Kohlrausch’s Law, also called the Law of Independent Migration of Ions, states that ^[1,3]

At infinite dilution, each ion contributes independently to the limiting molar conductivity of an electrolyte. 

At a given temperature and in a given solvent, this contribution is characteristic of the ion and is independent of the identity of the oppositely charged ion.

Molar Conductivity and Limiting Molar Conductivity

The molar conductivity, Λ_m, of an electrolyte is related to its conductivity, κ, and concentration. ^[1,2]

When concentration, c, is expressed in mol m^–3:

Λ_m = κ / c

The SI unit of molar conductivity is S m² mol^–1

When the numerical value of conductivity, κ, is expressed in S cm^–1 and concentration, C, in mol L^–1, the molar conductivity in S cm² mol^–1 is calculated as:

Λ_m = (1000 × κ) / C

As a solution is diluted, its molar conductivity generally increases and approaches a limiting value called the limiting molar conductivity, Λ_m°.

As c → 0, Λₘ → Λₘ°. This limiting condition is called infinite dilution.

Formula of Kohlrausch’s Law

For an electrolyte producing ν₊ cations and ν_– anions: ^[1,3]

Λ_m° = ν₊λ₊° + ν_–λ_–°

where:

Λ_m° = limiting molar conductivity of the electrolyte

λ₊° = limiting ionic molar conductivity of the cation

λ_–° = limiting ionic molar conductivity of the anion

ν₊ = number of cations produced

ν_– = number of anions produced

Examples

1. Sodium chloride (NaCl) dissociates as:

NaCl → Na⁺ + Cl^–

Therefore:

Λ_m° (NaCl) = λ° (Na⁺) + λ° (Cl^–)

2. Calcium chloride (CaCl₂) dissociates as:

CaCl₂ → Ca²⁺ + 2 Cl^–

Therefore:

Λ_m° (CaCl₂) = λ° (Ca²⁺) + 2 λ° (Cl^–)

The factor of 2 must be included because one mole of CaCl₂ produces two moles of chloride ions.

Why Is Infinite Dilution Important

At finite concentrations, ions interact with one another, which affects their movement through the solution. These interactions decrease as the solution is diluted. In the limit c → 0, each ion can therefore be assigned a characteristic limiting ionic molar conductivity. ^[1–3]

For example, the limiting contribution of Na⁺ is the same whether it is present in NaCl, NaNO₃, or another electrolyte containing Na⁺.

Applications of Kohlrausch’s Law

1. Limiting Molar Conductivity of Weak Electrolytes ^[2–4]

Weak electrolytes cannot usually be treated by simple extrapolation in the same way as strong electrolytes because their degree of ionization changes significantly with dilution. Their limiting molar conductivity can instead be calculated using suitable strong electrolytes.

For acetic acid:

Λ_m° (CH₃COOH) = λ° (H⁺) + λ° (CH₃COO^–)

Using HCl, CH₃COONa, and NaCl:

Λ_m° (CH₃COOH) = Λ_m° (HCl) + Λ_m° (CH₃COONa) − Λ_m° (NaCl)

This works because the contributions of Na⁺ and Cl^– cancel out, leaving only those of H⁺ and CH₃COO^–.

2. Degree of Ionization

For a dilute solution of a weak electrolyte, the degree of ionization can be estimated:

α = Λ_m / Λ_m°

where:

α = degree of ionization

Λ_m = molar conductivity at a particular concentration

Λ_m° = limiting molar conductivity

Percentage ionization is:

Percentage ionization = α × 100

3. Dissociation Constant

For a dilute solution of a weak monoprotic acid, using the concentration approximation:

K_a = cα² / (1 − α)

Therefore, the calculation follows this sequence:

Λ_m and Λ_m° → α → K_a

4. Solubility of Sparingly Soluble Salts

A saturated solution of a sparingly soluble salt is usually very dilute, so its molar conductivity can be approximated by its limiting value:

Λ_m ≈ Λ_m°

If κ is expressed in S cm^–1 and Λ_m° in mol L^–1, the molar solubility, s, can be estimated from:

s ≈ (1000 × κ_corr) / Λ_m°

where:

κ_corr = κ_solution − κ_solvent

Solved Example

At 25 °C, take the following limiting molar conductivities:

Λ_m° (HCl) = 426.2 S cm² mol^–1

Λ_m° (CH₃COONa) = 91.0 S cm² mol^–1

Λ_m° (NaCl) = 126.5 S cm² mol^–1

Calculate the limiting molar conductivity of acetic acid.

Using Kohlrausch’s Law:

Λ_m° (CH₃COOH) = 426.2 + 91.0 − 126.5 ≈ 390.7 S cm² mol^–1