Kohlrausch's Law of Independent Migration of Ions and Its Applications

Kohlrausch's Law states that at infinite dilution, each ion contributes independently to the total molar conductivity of an electrolyte. Therefore, the limiting molar conductivity of an electrolyte equals the sum of the limiting molar conductivities of its constituent ions.

More precisely: the molar conductivity at infinite dilution for any strong or weak electrolyte equals the sum of the molar conductivity of each ionisable cation and anion at infinite dilution, with each multiplied by the number of that ion present per formula unit.

General Expression

Λ_{m}^{°} = ν_{+} λ_{+}^{°} + ν_{-} λ_{-}^{°}

where:

Λ°_m = limiting molar conductivity of the electrolyte
λ°₊ = limiting molar conductivity of the cation
λ°₋ = limiting molar conductivity of the anion
ν₊ = number of cations per formula unit
ν₋ = number of anions per formula unit

Example 1: NaCl

NaCl → Na⁺ + Cl⁻

Λ°_m(NaCl) = λ°(Na⁺) + λ°(Cl⁻)

Example 2: Al₂(SO₄)₃

Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻

Λ°_m(Al₂(SO₄)₃) = 2λ°(Al³⁺) + 3λ°(SO₄²⁻)

Example 3: H₂CO₃

H₂CO₃ → 2H⁺ + CO₃²⁻

Λ°_m(H₂CO₃) = 2λ°(H⁺) + λ°(CO₃²⁻)

Application 1: Determination of Λ°_m of Weak Electrolytes

For weak electrolytes, Λ°_m cannot be measured directly because they do not dissociate completely even at very low concentrations — the conductivity curve does not approach a clear limiting value. Instead, Kohlrausch's Law is used indirectly via strong electrolytes.

Example: Finding Λ°_m(CH₃COOH)

Using the strong electrolytes HCl, NaCl, and CH₃COONa:

Λ°_m(HCl) = λ°(H⁺) + λ°(Cl⁻)

Λ°_m(NaCl) = λ°(Na⁺) + λ°(Cl⁻)

Λ°_m(CH₃COONa) = λ°(Na⁺) + λ°(CH₃COO⁻)

Therefore:

Λ°_m(CH₃COOH) = Λ°_m(HCl) + Λ°_m(CH₃COONa) − Λ°_m(NaCl)

This is an essential formula for numerical problems.

Application 2: Degree of Dissociation of a Weak Electrolyte

$α = \frac{Λ_{m}}{Λ_{m}^{°}}$

where Λ_m = molar conductivity at concentration C and Λ°_m = molar conductivity at infinite dilution.

Formula for Molar Conductivity

$Λ_{m} = κ \times \frac{1000}{C}$

where κ (kappa) = specific conductivity (S cm⁻¹), C = molarity (mol L⁻¹). Units of Λ_m: S cm² mol⁻¹.

Application 3: Determination of the Dissociation Constant (K_a) of a Weak Electrolyte

From Ostwald's dilution law, using α = Λ_m/Λ°_m:

$K_{a} = \frac{C α^{2}}{1 - α} = \frac{C {(\frac{Λ_{m}}{Λ_{m}^{°}})}^{2}}{1 - \frac{Λ_{m}}{Λ_{m}^{°}}}$

Application 4: Solubility of Sparingly Soluble Salts

For a salt such as AgCl, at infinite dilution:

Λ°_m = λ°(Ag⁺) + λ°(Cl⁻)

If κ is the measured specific conductivity of the saturated solution, the molar conductivity is Λ_m = κ × 1000/C. Since the concentration C equals the solubility S:

$S = \frac{κ \times 1000}{Λ_{m}^{°}}$

This is the standard method for calculating the solubility of sparingly soluble salts from conductivity measurements.

Importance of Kohlrausch's Law

Kohlrausch's law is used for:

Calculating the limiting molar conductivity of weak electrolytes indirectly
Determining the degree of dissociation (α)
Finding dissociation constants (K_a, K_b)
Calculating solubility of sparingly soluble salts
Determining ionic conductivities

Solved Numericals

Q1. Given: Λ°_m(NaCl) = 125 S cm² mol⁻¹, Λ°_m(CH₃COONa) = 90 S cm² mol⁻¹, Λ°_m(HCl) = 420 S cm² mol⁻¹. Find Λ°_m(CH₃COOH).

Λ°_m(CH₃COOH) = Λ°_m(HCl) + Λ°_m(CH₃COONa) − Λ°_m(NaCl)

= 420 + 90 − 125 = 385 S cm² mol⁻¹

Q2. Given: λ°(Na⁺) = 50 S cm² mol⁻¹, λ°(Cl⁻) = 76 S cm² mol⁻¹. Find Λ°_m(NaCl).

Λ°_m(NaCl) = λ°(Na⁺) + λ°(Cl⁻) = 50 + 76 = 126 S cm² mol⁻¹

FAQs

Q1. What are the limitations of Kohlrausch's Law?

The law is strictly valid only at infinite dilution. It is mainly applicable to aqueous solutions, does not apply where strong ion association occurs, and does not account for interionic interactions at finite concentrations.

Q2. Where is Kohlrausch's Law used?

It is used in designing batteries and electrochemical cells, in electroplating, in the determination of solubility products, and in calculating dissociation constants of weak electrolytes.

Q3. Is Kohlrausch's Law valid for only weak electrolytes?

No. The law applies to both strong and weak electrolytes. Its particular importance for weak electrolytes lies in allowing indirect determination of their limiting molar conductivity, which cannot be obtained by direct measurement.

Q4. Why is Kohlrausch's Law especially significant for weak electrolytes?

Weak electrolytes show a large increase in molar conductivity as dilution increases (because degree of ionisation increases), but they never reach a clearly measurable limiting value directly from experiment. Kohlrausch's Law provides an indirect route using strong electrolyte data.