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Concept 04 / 07Electrochemistry
Kohlrausch's Law of Independent Migration and its Applications
Kohlrausch's law of independent migration of ions, limiting molar ionic conductivities, calculation of limiting molar conductivity of weak electrolytes from strong electrolytes, and applications to degree of dissociation, dissociation constant and solubility of sparingly soluble salts.
Figure 1
Kohlrausch's law. At infinite dilution, when interionic interactions vanish, each ion migrates independently and contributes a definite amount to the molar conductivity of the electrolyte, regardless of the other ion:
Λm∘=ν+λ+∘+ν−λ−∘,
where ν+ and ν− are the numbers of cations and anions per formula unit and λ∘ are the limiting molar ionic conductivities (see figure). For CaCl2: Λm∘=λ∘(Ca2+)+2λ∘(Cl−)=119.0+2×76.3=271.6Scm2mol−1.
Evidence: the difference in Λm∘ between pairs of salts with a common ion is constant (KCl minus NaCl equals KBr minus NaBr, about 23.4).
H+ (349.6) and OH− (199.1) have exceptionally high values because they move by proton hopping (Grotthuss mechanism) through the hydrogen-bonded network of water.
Applications.
1. Λm∘ of weak electrolytes. It cannot be obtained by extrapolation, but it can be computed from strong electrolytes:
Λm∘(CH3COOH)=Λm∘(HCl)+Λm∘(CH3COONa)−Λm∘(NaCl)=425.9+91.0−126.4=390.5Scm2mol−1.
2. Degree of dissociation of a weak electrolyte: α=Λm∘Λmc.
3. Dissociation constant: Ka=1−αcα2.
4. Solubility of a sparingly soluble salt (AgCl, BaSO4): its saturated solution is so dilute that Λm≈Λm∘. Measure κ (after subtracting the conductivity of the water), then solubility s=Λm∘1000κ (molL−1), and Ksp follows.
5. Ionic product of water, using Λm∘(H2O)=λ∘(H+)+λ∘(OH−) and the conductivity of pure water.
Key Takeaways:
- Λm∘=ν+λ+∘+ν−λ−∘.
- Weak electrolytes: combine strong electrolytes (HCl + NaAc minus NaCl).
- α=Λm∘Λm; Ka=1−αcα2.
- Solubility of sparingly soluble salts: s=Λm∘1000κ.
- Pitfall: remember the stoichiometric factor, e.g. 2λ∘(Cl−) for CaCl2.
Brain-Teaser 1: The conductivity of 0.001028M acetic acid is 4.95×10−5Scm−1. Find its dissociation constant (Λm∘=390.5Scm2mol−1).
Solution: Λm=0.0010281000×4.95×10−5=48.15Scm2mol−1. α=390.548.15=0.1233. Ka=1−0.12330.001028×0.12332=1.78×10−5.
Brain-Teaser 2: Calculate Λm∘ for NH4OH given Λm∘ of Ba(OH)2, BaCl2 and NH4Cl as 457.6, 240.6 and 129.8Scm2mol−1.
Solution: Λ∘(NH4OH)=Λ∘(NH4Cl)+21Λ∘(Ba(OH)2)−21Λ∘(BaCl2)=129.8+228.8−120.3=238.3Scm2mol−1.
Brain-Teaser 3: A saturated solution of AgCl has conductivity 1.86×10−6Scm−1 after correcting for water. Given λ∘(Ag+)=61.9 and λ∘(Cl−)=76.3, find the solubility and Ksp.
Solution: Λm∘=138.2Scm2mol−1. s=138.21000×1.86×10−6=1.35×10−5molL−1. Ksp=s2=1.8×10−10.
