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MHT CET · Chemistry · Ionic Equilibrium

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

  1. A \(2 \times 10^{-2}\)
  2. B \(4 \times 10^{-4}\)
  3. C \(5 \times 10^{-5}\)
  4. D \(2.5 \times 10^{-4}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(5 \times 10^{-5}\)

Step-by-step Solution

Detailed explanation

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