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Answer the following · Q1

Q.i. What is Kohrausch law of independent migration of ions? How is it useful in obtaining molar conductivity at zero concentration of a weak electrolyte ? Explain with an example.

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Step 1. Kohlrausch's law of independent migration of ions states that at infinite dilution, each ion migrates independently of the co-ion it is associated with, and contributes a fixed share to the total molar conductivity of the electrolyte irrespective of the other ion present.

Step 2. Mathematically, Λ0=n+λ+0+n−λ−0\Lambda^0=n_+\lambda^0_++n_-\lambda^0_-, where λ+0,λ−0\lambda^0_+,\lambda^0_- are the limiting ionic molar conductivities of the cation and anion, and n+,n−n_+,n_- their numbers in the formula.

Step 3. For a STRONG electrolyte, the Λ\Lambda vs c\sqrt{c} plot is linear and can be extrapolated to c=0 to get Λ0\Lambda^0 directly -- but for a WEAK electrolyte this plot is sharply non-linear near c=0, so extrapolation fails.

Step 4. Kohlrausch's law solves this: since ions migrate independently, a weak electrolyte's Λ0\Lambda^0 can be assembled from the (extrapolation-obtainable) Λ0\Lambda^0 values of related STRONG electrolytes whose common ions cancel algebraically.

Step 5. Worked example: for acetic acid, Λ0(CH3COOH)=λH+0+λCH3COO−0\Lambda^0(CH_3COOH)=\lambda^0_{H^+}+\lambda^0_{CH_3COO^-}. Using Λ0(HCl)=λH+0+λCl−0\Lambda^0(HCl)=\lambda^0_{H^+}+\lambda^0_{Cl^-}, Λ0(CH3COONa)=λNa+0+λCH3COO−0\Lambda^0(CH_3COONa)=\lambda^0_{Na^+}+\lambda^0_{CH_3COO^-} and Λ0(NaCl)=λNa+0+λCl−0\Lambda^0(NaCl)=\lambda^0_{Na^+}+\lambda^0_{Cl^-}: adding the first two and subtracting the third cancels λNa+0\lambda^0_{Na^+} and λCl−0\lambda^0_{Cl^-}, leaving exactly Λ0(CH3COOH)=Λ0(HCl)+Λ0(CH3COONa)−Λ0(NaCl)\Lambda^0(CH_3COOH)=\Lambda^0(HCl)+\Lambda^0(CH_3COONa)-\Lambda^0(NaCl).

✓Final answer

Kohlrausch's law: ∧0 = n+λ0+ + n-λ0-. For acetic acid: ∧0(CH3COOH) = ∧0(HCl) + ∧0(CH3COONa) - ∧0(NaCl), since the common Na+ and Cl- terms cancel.

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