Electrochemistry · Lesson 7 of 13
Kohlrausch’s Law and Weak Electrolytes
“Combine independent ionic contributions, calculate weak-electrolyte limits and infer dissociation constants from conductivity.”
• State the independent-migration law with correct stoichiometric factors. • Construct limiting conductivity of a weak electrolyte from strong-electrolyte data. • Derive degree of dissociation and weak-acid equilibrium expressions. • Calculate Ka while identifying mobility and ideal-solution approximations.
Separate the contributions of individual ions
At very low concentration, ions are so far apart that each type contributes a characteristic amount to the limiting molar conductivity, largely independent of its counter-ion. For example, replacing Na⁺ by K⁺ produces nearly the same change whether the anion is chloride, bromide or iodide. This regularity is the basis of Kohlrausch’s law of independent migration of ions.
For NaCl, Λm° = λ°(Na⁺) + λ°(Cl⁻). For CaCl₂, two chloride ions are produced per formula unit, so Λm° = λ°(Ca²⁺) + 2λ°(Cl⁻). The charge magnitude is not an extra multiplier when the tabulated λ values are already molar ionic conductivities. For MgSO₄ one magnesium ion and one sulphate ion are counted, despite both being doubly charged.
| Ion in water at 298 K | λ° / S cm² mol⁻¹ |
|---|---|
| H⁺ | 349.6 |
| Na⁺ | 50.1 |
| K⁺ | 73.5 |
| Ca²⁺ | 119.0 |
| Mg²⁺ | 106.0 |
| OH⁻ | 199.1 |
| Cl⁻ | 76.3 |
| Br⁻ | 78.1 |
| CH₃COO⁻ | 40.9 |
| SO₄²⁻ | 160.0 |
The large H⁺ and OH⁻ values reflect exceptionally effective charge transport in water, including proton-transfer processes through the hydrogen-bonded solvent. Do not infer that bare proton size alone determines the numbers. Every ionic conductivity belongs to a specified solvent and temperature; the same ion need not have the same mobility in a different solvent.
Problem
Calculate Λm° for CaCl₂ and MgSO₄ from the ionic table.
- 1.CaCl₂ gives Ca²⁺ + 2Cl⁻, so add 119.0 + 2(76.3) = 271.6 S cm² mol⁻¹.
- 2.MgSO₄ gives Mg²⁺ + SO₄²⁻, so add 106.0 + 160.0 = 266.0 S cm² mol⁻¹.
- 3.The factors count ions per formula unit. Multiplying each divalent-ion entry by 2 again would double-count charge already reflected in its tabulated conductivity.
Find a weak-acid limit without extrapolating its curve
Ethanoic acid, also called acetic acid, is weakly ionised at ordinary concentrations, so its conductivity curve cannot be extrapolated using the strong-electrolyte straight-line rule. Instead combine limiting conductivities of strong electrolytes whose ions cancel algebraically. For HCl + CH₃COONa − NaCl, the Na⁺ and Cl⁻ contributions cancel, leaving H⁺ and CH₃COO⁻. This is an arithmetic construction of transport contributions, not an instruction to mix or subtract physical solutions.
Problem
Λm° values for HCl, sodium ethanoate and NaCl are 425.9, 91.0 and 126.4 S cm² mol⁻¹. Find the acid value.
- 1.Write each value as a sum of independent ionic contributions.
- 2.HCl + sodium ethanoate − NaCl leaves only H⁺ and ethanoate contributions.
- 3.Λm°(CH₃COOH) = 425.9 + 91.0 − 126.4 = 390.5 S cm² mol⁻¹. The result is consistent with 349.6 + 40.9 from the ionic table.
Conductivity as an estimate of degree of dissociation
Consider a dilute solution of a weak monobasic acid HA ⇌ H⁺ + A⁻. Let c be the analytical concentration of acid and α the fraction dissociated. Each mole added produces α moles each of H⁺ and A⁻. If their mobilities can be approximated by limiting values and other ionic contributions are negligible or corrected, Λm ≈ α[λ°(H⁺) + λ°(A⁻)] = αΛm°. Thus the ratio Λm/Λm° estimates α.
The equilibrium concentrations are approximately [H⁺] = cα, [A⁻] = cα and [HA] = c(1 − α), if water autoionisation and added common ions are negligible. Substitution into the concentration form of the acid equilibrium expression gives Ka,c = cα²/(1 − α). The thermodynamic Ka is dimensionless when activities are used; the concentration expression is commonly reported with mol L⁻¹ units in this approximation.
Problem
A 0.001028 M acid solution has κ = 4.95 × 10⁻⁵ S cm⁻¹ and Λm° = 390.5 S cm² mol⁻¹. Find α and Ka,c.
