Question

Define molar conductivity of a solution of an electrolyte and explain Kohlrausch law.

Hint:

Molar conductivity \(\Lambda_m = \kappa \times 1000 / c\); Kohlrausch: limiting molar conductivity is the sum of independent cation and anion contributions.

Answer 1

Step 1: Molar conductivity (definition).
Molar conductivity \(\Lambda_m\) of an electrolytic solution is the conductance of all the ions produced by one mole of the electrolyte, when the solution is placed between two electrodes that are 1 cm apart and large enough to hold the whole volume.
It is related to conductivity \(\kappa\) and molar concentration \(c\) (in mol L\(^{-1}\)) by:
\[ \Lambda_m = \frac{\kappa \times 1000}{c} \]
Its SI-based unit is S cm\(^2\) mol\(^{-1}\). \(\Lambda_m\) increases on dilution because the total volume containing one mole of electrolyte increases.

Step 2: Kohlrausch law of independent migration of ions.
At infinite dilution (when dissociation is complete), each ion migrates independently of the other ions. So the limiting molar conductivity \(\Lambda_m^0\) of an electrolyte is the sum of the individual limiting molar conductivities of its cation and anion:
\[ \Lambda_m^0 = \nu_+ \lambda_+^0 + \nu_- \lambda_-^0 \]
where \(\lambda_+^0\) and \(\lambda_-^0\) are the limiting molar conductivities of the cation and anion, and \(\nu_+, \nu_-\) are the numbers of each ion produced by one formula unit.

Step 3: Applications.
(a) It gives \(\Lambda_m^0\) of weak electrolytes (e.g. \(CH_3COOH\)) which cannot be found by extrapolation, using values of strong electrolytes.
(b) It gives the degree of dissociation \(\alpha = \dfrac{\Lambda_m}{\Lambda_m^0}\) and hence the dissociation constant of a weak electrolyte.