Search any question & find its solution
Question:
Answered & Verified by Expert
An organic monobasic acid has dissociation constant $2.25 \times 10^{-6}$. What is percent dissociation in its $0.01 \mathrm{M}$ solution?
Options:
Solution:
1359 Upvotes
Verified Answer
The correct answer is:
$1.5 \%$
$\mathrm{K}_{\mathrm{a}}=2.25 \times 10^{-6}$
For a weak monobasic acid,
$\begin{aligned}
\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{c}}}=\sqrt{\frac{2.25 \times 10^{-6}}{0.01}}=\sqrt{2.25 \times 10^{-4}}=0.015 \\
\therefore \quad \text { Percent dissociation }=\alpha \times 100 \\
=0.015 \times 100=1.5 \%
\end{aligned}$
For a weak monobasic acid,
$\begin{aligned}
\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{c}}}=\sqrt{\frac{2.25 \times 10^{-6}}{0.01}}=\sqrt{2.25 \times 10^{-4}}=0.015 \\
\therefore \quad \text { Percent dissociation }=\alpha \times 100 \\
=0.015 \times 100=1.5 \%
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.