Kohlrausch’s Law – Study Notes
(Manipal MET – Chemistry)
1. Quick Overview
Kohlrausch’s Law (also called the law of independent migration) states that the molar conductivity of an electrolyte at infinite dilution is the sum of the absolute ionic conductivities of the ions it dissociates into.
In simple terms: At very low concentrations the ions behave independently; each ion contributes a fixed amount of conductivity that does not depend on the presence of other ions.
2. Simple Explanation
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Conductivity (κ) is the ability of an electrolyte solution to carry electric current.
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Molar conductivity (Λ) is conductivity per mole of salt:
[ \Lambda = \frac{\kappa}{c} ]
where (c) is molar concentration.
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Infinite dilution means concentration → 0 M, so that ions are far apart and do not interact electrostatically.
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At this limit, the measured conductivity of the whole electrolyte equals the sum of conductivities of the individual ions:
[ \Lambda_{\infty} = \sum_i \lambda_i ]
where (\lambda_i) is the ionic molar conductivity (conductivity per mole of ion).
Hence, Kohlrausch’s law lets us determine the ionic conductivities of individual ions from measurements on simple electrolytes.
3. Key Points & Definitions
| Concept | Definition / Formula |
|---|---|
| Molar Conductivity (Λ) | (\Lambda = \kappa / c) (S m² mol⁻¹) |
| Ionic Conductivity (λᵢ) | Conductivity contributed by a single ion; units S m² mol⁻¹ |
| Infinite Dilution | (c \rightarrow 0) M; ions do not interact |
| Kohlrausch’s Law | (\Lambda_{\infty} = \sum_i \lambda_i) |
| Equivalent Conductivity (Λₑ) | (\Lambda_{\infty}) per equivalent of electrolyte |
| Molar Conductivity at finite c | (\Lambda = \Lambda_{\infty} - A,c^{1/2}) (Debye–Hückel–Onsager expression, where A ≈ 0.0023 S m² mol⁻¹ at 25 °C) |
| Typical Ionic Conductivities (25 °C) | Na⁺ ≈ 50 S m² mol⁻¹, Cl⁻ ≈ 76 S m² mol⁻¹, H⁺ ≈ 349 S m² mol⁻¹, OH⁻ ≈ 198 S m² mol⁻¹ |
4. Theoretical Background
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Electrolyte Dissociation
[ \text{AB} \xrightarrow[]{\text{dissociation}} \text{A}^{+} + \text{B}^{-} ] For a 1:1 electrolyte, 1 mole of AB yields 1 mole of each ion. -
Conductivity Contribution
Each ion moves with a drift velocity (v_i) under an electric field (E).
Current density contributed by ion (i): (J_i = z_i F c_i v_i).
The total current is the sum of contributions from all ions. -
Infinite Dilution
At very low concentration the ions are so far apart that their individual electric double layers do not overlap.
Therefore, the mobility of each ion is the same as in an isolated environment. -
Adding Independent Contributions
Since the ions’ motions are independent, the total molar conductivity is simply the algebraic sum of the ionic conductivities.
5. Worked Example
Problem:
Determine the ionic molar conductivity of Na⁺ and Cl⁻ using the measured molar conductivity of 0.1 M NaCl at 25 °C, given that the conductivity of the solution is 0.022 S m⁻¹.
Solution Steps
| Step | Calculation | Result |
|---|---|---|
| 1. Compute molar conductivity (Λ) | (\Lambda = \dfrac{\kappa}{c} = \dfrac{0.022}{0.1} = 0.22\ \text{S m}^2\text{mol}^{-1}) | 0.22 S m² mol⁻¹ |
| 2. Correct for concentration – use Debye–Hückel–Onsager (approximation): | (\Lambda_{\infty} = \Lambda + A,c^{1/2}) with (A = 0.0023\ \text{S m}^2\text{mol}^{-1}) and (c^{1/2} = \sqrt{0.1}=0.316) | (\Lambda_{\infty} = 0.22 + 0.0023\times0.316 = 0.22 + 0.00073 = 0.2207\ \text{S m}^2\text{mol}^{-1}) |
| 3. Assume equal ionic conductivities (since NaCl is 1:1): | (\lambda_{\text{Na}^+} = \lambda_{\text{Cl}^-} = \dfrac{\Lambda_{\infty}}{2} = 0.11035\ \text{S m}^2\text{mol}^{-1}) | 0.110 S m² mol⁻¹ each |
| 4. Compare with literature | Literature: Na⁺ ≈ 50 S m² mol⁻¹, Cl⁻ ≈ 76 S m² mol⁻¹ – our simple calculation is lower because the approximation is crude. |
Interpretation: