Question
How is the molar conductivity of a weak electrolyte at infinite dilution determined?
✓
Answer
The value of limiting molar conductivity of weak electrolytes is not obtained from the graph between $\Lambda_{ m }$ and $C ^{\frac{1}{2}}$, hence it can be calculated as follows :
For example :
$C H _3 C O O H$ (Acetic acid): To determine this, the limiting molar conductivities of $CH _3 COONa , HCl$ and NaCl are used which are already known, because all of them are strong electrolytes.
Hence, $\Lambda_{ m }^{\circ}\left( CH _3 COOH \right)=\lambda_{\left( CH _3 COO ^{-}\right)}^{\circ}+\lambda_{\left( H ^{+}\right)}^{\circ}$
Acetate ion $\left( CH _3 COO ^{-}\right)$is indicated as $Ac ^{-}$.
Therefore, from $NaCl , HCl$ and $CH _3 COONa$
$
\begin{aligned}
\Lambda_{m(NaAc)}^{\circ} & =\lambda_{Ac^{+}}^{\circ}+\lambda_{Na^{+}}^{\circ} \\
\Lambda_{m(HCl)}^{\circ} & =\lambda_{H^{+}}^{\circ}+\lambda_{Cl^{-}}^{\circ} \\
\text { and } \Lambda_{m(NaCl)}^{\circ} & =\lambda_{Na^{+}}^{\circ}+\lambda_{Cl^{-}}^{\circ}
\end{aligned}
$
$\begin{aligned} \Lambda_{ m \left( CH H _3 COOH \right)}^{\circ} & =\Lambda_{ m ( HCl )}^{\circ}+\Lambda_{ m ( NaAc )}^{\circ}-\Lambda_{ m ( NaCl )}^{\circ} \\ \Lambda_{ m ( HAc )}^{\circ} & =\lambda_{ H ^{+}}^{\circ}+\lambda_{ Cl ^{-}}^{\circ}+\lambda_{ Na ^{+}}^{\circ}+\lambda_{ Ac ^{-}}^{\circ}-\lambda_{ Na ^{+}}^{\circ}-\lambda_{ Cl ^{-}}^{\circ} \\ \Lambda_{ m ( HAc )}^{\circ} & =\lambda_{\left( Ac ^{-}\right)}^{\circ}+\lambda_{ H ^{+}}^{\circ} \\ \text { or } \quad \Lambda_{ m ( HAc )}^{\circ} & =\lambda_{\left( CH _3 COO ^{-}\right)}^{\circ}+\lambda_{ H ^{+}}^{\circ}\end{aligned}$
For example :
$C H _3 C O O H$ (Acetic acid): To determine this, the limiting molar conductivities of $CH _3 COONa , HCl$ and NaCl are used which are already known, because all of them are strong electrolytes.
Hence, $\Lambda_{ m }^{\circ}\left( CH _3 COOH \right)=\lambda_{\left( CH _3 COO ^{-}\right)}^{\circ}+\lambda_{\left( H ^{+}\right)}^{\circ}$
Acetate ion $\left( CH _3 COO ^{-}\right)$is indicated as $Ac ^{-}$.
Therefore, from $NaCl , HCl$ and $CH _3 COONa$
$
\begin{aligned}
\Lambda_{m(NaAc)}^{\circ} & =\lambda_{Ac^{+}}^{\circ}+\lambda_{Na^{+}}^{\circ} \\
\Lambda_{m(HCl)}^{\circ} & =\lambda_{H^{+}}^{\circ}+\lambda_{Cl^{-}}^{\circ} \\
\text { and } \Lambda_{m(NaCl)}^{\circ} & =\lambda_{Na^{+}}^{\circ}+\lambda_{Cl^{-}}^{\circ}
\end{aligned}
$
$\begin{aligned} \Lambda_{ m \left( CH H _3 COOH \right)}^{\circ} & =\Lambda_{ m ( HCl )}^{\circ}+\Lambda_{ m ( NaAc )}^{\circ}-\Lambda_{ m ( NaCl )}^{\circ} \\ \Lambda_{ m ( HAc )}^{\circ} & =\lambda_{ H ^{+}}^{\circ}+\lambda_{ Cl ^{-}}^{\circ}+\lambda_{ Na ^{+}}^{\circ}+\lambda_{ Ac ^{-}}^{\circ}-\lambda_{ Na ^{+}}^{\circ}-\lambda_{ Cl ^{-}}^{\circ} \\ \Lambda_{ m ( HAc )}^{\circ} & =\lambda_{\left( Ac ^{-}\right)}^{\circ}+\lambda_{ H ^{+}}^{\circ} \\ \text { or } \quad \Lambda_{ m ( HAc )}^{\circ} & =\lambda_{\left( CH _3 COO ^{-}\right)}^{\circ}+\lambda_{ H ^{+}}^{\circ}\end{aligned}$
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