Q7 / 100 · BITSAT 2023 (Memory Based Paper 1) · Chemistry
The conductivity of a weak acid HA of concentration $0.001 \mathrm{~mol} \mathrm{~L}^{-1}$ is $2.0 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$.
If $\Lambda_{\mathrm{m}}^{\circ}(\mathrm{HA})=190 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, the ionization constant $\left(\mathrm{K}_{\mathrm{a}}\right)$ of $\mathrm{HA}$ is equal to _______ x $10^{-6}$.
If $\Lambda_{\mathrm{m}}^{\circ}(\mathrm{HA})=190 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, the ionization constant $\left(\mathrm{K}_{\mathrm{a}}\right)$ of $\mathrm{HA}$ is equal to _______ x $10^{-6}$.
- A24
- B48
- Correct answerC12
- D45
Solution
$\Lambda_{\mathrm{m}}=1000 \times \frac{\mathrm{K}}{\mathrm{M}}$
$=1000 \times \frac{2 \times 10^{-5}}{0.001}=20 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\Rightarrow \alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{20}{190}=\left(\frac{2}{19}\right)$
$\underset{0.001(1-\alpha)}{\mathrm{HA}} \rightleftharpoons \underset{0.001 \alpha}{\mathrm{H}^{+}}+\underset{0.001 \alpha}{\mathrm{A}^{-}}$
$K_{\mathrm{a}}=0.001\left(\frac{\alpha^2}{1-\alpha}\right)=\frac{0.001 \times\left(\frac{2}{19}\right)^2}{1-\left(\frac{2}{19}\right)}$
$=12.3 \times 10^{-6}$
$=1000 \times \frac{2 \times 10^{-5}}{0.001}=20 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\Rightarrow \alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{20}{190}=\left(\frac{2}{19}\right)$
$\underset{0.001(1-\alpha)}{\mathrm{HA}} \rightleftharpoons \underset{0.001 \alpha}{\mathrm{H}^{+}}+\underset{0.001 \alpha}{\mathrm{A}^{-}}$
$K_{\mathrm{a}}=0.001\left(\frac{\alpha^2}{1-\alpha}\right)=\frac{0.001 \times\left(\frac{2}{19}\right)^2}{1-\left(\frac{2}{19}\right)}$
$=12.3 \times 10^{-6}$
Hence, the correct answer is option: 'C'.