CSIR-NET Electrochemistry — Conductance and Kohlrausch's Law
Conductance is one of the most reliably scoring corners of physical chemistry for CSIR-NET Chemical Sciences. The physics is small, the formulas are few, and almost every question reduces to converting between four quantities — resistance, conductivity, molar conductivity and degree of dissociation. What separates a correct answer from a wasted three minutes is unit discipline. This article fixes the definitions, then works three numericals end to end.
The four quantities, in order
Start from resistance and build upwards. A conductivity cell of electrode area A and separation l filled with electrolyte has resistance R.
Conductivity κ = G × (l/A) = (1/R) × cell constant (S cm⁻¹)
Molar conductivity Λm = κ/c
• with c in mol cm⁻³ → Λm in S cm² mol⁻¹
• in practice, with c in mol L⁻¹: Λm = 1000 κ / c
The factor of 1000 is the single most common source of a wrong answer in this topic. It comes from 1 L = 1000 cm³ and nothing else. Molar conductivity is conductivity normalised for how much electrolyte is present, which is why it rises on dilution while conductivity falls: κ falls because there are fewer ions per cm³, but Λm rises because each mole of electrolyte is carrying current more effectively.
Kohlrausch's two statements
Students often merge two different results under one name. Keep them separate.
A plot of Λm against √c is linear; the intercept gives Λm° directly.
(2) Law of independent migration of ions (all electrolytes): Λm° = ν₊λ₊° + ν₋λ₋°
Each ion contributes a fixed limiting conductivity, independent of its partner.
Statement (2) is what lets you obtain Λm° for a weak electrolyte, where extrapolation of a √c plot is impossible because the curve shoots up steeply near infinite dilution. The Debye–Hückel–Onsager treatment explains the √c dependence physically: K = A + BΛm°, where the two terms are the electrophoretic effect (the ionic atmosphere drags solvent the wrong way) and the relaxation effect (the atmosphere takes time to re-form behind a moving ion).
| Ion (298 K, aqueous) | λ° / S cm² mol⁻¹ | Ion | λ° / S cm² mol⁻¹ |
|---|---|---|---|
| H⁺ | 349.6 | OH⁻ | 199.1 |
| K⁺ | 73.5 | Cl⁻ | 76.3 |
| Na⁺ | 50.1 | CH₃COO⁻ | 40.9 |
| NH₄⁺ | 73.5 | NO₃⁻ | 71.4 |
Note how far H⁺ and OH⁻ sit above everything else. That is not because they are small — it is the Grotthuss mechanism, in which a proton hops along a hydrogen-bonded chain of water molecules instead of physically migrating. A question asking "why is λ°(H⁺) about five times λ°(K⁺)?" is asking for that one word.
Worked example 1 — cell constant from a KCl standard
Q. A conductivity cell filled with 0.0100 M KCl (κ = 0.001413 S cm⁻¹ at 298 K) has resistance 484.1 Ω. The same cell filled with 0.00500 M of a salt gives R = 1050 Ω. Find the cell constant and Λm of the salt.
Step 1 — cell constant. κ = (1/R) × (l/A), so
l/A = κ × R = 0.001413 × 484.1 = 0.6841 cm⁻¹.
Step 2 — conductivity of the unknown.
κ = 0.6841 / 1050 = 6.515 × 10⁻⁴ S cm⁻¹.
Step 3 — molar conductivity.
Λm = 1000 × 6.515 × 10⁻⁴ / 0.00500 = 0.6515 / 0.00500 =
130.3 S cm² mol⁻¹.
Worked example 2 — Λm° for a weak acid
Q. Using the ionic values above, find Λm° for acetic acid, first directly and then from the conductivities of HCl, CH₃COONa and NaCl.
Direct route.
Λm°(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻) = 349.6 + 40.9 =
390.5 S cm² mol⁻¹.
Salt-cycle route (the version usually asked, because those three salts are
strong electrolytes whose Λm° can be measured by extrapolation):
Λm°(HCl) = 349.6 + 76.3 = 425.9
Λm°(CH₃COONa) = 50.1 + 40.9 = 91.0
Λm°(NaCl) = 50.1 + 76.3 = 126.4
Λm°(CH₃COOH) = 425.9 + 91.0 − 126.4 = 390.5 S cm² mol⁻¹ ✓
The two routes agree exactly, which is the whole content of the law of independent migration: Na⁺ and Cl⁻ cancel because each ion's contribution does not depend on its partner.
Worked example 3 — degree of dissociation and Ka
Q. 0.00100 M acetic acid has Λm = 49.0 S cm² mol⁻¹ at 298 K. Calculate α and Ka.
Step 1 — degree of dissociation. For a weak electrolyte, Arrhenius' relation applies:
α = Λm / Λm° = 49.0 / 390.5 = 0.1255 (12.55 %).
Step 2 — Ostwald dilution law.
Ka = cα²/(1 − α) = (0.00100 × 0.1255²) / (1 − 0.1255)
= (0.00100 × 0.015750) / 0.8745 = 1.575 × 10⁻⁵ / 0.8745 =
1.80 × 10⁻⁵.
That is the textbook value of Ka for acetic acid — a good sign that the arithmetic is right. Note that dropping the (1 − α) term would have given 1.58 × 10⁻⁵, a 12 % error; at α above about 5 % the approximation is no longer safe.
Transport numbers, in one line
Ionic mobility: u = λ / F, where F = 96485 C mol⁻¹
For HCl, t₊ = 349.6 / (349.6 + 76.3) = 349.6/425.9 = 0.821. Over 80 % of the current in hydrochloric acid is carried by the proton — again, the Grotthuss mechanism.
Common mistakes that cost marks
- Applying the √c law to a weak electrolyte. The steep rise of Λm for acetic acid on dilution is increasing dissociation, not an ion-atmosphere effect. Extrapolating that curve gives nonsense; use independent migration.
- Losing the factor 1000. Λm = κ/c only when c is in mol cm⁻³. With mol L⁻¹ you must write Λm = 1000κ/c.
- Forgetting stoichiometric coefficients. For CaCl₂, Λm° = λ°(Ca²⁺) + 2λ°(Cl⁻), not a simple sum of one of each.
- Confusing κ falling with Λm falling. On dilution κ always decreases and Λm always increases. Questions are often written to catch this.
- Dropping (1 − α) in Ostwald's law. Safe only for α below roughly 5 %.
Summary table
| Quantity | Formula | SI-practical unit | Behaviour on dilution |
|---|---|---|---|
| Conductance G | 1/R | S | Decreases |
| Conductivity κ | (1/R)(l/A) | S cm⁻¹ | Decreases |
| Molar conductivity Λm | 1000κ/c | S cm² mol⁻¹ | Increases |
| Limiting Λm° | ν₊λ₊° + ν₋λ₋° | S cm² mol⁻¹ | Constant (a limit) |
| Degree of dissociation α | Λm/Λm° | dimensionless | Increases |
Check your electrochemistry arithmetic instantly. The Nernst Equation calculator handles cell potentials, concentration dependence and the 0.0592/n term at 298 K — the companion calculation to almost every conductance question in Part C.
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