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Electrochemistry - Kohlrausch's Law

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Kohlrausch's Law of Independent Migration of Ions states that at infinite dilution, when dissociation is complete, each ion makes a definite contribution towards the molar conductivity of the electrolyte, irrespective of the nature of the other ion with which it is associated.

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The limiting molar conductivity (Λm∘\Lambda_m^\circ) of an electrolyte AxByA_x B_y is the sum of the individual molar ionic conductivities: Λm∘=xλ+∘+yλ−∘\Lambda_m^\circ = x \lambda_+^\circ + y \lambda_-^\circ.

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Application 1: Calculation of Λm∘\Lambda_m^\circ for weak electrolytes like CH3COOHCH_3COOH or NH4OHNH_4OH, which cannot be determined by extrapolation of Λm\Lambda_m vs C\sqrt{C} graphs.

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Application 2: Calculation of the degree of dissociation (α\alpha) for weak electrolytes using the relation α=ΛmcΛm∘\alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ}, where Λmc\Lambda_m^c is the molar conductivity at concentration CC.

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Application 3: Calculation of the dissociation constant (KcK_c) for weak acids/bases: Kc=Cα21−αK_c = \frac{C \alpha^2}{1 - \alpha}.

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Application 4: Determination of the solubility product (KspK_{sp}) of sparingly soluble salts like AgClAgCl or BaSO4BaSO_4. Since the solution is saturated and extremely dilute, Λm≈Λm∘=κ×1000Solubility\Lambda_m \approx \Lambda_m^\circ = \frac{\kappa \times 1000}{\text{Solubility}}.

📐Formulae

Λm∘=ν+λ+∘+ν−λ−∘\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ

α=ΛmcΛm∘\alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ}

Kc=Cα21−α=C(Λmc)2Λm∘(Λm∘−Λmc)K_c = \frac{C \alpha^2}{1 - \alpha} = \frac{C (\Lambda_m^c)^2}{\Lambda_m^\circ (\Lambda_m^\circ - \Lambda_m^c)}

Solubility (mol/L)=κ×1000Λm∘\text{Solubility (mol/L)} = \frac{\kappa \times 1000}{\Lambda_m^\circ}

💡Examples

Problem 1:

The molar conductivities at infinite dilution for NaClNaCl, HClHCl, and CH3COONaCH_3COONa are 126.4126.4, 425.9425.9, and 91.0 S cm2 mol−191.0 \text{ S cm}^2 \text{ mol}^{-1} respectively. Calculate the limiting molar conductivity (Λm∘\Lambda_m^\circ) for acetic acid (CH3COOHCH_3COOH).

Solution:

Λm∘(CH3COOH)=λ∘(CH3COO−)+λ∘(H+)=[λ∘(CH3COO−)+λ∘(Na+)]+[λ∘(H+)+λ∘(Cl−)]−[λ∘(Na+)+λ∘(Cl−)]=Λm∘(CH3COONa)+Λm∘(HCl)−Λm∘(NaCl)=91.0+425.9−126.4=390.5 S cm2 mol−1\Lambda_m^\circ(CH_3COOH) = \lambda^\circ(CH_3COO^-) + \lambda^\circ(H^+) \\ = [\lambda^\circ(CH_3COO^-) + \lambda^\circ(Na^+)] + [\lambda^\circ(H^+) + \lambda^\circ(Cl^-)] - [\lambda^\circ(Na^+) + \lambda^\circ(Cl^-)] \\ = \Lambda_m^\circ(CH_3COONa) + \Lambda_m^\circ(HCl) - \Lambda_m^\circ(NaCl) \\ = 91.0 + 425.9 - 126.4 = 390.5 \text{ S cm}^2 \text{ mol}^{-1}.

Explanation:

Kohlrausch's law allows us to determine the limiting molar conductivity of weak electrolytes by combining the limiting molar conductivities of strong electrolytes containing the same ions.

Problem 2:

The molar conductivity of 0.025 mol L−10.025 \text{ mol L}^{-1} methanoic acid is 46.1 S cm2 mol−146.1 \text{ S cm}^2 \text{ mol}^{-1}. Calculate its degree of dissociation (α\alpha) if λ∘(H+)=349.6 S cm2 mol−1\lambda^\circ(H^+) = 349.6 \text{ S cm}^2 \text{ mol}^{-1} and λ∘(HCOO−)=54.6 S cm2 mol−1\lambda^\circ(HCOO^-) = 54.6 \text{ S cm}^2 \text{ mol}^{-1}.

Solution:

Step 1: Calculate Λm∘(HCOOH)=λ∘(H+)+λ∘(HCOO−)=349.6+54.6=404.2 S cm2 mol−1\Lambda_m^\circ(HCOOH) = \lambda^\circ(H^+) + \lambda^\circ(HCOO^-) = 349.6 + 54.6 = 404.2 \text{ S cm}^2 \text{ mol}^{-1}. Step 2: Calculate α=ΛmcΛm∘=46.1404.2≈0.114\alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ} = \frac{46.1}{404.2} \approx 0.114.

Explanation:

The degree of dissociation is found by dividing the observed molar conductivity at a specific concentration by the theoretical molar conductivity at infinite dilution (calculated using Kohlrausch's Law).