Kohlrausch's Law - Practice Questions & MCQ

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The molar conductivities $\Lambda ^{\circ}{_{NaOAc}}\: and \: \Lambda ^{\circ}{_{HCl}}$ at infinite dilution in water at $25^{\circ}C$ are 91.0 and 426.2 S cm²/mol respectively. To calculate $\Lambda ^{\circ}{_{HOAc}}$ , the additional value required is

$\Lambda ^{\circ}{_{H_{2}O}}$

$\Lambda ^{\circ}{_{KCl}}$

$\Lambda ^{\circ}{_{NaOH}}$

$\Lambda ^{\circ}{_{NaCl}}$

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The equivalent conductances of two strong electrolytes at infinite dilution in $\mathrm{H_{2}O}$ (where ions move freely through a solution ) at 25°C are given below:

$\mathrm{ \Lambda ^{\circ}{_{CH_{3}COONa}}= 91.0 ~S cm^{2}/equiv.}$

$\mathrm{\Lambda ^{\circ}{_{HCl}}=426.2 S cm^{2}/equiv.}$

What additional information/quantity one needs to calculate $\Lambda ^{\circ}$ of an aqueous solution of acetic acid?

$\Lambda ^{\circ}$ of chloroacetic acid $\mathrm{\left ( ClCH_{2}COOH \right )}$

$\mathrm{\Lambda^{\circ} \text { of } NaCl}$

$\mathrm{\Lambda^{\circ}~ of~ CH _{3} COOK}$

The limiting equivalent conductance of $H^{+}\left ( \lambda ^{\circ} {_{H^{+}}}\right )$

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Concepts Covered - 1

Kohlrausch's Law

Kohlrausch examined Λ^oor Λ^∞ values for a number of strong electrolytes and observed certain regularities. He noted that the difference in Λ^oof the electrolytes NaX and KX for any X in nearly constant.
On the basis of these observations he introduced Kohlrausch law of Independent Migration of ions. The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte that is, at infinite dilution, the contribution of any ion towards equivalent conductance is constant; it does not depend upon presence of any ion.

$\begin{array}{l}{\text {For any electrolyte: }} \\\\ {\mathrm{P}_{\mathrm{x}} \mathrm{Q}_{\mathrm{Y}}=\mathrm{XP}^{\mathrm{t} y}+\mathrm{YQ}^{-\mathrm{X}}} \\\\ {\Lambda^{\circ}=\mathrm{X} \lambda^{\circ}+\mathrm{Y} \lambda^{\circ}} \\\\ {\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+}} \\\\ {\Lambda_{\mathrm{m}}^{\infty}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=\left(\Lambda_{\mathrm{H}}^{\infty}+\Lambda_{\mathrm{Cr}}^{\infty}\right)+\left(\Lambda_{\mathrm{CH}, \mathrm{COO}^{-}}^{\infty}+\Lambda_{\mathrm{Na}^{\prime}}^{\infty}\right)} \\\\ {-\left(\Lambda_{\mathrm{Na}^{-}}^{\infty}-\Lambda_{\mathrm{Cr}}^{\infty}\right)} \\\\ {=\Lambda_{\mathrm{HCl}}^{\infty}+\Lambda_{\mathrm{CH}, \mathrm{COONa}}^{\infty}-\Lambda_{\mathrm{Nad}}^{\infty}}\end{array}$

Application of Kohlrausch's Law

Determination of Λ^oB of a weak electrolyte:
In case of weak electrolytes, the degree of ionization increases which increases the value of Λm. However, it cannot be obtained by extrapolating the graph. The limiting value, Λm^∞, for weak electrolytes can be obtained by Kohlrausch law.
To determine degree of dissociation and equilibrium constant of weak electrolyte:
$\mathrm{CH}_{3} \mathrm{COOH}\rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\quad \mathrm{H}^{+}$
C 0 0
C-C $\alpha$ C $\alpha$ C $\alpha$

$\begin{array}{l}{\text {Here } \mathrm{C}=\text { Initial concentration }} \\\\ {\alpha=\text { Degree of dissociation }} \\\\ {\alpha=\frac{\Lambda^{\circ}}{\Lambda_{\mathrm{M}}}}\end{array}$

$\begin{array}{l}{\text { Here } \Lambda^{\circ} \text { or } \Lambda^{\infty}=\text { Molar conductance at infinite }} {\text {dilution or zero concentration. }}\end{array}$

$\begin{array}{l}{\Lambda_{\mathrm{M}}=\text { Molar conductance at given conc. } \mathrm{C}} \\\\ {\mathrm{K}=\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}} \\\\ {\mathrm{K}=\frac{\mathrm{C} \alpha \cdot \mathrm{C} \alpha}{\mathrm{C}(1-\alpha)}}\end{array}$

$\begin{array}{l}{\mathrm{K}=\frac{\mathrm{C} \alpha^{2}}{1-\alpha}} \\\\ {\mathrm{K}=\frac{\mathrm{C} \cdot\left(\Lambda^{\circ} / \Lambda_{\mathrm{M}}\right)^{2}}{\left(1-\Lambda^{\circ} / \Lambda_{\mathrm{M}}\right)^{2}}} \\\\ {=\frac{\mathrm{C}\left(\Lambda^{\circ}\right)^{2}}{\Lambda^{\circ}\left(\Lambda^{\circ}-\Lambda_{\mathrm{m}}\right)}} \\\\ {\text {These are Ostwald's relations. }}\end{array}$
To determine solubility of salt and K_sp:
$\begin{array}{l}{\mathrm{AgCl}(\mathrm{s})=\mathrm{Ag}^{+}+\mathrm{Cl}^{-}} \\\\ {\Lambda^{\circ}=\frac{1000 \mathrm{K}}{\mathrm{M}}} \\\\ {\Lambda^{\circ}=\lambda^{\circ} \mathrm{Ag}^{+}+\lambda^{\circ} \mathrm{Cl}^{-}} \\\\ {\mathrm{M}=\frac{1000 \mathrm{K}}{\Lambda^{\circ}}} \\\\ {\text {Here } \mathrm{M}=\text {Solubility of } \mathrm{AgCl}}\end{array}$

$\begin{array}{l}{\text {Solubility product: }} \\\\ {\mathrm{Ksp}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-1}\right]} \\\\ {\text { As }\left[\mathrm{Ag}^{+}\right]=\left[\mathrm{Cl}^{-}\right]} \\\\ {\mathrm{Ksp}=\frac{1000 \mathrm{K}}{\Lambda^{\circ}} \times \frac{1000 \mathrm{K}}{\Lambda^{\circ}}} \\\\ {\mathrm{Ksp}=\left(1000 \mathrm{K} / \Lambda^{\circ}\right)^{2}}\end{array}$