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What is the molar conductivity of $\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}$ at infinite dilution?
Given that,
$\wedge_{\mathrm{m}}^{\mathrm{o}}\left(\mathrm{CH}_3 \mathrm{CO}_2\right)_2 \mathrm{Ba}=\mathrm{x}_1 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\begin{aligned} & \wedge_{\mathrm{m}}^{\mathrm{o}}\left(\mathrm{BaCl}_2\right)=\mathrm{x}_2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \wedge_{\mathrm{m}}^{\mathrm{o}}(\mathrm{HCl})=\mathrm{x}_3 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}$
Options:
Given that,
$\wedge_{\mathrm{m}}^{\mathrm{o}}\left(\mathrm{CH}_3 \mathrm{CO}_2\right)_2 \mathrm{Ba}=\mathrm{x}_1 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\begin{aligned} & \wedge_{\mathrm{m}}^{\mathrm{o}}\left(\mathrm{BaCl}_2\right)=\mathrm{x}_2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \wedge_{\mathrm{m}}^{\mathrm{o}}(\mathrm{HCl})=\mathrm{x}_3 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}$
Solution:
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Verified Answer
The correct answer is:
$\frac{x_1-x_2}{2}+x_3$
$\begin{aligned} & \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} ; \mathrm{HCl} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-} \\ & \left(\mathrm{CH}_3 \mathrm{COO}\right)_2 \mathrm{Ba} \longrightarrow 2 \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{Ba}^{2+} \\ & \mathrm{BaCl}_2 \rightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}\end{aligned}$
$\begin{gathered}2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{CH}_3 \mathrm{COOH}}=2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{CH}_3 \mathrm{COO}^{-}}+2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{H}^{+}} \\ =\left[2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{CH}_3 \mathrm{COO}^{-}}+\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Ba}^{2+}}\right]+ \\ {\left[2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{H}^{+}}+2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Cl}^{-}}\right]-} \\ \left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Ba}^{2+}}-2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Cl}^{-}}\end{gathered}$
$\begin{aligned} & =\left(\Lambda^{\circ}{ }_{\mathrm{m}}\right)_{\left(\mathrm{CH}_3 \mathrm{COO}\right)_2 \mathrm{Ba}}+2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{HCl}}-\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{BaCl}_2} \\ & =\mathrm{x}_1+2 \mathrm{x}_3-\mathrm{x}_2 \\ & \therefore\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{CH}_3 \mathrm{COOH}}=\frac{\mathrm{x}_1-\mathrm{x}_2}{2}+\mathrm{x}_3\end{aligned}$
$\begin{gathered}2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{CH}_3 \mathrm{COOH}}=2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{CH}_3 \mathrm{COO}^{-}}+2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{H}^{+}} \\ =\left[2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{CH}_3 \mathrm{COO}^{-}}+\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Ba}^{2+}}\right]+ \\ {\left[2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{H}^{+}}+2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Cl}^{-}}\right]-} \\ \left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Ba}^{2+}}-2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{Cl}^{-}}\end{gathered}$
$\begin{aligned} & =\left(\Lambda^{\circ}{ }_{\mathrm{m}}\right)_{\left(\mathrm{CH}_3 \mathrm{COO}\right)_2 \mathrm{Ba}}+2\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{HCl}}-\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{BaCl}_2} \\ & =\mathrm{x}_1+2 \mathrm{x}_3-\mathrm{x}_2 \\ & \therefore\left(\Lambda_{\mathrm{m}}^{\circ}\right)_{\mathrm{CH}_3 \mathrm{COOH}}=\frac{\mathrm{x}_1-\mathrm{x}_2}{2}+\mathrm{x}_3\end{aligned}$
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