A monobasic acid is $5 \%$ dissociated in its 0.02 M solution. Calculate the dissociation constant of acid.

A monobasic acid is $5 \%$ dissociated in its 0.02 M solution. Calculate the dissociation constant of acid.
  1. $2 \times 10^{-2}$
  2. $4 \times 10^{-4}$
  3. $5 \times 10^{-5}$
  4. $2.5 \times 10^{-4}$

Solution

$\begin{aligned} & \alpha=\frac{\text { Percent dissociation }}{100}=\frac{5}{100}=5 \times 10^{-2} \\ & c=0.02 \mathrm{M}=2 \times 10^{-2} \mathrm{M} \end{aligned}$
Using formula, $\mathrm{K}_{\mathrm{a}}=\alpha^2 \mathrm{c}$ $\mathrm{K}_{\mathrm{a}}=\left(5 \times 10^{-2}\right)^2 \times 2 \times 10^{-2}=5.0 \times 10^{-5}$

Asked in: MHT CET 2024 (11 May Shift 1)

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