GATE Conductance and Ion Transport — Kohlrausch, Transport Numbers and Mobility

By Aniket Bhardwaj · 17 September 2026 · GATE Chemistry

Conductance is one of the most reliable scoring areas in GATE physical chemistry, because the chain of reasoning is fixed. You measure a resistance, convert it to conductivity using the cell constant, convert that to molar conductivity using the concentration, and then use molar conductivity to get a degree of dissociation, a dissociation constant, a transport number or an ionic mobility. Nothing in that chain is guesswork. This guide walks through every link with the units written out, and gives four worked numericals with the arithmetic shown step by step.

Confirm the examinable syllabus from the official GATE notification for your year rather than from any website, this one included.

The chain of quantities — and their units

G = 1 / R (conductance, siemens S)
κ = G × (l / A) = G × cell constant (conductivity, S cm⁻¹)
Λ_m = 1000 κ / C (molar conductivity, S cm² mol⁻¹, with C in mol L⁻¹)

R — the measured resistance of the solution in ohms, taken with an alternating current so that the electrodes do not polarise.
l / A — the cell constant in cm⁻¹: the distance between the electrodes divided by their area. It is never measured directly; it is found by calibrating the cell with a solution of known conductivity, almost always standard KCl.
κ — conductivity (older books say specific conductance): the conductance of a 1 cm cube of solution.
Λ_m — molar conductivity: the conducting power of all the ions produced by one mole of electrolyte. The factor of 1000 converts litres to cm³.

The single most important conceptual point in this topic: on dilution, κ falls but Λ_m rises. There are fewer ions per cm³, so the conductivity drops; but Λ_m is defined per mole of electrolyte, and dilution reduces inter-ionic interference, so each mole conducts better. Students who mix these two up lose marks on a question that is otherwise free.

Worked example 1 — cell constant, then molar conductivity

A conductivity cell filled with 0.100 M KCl (κ = 0.0129 S cm⁻¹ at 298 K) has a resistance of 100 Ω. The same cell filled with 0.0200 M of an electrolyte has a resistance of 520 Ω. Find the cell constant, the conductivity and the molar conductivity of the second solution.

Step 1 — cell constant. Since κ = G × (l/A) = (l/A) ÷ R,
cell constant (l/A) = κ × R = 0.0129 × 100 = 1.29 cm⁻¹

Step 2 — conductivity of the unknown.
κ = (l/A) ÷ R = 1.29 ÷ 520 = 2.481 × 10⁻³ S cm⁻¹

Step 3 — molar conductivity.
Λ_m = 1000 κ ÷ C = (1000 × 2.481 × 10⁻³) ÷ 0.0200
= 2.481 ÷ 0.0200 = 124.0 S cm² mol⁻¹

Unit check: (cm³ L⁻¹ × S cm⁻¹) ÷ (mol L⁻¹) gives S cm² mol⁻¹. If your answer comes out around 0.12 or around 124,000, you have misplaced the factor of 1000.

Kohlrausch's laws — two separate statements

Students often merge these into one; GATE tests them separately.

Concentration dependence (strong electrolytes): Λ_m = Λ°_m − A√C
Independent migration of ions: Λ°_m = ν₊λ°₊ + ν₋λ°₋

The first is an empirical square-root law. Plot Λ_m against √C for a strong electrolyte and you get a straight line whose intercept is Λ°_m, the molar conductivity at infinite dilution. The theoretical justification is the Debye–Hückel–Onsager equation, which explains the slope in terms of two effects: the relaxation (asymmetry) effect, where the ionic atmosphere lags behind a moving ion and pulls it back, and the electrophoretic effect, where the atmosphere and its solvent drag move the opposite way.

The second law says that at infinite dilution each ion contributes independently. It is what lets you obtain Λ°_m for a weak electrolyte, which the √C extrapolation cannot give — because a weak electrolyte's Λ_m rises very steeply near zero concentration and the plot is not linear. For acetic acid:

Λ°_m(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻) = 349.6 + 40.9 = 390.5 S cm² mol⁻¹

You can reach the same value by combining strong electrolytes: Λ°(HCl) + Λ°(CH₃COONa) − Λ°(NaCl), which is the classic exam route.

Worked example 2 — degree of dissociation and K_a

A 0.0100 M solution of acetic acid has Λ_m = 16.30 S cm² mol⁻¹ at 298 K. Given Λ°_m = 390.5 S cm² mol⁻¹, find α and K_a.

Step 1 — degree of dissociation.
α = Λ_m ÷ Λ°_m = 16.30 ÷ 390.5 = 0.04174 (about 4.2%)

Step 2 — Ostwald's dilution law.
K_a = Cα² ÷ (1 − α)
α² = 0.04174² = 1.7422 × 10⁻³
Cα² = 0.0100 × 1.7422 × 10⁻³ = 1.7422 × 10⁻⁵
1 − α = 1 − 0.04174 = 0.95826
K_a = 1.7422 × 10⁻⁵ ÷ 0.95826 = 1.82 × 10⁻⁵

Cross-check: the accepted K_a of acetic acid is about 1.8 × 10⁻⁵, so the conductance route and the pH route agree — which is exactly the point of the experiment. Note also that if α is small, the approximation K_a ≈ Cα² gives 1.74 × 10⁻⁵, within 5% of the exact answer. Use the approximation only when the question allows it.

Transport numbers

Cations and anions do not carry equal shares of the current. The fraction carried by each is its transport (transference) number.

t₊ = λ₊ / (λ₊ + λ₋) t₋ = λ₋ / (λ₊ + λ₋) t₊ + t₋ = 1

Transport numbers are measured by the Hittorf method (analysing concentration changes around each electrode after electrolysis) or the moving boundary method (watching a visible boundary between two electrolytes move a measured distance). The moving boundary method is generally the more accurate.

