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Kohlrausch's Law is considered one of the most asked concept.
9 Questions around this concept.
The molar conductivities at infinite dilution in water at are 91.0 and 426.2 S cm2/mol respectively. To calculate , the additional value required is
The equivalent conductances of two strong electrolytes at infinite dilution in (where ions move freely through a solution ) at 25°C are given below:
What additional information/quantity one needs to calculate of an aqueous solution of acetic acid?
Kohlrausch examined Λo or Λ∞ values for a number of strong electrolytes and observed certain regularities. He noted that the difference in Λo of 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.
For any electrolyte:
$
\begin{aligned}
& \mathrm{P}_{\mathrm{X}} \mathrm{QY}_{\mathrm{Y}} \rightarrow \mathrm{XP}^{+\mathrm{Y}}+\mathrm{YQ}^{-\mathrm{X}} \\
& \Lambda^{\circ}\left(P_X Q_Y\right)=\mathrm{X}_{P^{+Y}}^{\circ}+\mathrm{Y} \lambda_{Q^{-X}}^{\circ} \\
& \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}
\end{aligned}
$
$\begin{aligned} & \Lambda_{\mathrm{m}}^{\infty}\left(\mathrm{CH}_3 \mathrm{COOH}\right)=\left(\lambda_{\mathrm{H}^{+}}^{\infty}+\lambda_{\mathrm{Cl}^{-}}^{\infty}\right)+\left(\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\infty}+\lambda_{\mathrm{Na}^{+}}^{\infty}\right)-\left(\lambda_{\mathrm{Na}^{+}}^{\infty}-\lambda_{\mathrm{Cl}^{-}}^{\infty}\right) \\ & =\Lambda_{\mathrm{HCl}}^{\infty}+\Lambda_{\mathrm{CH}, \mathrm{COONa}}^{\infty}-\Lambda_{\mathrm{NaCl}}^{\infty}\end{aligned}$
Application of Kohlrausch's Law
Here $\mathrm{C}=$ Initial concentration
$
\begin{aligned}
& \alpha=\text { Degree of dissociation } \\
& \alpha=\frac{\Lambda_M}{\Lambda_{\mathrm{M}}}
\end{aligned}
$
$
\mathrm{K}=\frac{\mathrm{C} \alpha^2}{1-\alpha}=\frac{\mathrm{C} \cdot\left(\Lambda / \Lambda_{\mathrm{M}}^o\right)^2}{\left(1-\Lambda / \Lambda_{\mathrm{M}}^o\right)^2}=\frac{\mathrm{C} \Lambda_M^2}{\Lambda^0\left(\Lambda^0-\Lambda_{\mathrm{m}}\right)}
$
These are Ostwald's relations.
$\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-}$
If the solubility of AgCl be M and the K and $\Lambda^o$ have values in $\mathrm{S} \mathrm{cm}^{-1}$ and $S \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, then
$
\begin{aligned}
& \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}}
\end{aligned}
$
Here $\mathrm{M}=$ Solubility of AgCl
Solubility product:
$
\begin{gathered}
\mathrm{Ksp}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-1}\right] \\
\mathrm{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{gathered}
$
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