Calculate dissociation constant of a weak monobasic acid if it is $0.05 \%$ dissociated in 0.02 M solution.

Calculate dissociation constant of a weak monobasic acid if it is $0.05 \%$ dissociated in 0.02 M solution.
  1. $2.0 \times 10^{-9}$
  2. $3.0 \times 10^{-9}$
  3. $4.0 \times 10^{-9}$
  4. $5.0 \times 10^{-9}$

Solution

$\begin{aligned} & \mathrm{HA} \rightleftharpoons \mathrm{H}_{\text {(aq) }}^{\oplus}+\mathrm{A}_{(\mathrm{aq})}^{\Theta} \\ & \begin{aligned} & \begin{aligned} & =0.05 \%\end{aligned}=0.05 \times 10^{-2} \\ & \mathrm{~K}_{\mathrm{a}}= \alpha^2 \mathrm{C} \\ &=\left(0.05 \times 10^{-2}\right)^2 \times 0.02 \mathrm{M} \\ &=5.0 \times 10^{-9}\end{aligned}\end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 2)

Practice more Ionic Equilibria questions on Aicharya