MHT CET · Chemistry · Ionic Equilibrium
A monobasic acid is \(5 \%\) dissociated in its 0.02 M solution. Calculate the dissociation constant of acid.
- A \(2 \times 10^{-2}\)
- B \(4 \times 10^{-4}\)
- C \(5 \times 10^{-5}\)
- D \(2.5 \times 10^{-4}\)
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, \(\mathrm{K}_{\mathrm{a}}=\alpha^2 \mathrm{c}\)
\(\mathrm{K}_{\mathrm{a}}=\left(5 \times 10^{-2}\right)^2 \times 2 \times 10^{-2}=5.0 \times 10^{-5}\)
& \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, \(\mathrm{K}_{\mathrm{a}}=\alpha^2 \mathrm{c}\)
\(\mathrm{K}_{\mathrm{a}}=\left(5 \times 10^{-2}\right)^2 \times 2 \times 10^{-2}=5.0 \times 10^{-5}\)
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