State Kohlrausch's law. How is it useful to determine the molar conductivity of a weak electrolyte at infinite

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Q. State Kohlrausch's law. How is it useful to determine the molar conductivity of a weak electrolyte at infinite dilution?

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Concept understanding — Molar Conductivity and Kohlrausch's Law

Because different concentrations of the same electrolyte contain different numbers of ions per unit volume, raw specific conductance κ cannot fairly compare how well different electrolytes conduct. Molar conductivity, $\Lambda_m$ , fixes this by standardising to a fixed AMOUNT of electrolyte — one mole — giving $\Lambda_m\ (\text{S m}^2\text{mol}^{-1}) = \dfrac{\kappa\times10^{-3}}{M}$ ; equivalent conductance does the same using gram-equivalents and normality instead of moles and molarity.

Kohlrausch found experimentally that molar conductance always rises on dilution, following the empirical relationship $\Lambda_m = \Lambda_m^{o} - k\sqrt{C}$ for strong electrolytes — a straight line in $\Lambda_m$ vs $\sqrt{C}$ whose intercept, $\Lambda_m^{o}$ , is the limiting molar conductivity at infinite dilution, where ion-ion interactions vanish (the Debye–Hückel–Onsager equation supplies the theoretical basis for exactly why this line is straight). A weak electrolyte's plot is not linear — it stays flat then rises sharply only very close to zero concentration, because dilution there increases the actual DEGREE OF DISSOCIATION rather than merely spreading already-free ions apart — so its $\Lambda_m^{o}$ cannot be read off by extrapolation.

Kohlrausch's law of independent ionic migration solves this: at infinite dilution, each ion contributes a fixed, additive amount to $\Lambda_m^{o}$ regardless of its partner ion, so $\Lambda_m^{o}(A_xB_y) = x\lambda^{o}(A^{y+}) + y\lambda^{o}(B^{x-})$ . This single law delivers three practical results: (1) a weak electrolyte's $\Lambda_m^{o}$ can be built from three STRONG electrolytes sharing its ions (e.g. acetic acid's $\Lambda_m^o$ from HCl + CH₃COONa − NaCl); (2) the degree of dissociation at any concentration is $\alpha = \Lambda_m/\Lambda_m^{o}$ , which combines with Ostwald's dilution law to give $K_a = \dfrac{C\Lambda_m^2}{\Lambda_m^{o}(\Lambda_m^{o}-\Lambda_m)}$ ; and (3) the solubility product of a sparingly soluble 1:1 salt follows from a single conductivity measurement as $K_{sp} = \left(\dfrac{\kappa\times10^{-3}}{\Lambda_m^{o}}\right)^{2}$ .

Step 1. Kohlrausch's law states that, at infinite dilution, the limiting molar conductivity of an electrolyte equals the sum of the independent contributions of its constituent cation and anion, since ion-ion interference vanishes at that dilution: $\Lambda_m^{o}(A_xB_y) = x\lambda_m^{o}(A^{y+}) + y\lambda_m^{o}(B^{x-})$ .

Step 2. For a strong electrolyte, $\Lambda_m^{o}$ can be measured directly by plotting $\Lambda_m$ against $\sqrt{C}$ and extrapolating the (linear) plot to $\sqrt{C}=0$ . For a WEAK electrolyte, this plot is not linear — it rises sharply only very close to zero concentration — so no reliable extrapolation is possible, and $\Lambda_m^{o}$ cannot be measured directly.

Step 3. Kohlrausch's law gets around this: choose three STRONG electrolytes whose ions, taken together, exactly reconstruct the weak electrolyte's ions. For acetic acid, combine HCl, sodium acetate and NaCl: $\Lambda_m^{o}(\text{HOAc}) = \Lambda_m^{o}(\text{HCl}) + \Lambda_m^{o}(\text{CH}_3\text{COONa}) - \Lambda_m^{o}(\text{NaCl})$ — the common Na⁺ and Cl⁻ contributions cancel exactly, leaving only H⁺ and CH₃COO⁻, the ions of acetic acid.

✓Final answer

Kohlrausch's law (independent migration of ions) lets a weak electrolyte's limiting molar conductivity be calculated as an algebraic combination of three strong electrolytes' Λm° values, entirely avoiding the direct extrapolation that is impossible for a weak electrolyte's non-linear dilution curve.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$0.5\ mol\ dm^{-3}$
(a) $0.086\ S\,cm^2\,mol^{-1}$
(b) $2.88\ S\,cm^2\,mol^{-1}$
(c) $28.8\ S\,cm^2\,mol^{-1}$
(d) $11.52\ S\,cm^2\,mol^{-1}$
›Reveal solutionSolution

Molar conductivity is obtained from the specific conductance by dividing by the molar concentration (after converting volume units from litres to cubic centimetres via the factor 1000); substituting the given values directly gives the answer.

Formula: $\Lambda_m = \dfrac{\kappa \times 1000}{C}$ where $\kappa$ is the specific conductance (in $S\,cm^{-1}$ ) and $C$ is the molar concentration (in $mol\,dm^{-3} = mol\,L^{-1}$ ), giving $\Lambda_m$ in $S\,cm^2\,mol^{-1}$ .

Substituting the given values: $\kappa = 5.76\times10^{-3}\ S\,cm^{-1}$ , $C = 0.5\ mol\,dm^{-3}$ : $\Lambda_m = \dfrac{5.76\times10^{-3}\times1000}{0.5} = \dfrac{5.76}{0.5} = 11.52\ S\,cm^2\,mol^{-1}$

✓Final answer
$11.52\ S\,cm^2\,mol^{-1}$

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Q.State Kohlrausch's law. How is it useful to determine the molar conductivity of a weak electrolyte at infinite dilution?

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