Question

Explain Kohlrausch Law. Limiting molar conductances (\( \lambda^0 \)) of \( \text{Ca}^{2+} \) and \( \text{Cl}^- \) ions in water at 298 K are 119.0 S cm\(^2\) mol\(^{-1}\) and 76.3 S cm\(^2\) mol\(^{-1}\) respectively. Calculate \( \lambda^0_m \) of \( \text{CaCl}_2 \).

Hint:

Kohlrausch's Law applies to electrolytes in dilute solutions. It allows us to determine the conductivity of an electrolyte by summing the conductivities of its ions.

Answer 1

Kohlrausch's Law:
Kohlrausch's Law of limiting molar conductivity states that the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductances of the individual ions that make up the electrolyte. Mathematically, it is expressed as: \[ \lambda^0_m = \lambda^0_{\text{cation}} + \lambda^0_{\text{anion}} \] Where: - \( \lambda^0_m \) is the limiting molar conductivity of the electrolyte,
- \( \lambda^0_{\text{cation}} \) is the limiting molar conductivity of the cation,
- \( \lambda^0_{\text{anion}} \) is the limiting molar conductivity of the anion.
For \( \text{CaCl}_2 \), the cation is \( \text{Ca}^{2+} \) and the anion is \( \text{Cl}^- \). From the given data:
- \( \lambda^0_{\text{Ca}^{2+}} = 119.0 \, \text{S} \, \text{cm}^{-1} \, \text{mol}^{-1} \),
- \( \lambda^0_{\text{Cl}^-} = 76.3 \, \text{S} \, \text{cm}^{-1} \, \text{mol}^{-1} \).
Using Kohlrausch's Law: \[ \lambda^0_m(\text{CaCl}_2) = \lambda^0_{\text{Ca}^{2+}} + \lambda^0_{\text{Cl}^-} \] Substituting the values: \[ \lambda^0_m(\text{CaCl}_2) = 119.0 + 76.3 = 195.3 \, \text{S} \, \text{cm}^{-1} \, \text{mol}^{-1} \] Thus, the limiting molar conductivity of \( \text{CaCl}_2 \) is 195.3 S cm\(^{-1}\) mol\(^{-1}\).