Worked example 3 — transport numbers and ionic mobility

At 298 K, λ°(Na⁺) = 50.1 and λ°(Cl⁻) = 76.3 S cm² mol⁻¹. Find Λ°_m(NaCl), the two transport numbers, and the ionic mobility of Na⁺.

Step 1 — limiting molar conductivity.
Λ°_m(NaCl) = 50.1 + 76.3 = 126.4 S cm² mol⁻¹

Step 2 — transport numbers.
t₊ = 50.1 ÷ 126.4 = 0.396
t₋ = 76.3 ÷ 126.4 = 0.604
Check: 0.396 + 0.604 = 1.000 ✓

Step 3 — ionic mobility. Mobility u is drift velocity per unit field, and u = λ ÷ F.
u(Na⁺) = 50.1 ÷ 96,485 = 5.19 × 10⁻⁴ cm² V⁻¹ s⁻¹

Unit check: S cm² mol⁻¹ ÷ C mol⁻¹ = S cm² C⁻¹. Since S = A V⁻¹ and C = A s, this becomes cm² V⁻¹ s⁻¹ — a velocity per unit field, as it should be.

Why H⁺ and OH⁻ are anomalously fast

Compare the limiting ionic conductivities: H⁺ is 349.6 and OH⁻ about 199.1 S cm² mol⁻¹, while ordinary ions such as Na⁺ (50.1), K⁺ (73.5) and Cl⁻ (76.3) cluster far below. The mobility of H⁺ works out as 349.6 ÷ 96,485 = 3.62 × 10⁻³ cm² V⁻¹ s⁻¹, roughly seven times that of Na⁺.

The reason is not that the proton is small — a bare proton does not exist in water; it is hydrated as H₃O⁺ and larger clusters, which are not especially small. The reason is the Grotthuss mechanism: instead of the ion itself pushing through the solvent, a proton hops along a hydrogen-bonded chain of water molecules. Charge moves without matter moving very far. OH⁻ conducts by the same kind of structural hopping in the opposite direction. This is a favourite conceptual question, and the wrong answer ("because H⁺ has the smallest radius") is the one most students give.

A related trap: among the alkali metal cations, the order of limiting conductivity is Li⁺ < Na⁺ < K⁺ < Rb⁺, the opposite of the order of bare ionic radii. The small Li⁺ has the highest charge density, so it drags the largest hydration shell and moves most slowly. What matters is the hydrated radius, and Stokes' law (u = ze / 6πηr) then applies to that effective radius.

Conductometric titrations — read the V shape

A conductometric titration is followed by plotting conductance against titrant volume and finding the intersection of two straight lines. No indicator is needed, and it works for coloured or very dilute solutions where a visual endpoint would fail.

Titration	Shape of the plot	Why
Strong acid vs strong base	Sharp V — steep fall, then steep rise	Fast H⁺ replaced by slower Na⁺, then excess fast OH⁻ is added
Weak acid vs strong base	Slight initial dip or flat, then rise, then steeper rise	Few ions at first; salt formation raises conductance; excess OH⁻ raises it faster
Strong acid vs weak base	Fall, then nearly flat	H⁺ removed; excess weak base adds few ions
Weak acid vs weak base	Rise, then flat	Salt is formed; excess weak base contributes little

The unifying idea: conductance changes because one ion is being replaced by another of different mobility. Always ask which ion is leaving and which is arriving.

Common mistakes that cost marks

Dropping the factor of 1000. Λ_m = 1000κ/C when C is in mol L⁻¹. If C is in mol cm⁻³, the factor disappears. Decide which you are using before you start.
Saying molar conductivity falls on dilution. It rises; conductivity falls.
Extrapolating a weak electrolyte's √C plot to get Λ°_m. That only works for strong electrolytes. Use Kohlrausch's law of independent migration instead.
Using α = Λ_m/Λ°_m for a strong electrolyte. A strong electrolyte is fully dissociated; the ratio there reflects inter-ionic interference, not incomplete dissociation, and is called the conductivity ratio.
Explaining the mobility of H⁺ by its small size. The mechanism is proton hopping along hydrogen bonds, not motion of a small ion.
Forgetting that the cell constant must be determined, not assumed. Every conductance numerical starts with a KCl calibration for a reason.
Using direct current. Conductance is measured with AC to avoid electrolysis and electrode polarisation — a common one-line question.

Preparation map for this unit

Sub-topic	Question style	Priority
Cell constant, κ, Λ_m	Direct three-step numerical	High — near-certain marks
Kohlrausch's law of independent migration	Λ°_m of a weak acid from strong electrolytes	High
Ostwald dilution law	α and K_a from Λ_m	High
Transport numbers	From λ° values, or from a Hittorf experiment	Medium
Ionic mobility and Stokes' law	Ordering ions by hydrated radius	Medium
Debye–Hückel–Onsager	Which effect explains the √C slope	Medium — conceptual
Conductometric titrations	Identify the curve shape	Medium

Every conductance problem begins with G = 1/R. Before the cell constant and the factor of 1000 come into it, you are simply converting between resistance, conductance, current and voltage — and that is where careless slips start. The free Ohm's Law calculator handles V, I and R, so you can confirm the resistance step of a conductance-cell problem quickly before moving on to κ and Λ_m.

Open the Ohm's Law Calculator →

Preparing for GATE, IIT-JAM, CSIR-NET or CUET-PG chemistry? ABC Chemistry runs dedicated competitive-exam batches at its coaching centre and online for students across India — details at abcchemistry.in.