JEE Advance - Chemistry (2023 - Paper 1 Online - No. 5)
Plotting $1 / \Lambda_{\mathrm{m}}$ against $\mathrm{c} \Lambda_{\mathrm{m}}$ for aqueous solutions of a monobasic weak acid $(\mathrm{HX})$ resulted in a straight line with $\mathrm{y}$-axis intercept of $\mathrm{P}$ and slope of $\mathrm{S}$. The ratio $\mathrm{P} / \mathrm{S}$ is
$$ \begin{aligned} & {\left[\Lambda_{\mathrm{m}}=\right.\text { molar conductivity }} \\\\ & \Lambda_{\mathrm{m}}^{\mathrm{o}}=\text { limiting molar conductivity } \\\\ & \mathrm{c}=\text { molar concentration } \\\\ & \left.\mathrm{K}_{\mathrm{a}}=\text { dissociation constant of } \mathrm{HX}\right] \end{aligned} $$
$$ \begin{aligned} & {\left[\Lambda_{\mathrm{m}}=\right.\text { molar conductivity }} \\\\ & \Lambda_{\mathrm{m}}^{\mathrm{o}}=\text { limiting molar conductivity } \\\\ & \mathrm{c}=\text { molar concentration } \\\\ & \left.\mathrm{K}_{\mathrm{a}}=\text { dissociation constant of } \mathrm{HX}\right] \end{aligned} $$
$\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}}$
$\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}} / 2$
$2 \mathrm{~K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}}$
$1 /\left(\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}}\right)$
Explanation
For weak acid, $\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}$
$$ \begin{aligned} & \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \Rightarrow \mathrm{K}_{\mathrm{a}}(1-\alpha)=\mathrm{C} \alpha^2 \\\\ & \Rightarrow \mathrm{K}_{\mathrm{a}}\left(1-\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}\right)=\mathrm{C}\left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}\right)^2 \\\\ & \Rightarrow \mathrm{K}_{\mathrm{a}}-\frac{\Lambda_{\mathrm{m}} \mathrm{K}_{\mathrm{a}}}{\Lambda_0}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}^2}{\left(\Lambda_0\right)^2} \end{aligned} $$
Divide by ' $\Lambda_{\mathrm{m}}$ '
$$ \begin{aligned} & \Rightarrow \frac{\mathrm{K}_{\mathrm{a}}}{\Lambda_{\mathrm{m}}}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}}{\left(\Lambda_0\right)^2}+\frac{\mathrm{K}_{\mathrm{a}}}{\Lambda_0} \\\\ & \Rightarrow \frac{1}{\Lambda_{\mathrm{m}}}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}}{\mathrm{K}_{\mathrm{a}}\left(\Lambda_0\right)^2}+\frac{1}{\Lambda_0} \end{aligned} $$
Plot $\frac{1}{\Lambda_{\mathrm{m}}}$ vs $\mathrm{C} \Lambda_{\mathrm{m}}$ has
Slope $=\frac{1}{\mathrm{~K}_{\mathrm{a}}\left(\Lambda_0\right)^2}=\mathrm{S}$
$\mathrm{y}$-intercept $=\frac{1}{\Lambda_0}=\mathrm{P}$
Then, $\frac{\mathrm{P}}{\mathrm{S}}=\frac{\frac{1}{\Lambda_0}}{\frac{1}{\mathrm{~K}_{\mathrm{a}}\left(\Lambda_0\right)^2}}=\mathrm{K}_{\mathrm{a}} \Lambda_0$
$$ \begin{aligned} & \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \Rightarrow \mathrm{K}_{\mathrm{a}}(1-\alpha)=\mathrm{C} \alpha^2 \\\\ & \Rightarrow \mathrm{K}_{\mathrm{a}}\left(1-\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}\right)=\mathrm{C}\left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}\right)^2 \\\\ & \Rightarrow \mathrm{K}_{\mathrm{a}}-\frac{\Lambda_{\mathrm{m}} \mathrm{K}_{\mathrm{a}}}{\Lambda_0}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}^2}{\left(\Lambda_0\right)^2} \end{aligned} $$
Divide by ' $\Lambda_{\mathrm{m}}$ '
$$ \begin{aligned} & \Rightarrow \frac{\mathrm{K}_{\mathrm{a}}}{\Lambda_{\mathrm{m}}}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}}{\left(\Lambda_0\right)^2}+\frac{\mathrm{K}_{\mathrm{a}}}{\Lambda_0} \\\\ & \Rightarrow \frac{1}{\Lambda_{\mathrm{m}}}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}}{\mathrm{K}_{\mathrm{a}}\left(\Lambda_0\right)^2}+\frac{1}{\Lambda_0} \end{aligned} $$
Plot $\frac{1}{\Lambda_{\mathrm{m}}}$ vs $\mathrm{C} \Lambda_{\mathrm{m}}$ has
Slope $=\frac{1}{\mathrm{~K}_{\mathrm{a}}\left(\Lambda_0\right)^2}=\mathrm{S}$
$\mathrm{y}$-intercept $=\frac{1}{\Lambda_0}=\mathrm{P}$
Then, $\frac{\mathrm{P}}{\mathrm{S}}=\frac{\frac{1}{\Lambda_0}}{\frac{1}{\mathrm{~K}_{\mathrm{a}}\left(\Lambda_0\right)^2}}=\mathrm{K}_{\mathrm{a}} \Lambda_0$
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