MHT CET · Chemistry · Electrochemistry
An organic monobasic acid has dissociation constant \(2.25 \times 10^{-6}\). What is percent dissociation in its \(0.01 \mathrm{M}\) solution?
- A \(1.5 \%\)
- B \(15 \%\)
- C \(5 \%\)
- D \(0.5 \%\)
Answer & Solution
Correct Answer
(A) \(1.5 \%\)
Step-by-step Solution
Detailed explanation
\(\mathrm{K}_{\mathrm{a}}=2.25 \times 10^{-6}\)
For a weak monobasic acid,
\(\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{c}}}=\sqrt{\frac{2.25 \times 10^{-6}}{0.01}}=\sqrt{2.25 \times 10^{-4}}=\) \(0.015 \)
\( \therefore \text {Percent dissociation }=\alpha \times 100 \)
\( =0.015 \times 100=1.5 \%\)
For a weak monobasic acid,
\(\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{c}}}=\sqrt{\frac{2.25 \times 10^{-6}}{0.01}}=\sqrt{2.25 \times 10^{-4}}=\) \(0.015 \)
\( \therefore \text {Percent dissociation }=\alpha \times 100 \)
\( =0.015 \times 100=1.5 \%\)
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