm{M}$ solution.
  1. $7.92 \%$
  2. $3.95 \%$
  3. $6.25 \%$
  4. $2.51 \%$

Solution

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

Asked in: MHT CET 2021 (23 Sep Shift 1)

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