- 1.First calculate Λm = 1000κ/c = 1000(4.95 × 10⁻⁵)/0.001028 = 48.1518 S cm² mol⁻¹.
- 2.Then α ≈ 48.1518/390.5 = 0.12331, or 12.33%. The finite-concentration molar conductivity is much smaller than its fully ionised limit.
- 3.Ka,c ≈ 0.001028(0.12331)²/(1 − 0.12331) = 1.78 × 10⁻⁵ mol L⁻¹. This uses equilibrium concentrations inferred from transport, with mobility and ideality assumptions stated.
Problem
A 0.025 M methanoic-acid solution has Λm = 46.1 S cm² mol⁻¹. Ionic limiting values are H⁺: 349.6 and HCOO⁻: 54.6 in the same units.
- 1.The limiting acid conductivity is 349.6 + 54.6 = 404.2 S cm² mol⁻¹.
- 2.α ≈ 46.1/404.2 = 0.11405, or 11.405%.
- 3.Ka,c ≈ 0.025(0.11405)²/(1 − 0.11405) = 3.67 × 10⁻⁴ mol L⁻¹. The factor cα² follows from multiplying both ionic concentrations.
A useful water-ion construction
The sum λ°(H⁺) + λ°(OH⁻) is 548.7 S cm² mol⁻¹. It can be constructed as Λm°(HCl) + Λm°(NaOH) − Λm°(NaCl). This sum describes transport by one mole each of H⁺ and OH⁻ in the dilute-ion limit; it must not be mistaken for the molar conductivity obtained by dividing water conductivity by the roughly 55.5 M concentration of all solvent molecules.
At 25 °C, ideal pure water has [H⁺] = [OH⁻] ≈ 10⁻⁷ M. Its approximate conductivity is 10⁻⁷(349.6 + 199.1)/1000 = 5.487 × 10⁻⁸ S cm⁻¹ = 0.05487 μS cm⁻¹. This connects autoionisation, independent ionic transport and the very small measured conductivity. At extremely low added-acid concentration, this water background cannot be ignored.
If Λm/Λm° is greater than 1 by a significant amount, do not report more than 100% dissociation. Check units, temperature, background conductivity, concentration and the applicability of the mobility approximation.
Quick Quiz
Quiz
Which expression correctly gives Λm° for CaCl₂?
One formula unit gives one calcium ion and two chloride ions. Use stoichiometric ion counts.
Which combination isolates the limiting ionic contributions of ethanoic acid?
The first combination cancels sodium and chloride contributions, leaving H⁺ and ethanoate.
A weak acid has Λm = 40 and Λm° = 400 in the same units. What is α approximately?
The transport approximation gives α = 40/400 = 0.10, meaning 10% dissociation.
For HA with analytical concentration c and dissociation α, which concentration expression is correct?
The two ionic concentrations are each cα and the undissociated concentration is c(1 − α).
Why is very dilute weak-acid conductivity difficult to interpret?
The electrolyte signal becomes small enough that background conduction and experimental error matter.
Practice Problems
- Calculate Λm° for KBr and NaCl using the ionic table, and compare their difference with the component changes.
- The conductivity of 0.00241 M ethanoic acid is 7.896 × 10⁻⁵ S cm⁻¹. Use Λm° = 390.5 S cm² mol⁻¹ to find Λm, α and Ka,c.
- Construct Λm° for NaOH from HCl = 425.9, NaCl = 126.4 and the H⁺ + OH⁻ ionic sum = 548.7, all in S cm² mol⁻¹.
- Explain why molar conductivity rises more sharply on dilution for ethanoic acid than for KCl.
- A calculated dissociation fraction is 1.30. Diagnose the result before interpreting the chemistry.
1. KBr: 73.5 + 78.1 = 151.6 S cm² mol⁻¹. NaCl: 50.1 + 76.3 = 126.4 S cm² mol⁻¹. 2. The difference is 25.2. Replacing Na⁺ by K⁺ adds 23.4 and replacing Cl⁻ by Br⁻ adds 1.8. 3. These independent differences add to 25.2, illustrating the ionic-contribution idea.
Key Takeaways
• Kohlrausch’s law adds independent ionic contributions at infinite dilution. • Count ions per formula unit, not an extra factor for their charge. • Suitable strong-electrolyte combinations provide weak-electrolyte limiting values. • The ratio Λm/Λm° estimates weak-electrolyte dissociation under suitable assumptions. • For a monobasic acid, Ka,c ≈ cα²/(1 − α). • Concentration equilibrium expressions approximate activity-based thermodynamics. • Water autoionisation and contamination matter at extreme dilution. • A dissociation fraction must remain between zero and one.