Kohlrausch's Law
Independent ionic contributions at infinite dilution
Lesson 2089 of 4,500 · Electrochemistry
Learning objectives
- Apply limiting ionic molar conductivity sums
- Infer a weak electrolyte's limiting molar conductivity
Introduction
At very low concentration, ions are far enough apart that their motions become approximately independent. Kohlrausch's law expresses the limiting molar conductivity of an electrolyte as the sum of contributions from the ions produced by one formula unit. This principle gives a practical route to limiting values for weak electrolytes, whose steep dilution curves resist simple extrapolation.
Core explanation
For a binary 1:1 electrolyte AB that dissociates into A⁺ and B⁻, Λ°m(AB) = λ°(A⁺) + λ°(B⁻). Each λ° is a limiting ionic molar conductivity at the specified solvent and temperature. For CaCl₂, one formula unit produces one Ca²⁺ and two Cl⁻, so Λ°m(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻). The coefficients count ions from the chemical formula. The ionic charge already influences each ion's contribution, so do not insert an extra factor of two just because calcium has charge +2.
The law applies at the limiting condition of infinite dilution, where ion interactions that complicate finite-concentration behavior become negligible in the extrapolated model. It does not state that the measured molar conductivity of any concentrated mixture is an exact sum of fixed ionic values. Ionic mobilities depend on temperature and solvent, and real solution interactions change them with concentration. Tables therefore need matching conditions.
An elegant calculation derives Λ°m for a weak electrolyte from strong-electrolyte limiting data. For acetic acid, combine HCl, sodium acetate, and NaCl: Λ°m(CH₃COOH) = Λ°m(HCl) + Λ°m(CH₃COONa) − Λ°m(NaCl). Expanding ionic contributions shows why: [λ°(H⁺)+λ°(Cl⁻)] + [λ°(Na⁺)+λ°(CH₃COO⁻)] − [λ°(Na⁺)+λ°(Cl⁻)] leaves λ°(H⁺)+λ°(CH₃COO⁻). It is algebraic cancellation of ion contributions, not a claim that these three solutions are physically mixed to manufacture acetic acid.
Once a weak electrolyte's Λ°m is known, the ratio Λm/Λ°m can approximate its degree of ionization under an introductory idealized treatment. This use assumes the fully ionized ions in a dilute weak-electrolyte solution have approximately the limiting ionic mobility. At higher concentration, activity and mobility effects complicate the interpretation. The law is a foundation for such estimates but does not remove the need to state assumptions.
The limiting symbol ° here marks infinite-dilution molar conductivity at a specified temperature, not a standard-state cell potential. The two subjects share electrochemical measurements but represent different physical quantities. Conductivity is measured in S cm² mol⁻¹ or S m² mol⁻¹, whereas electrode potential is measured in volts.
Step-by-step reasoning
1. Write the electrolyte's dissociation and count each ionic species. 2. Add limiting ionic contributions with stoichiometric coefficients. 3. For a weak electrolyte, choose strong electrolytes whose unwanted ions cancel. 4. Keep temperature, solvent, and conductivity units consistent.
Visual explanation
Draw three ionic-contribution bars for HCl, sodium acetate, and NaCl. Add the first two bars and subtract the third, crossing out Na⁺ and Cl⁻ to leave H⁺ plus acetate.
Real-world analogy
If two invoices each contain a shared service charge, subtracting an invoice for the shared items isolates the desired components. Ionic-contribution algebra similarly cancels ions that are not part of the target electrolyte.
Real-world example
A weak-acid dilution curve becomes steep near low concentration. A laboratory can use well-characterized strong-electrolyte limiting data to calculate the acid's limiting molar conductivity more reliably than by extending a short steep curve.
Why?
Why do ionic contributions add at infinite dilution? Each ion's migration has little interaction with distant ions in the limiting model, so the separate current contributions combine approximately independently.
Common misconception
“CaCl₂ has two chloride ions, so every ionic term should be doubled.” Only the chloride contribution gets coefficient two; one calcium ion comes from each CaCl₂ formula unit.
Worked example
Suppose at one temperature Λ°m(HCl) = 426, Λ°m(CH₃COONa) = 91, and Λ°m(NaCl) = 126 S cm² mol⁻¹. Then Λ°m(CH₃COOH) = 426 + 91 − 126 = 391 S cm² mol⁻¹. This result represents the limiting sum of H⁺ and acetate contributions. The numerical values are illustrative; use the values supplied for the same conditions in an actual problem.
Quick check
1. Write the ionic-contribution expression for Λ°m(MgCl₂). Answer: λ°(Mg²⁺) + 2λ°(Cl⁻).
Exam focus
Show the ion cancellation explicitly for weak electrolytes. Apply formula-unit coefficients, and avoid mixing finite-concentration Λm data with limiting Λ°m values.
Advanced insight
Limiting ionic contributions also help estimate transport properties of ions that cannot be measured in isolation because electroneutral solutions necessarily contain counterions. Charge balance is preserved throughout measurement.
Summary
Kohlrausch's law adds ionic limiting contributions according to electrolyte stoichiometry. Suitable sums and differences of strong-electrolyte limits can yield a weak electrolyte's otherwise hard-to-extrapolate limit.
Practice questions
1. Write Λ°m(K₂SO₄) using ionic contributions. Answer: 2λ°(K⁺) + λ°(SO₄²⁻). 2. Why is the law not exact at arbitrary concentration? Answer: Interionic effects alter ionic mobility away from the infinite-dilution limit. 3. In the acetic-acid combination, which contributions cancel? Answer: Sodium and chloride limiting ionic contributions cancel.