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Kohlrausch’s Law: Definition, Formula, Applications, Limitations, and Examples in Electrochemistry
Physical Chemistry · Electrochemistry

Kohlrausch’s Law of Independent Migration of Ions

Every ion migrates at its own fixed speed, contributing its own fixed share of conductivity, regardless of which partner ion accompanies it. Dial in real or custom ions below and watch the field respond — then see the two competing transport mechanisms behind the numbers, and why the law rescues chemists from an impossible extrapolation.

1 Migration Simulator

Choose any cation/anion pair — or override with a custom mobility — and adjust temperature. The animation speed and the computed Λmo update live. Hover any ion in the tube for its transport mechanism.

+
Crossing time is drawn to scale — twice the mobility means half the crossing time.
349.6
199.1
25 °C
Λm° = 349.6 + 199.1 = 548.7 S cm² mol⁻&sup9;

2 The Law, Precisely

At infinite dilution (c → 0), interionic electrostatic interactions vanish and every ion moves as if no other ion existed. Under these conditions each ion contributes a fixed, additive amount to the electrolyte’s limiting molar conductivity:

Λmo = ν+λ+o + νλo ν± = number of ions per formula unit · λ±° = limiting ionic molar conductivity
Formally, this holds because at c → 0 the excess chemical potential from ion–ion interaction (captured, at finite c, by the Debye–Hückel ionic-cloud picture) drops to zero, leaving each ion’s mobility a genuine single-ion transport property rather than something that depends on its neighbours.

3 Additivity in Action — Assembling Λmo(CH₃COOH)

Acetic acid is weak and cannot be extrapolated directly (Section 5). Kohlrausch’s Law builds its value instead from three strong-electrolyte values, chosen so the unwanted Na⁺ and Cl⁻ terms cancel exactly:

Λ°(HCl)
H⁺ 349.6
Cl⁻
425.9
+ Λ°(CH₃COONa)
Na⁺ 50.1
CH₃COO⁻
91.0
− Λ°(NaCl)
Na⁺ 50.1
Cl⁻ 76.3
126.4
= Λ°(CH₃COOH)
H⁺ + CH₃COO⁻
390.5
425.9 + 91.0 − 126.4 = 390.5 S cm² mol⁻&sup9;, matching the literature value of ≈ 390.7 to within experimental precision. Na⁺ and Cl⁻ each appear once with a + sign and once with a − sign and cancel identically — direct proof of additivity.

4 Two Transport Mechanisms — Why H⁺ and OH⁻ Break the Pattern

H⁺ (349.6) and OH⁻ (199.1) are anomalously fast — roughly 3–7× any other singly-charged ion. They aren’t simply “smaller”; they use a fundamentally different transport mechanism.

Grotthuss (relay) mechanism — H⁺
H₂O
H₂O
H₂O
H₂O
H⁺
A proton hops between hydrogen-bonded water molecules by a rapid sequence of bond-breaking / bond-forming steps; net H⁺ charge moves without any single proton physically crossing the whole tube.
Vehicular (hydrodynamic) mechanism — Na⁺
Na⁺
A “normal” ion drags its entire hydration shell physically through the solvent, governed by Stokes–Einstein drag — inherently slower than a relay of bond rearrangements.
OH⁻ conducts by the mirror-image process: a proton hops onto a neighbouring water molecule from the opposite direction, effectively moving the “missing proton” (i.e. OH⁻) forward. Both mechanisms are why pure water’s tiny self-ionisation still gives it measurable conductivity.

5 The Extrapolation Problem

For a strong electrolyte, the Debye–Hückel–Onsager relation Λm = Λmo − (A + BΛmo)√c predicts a near-linear fall as concentration rises — so a short linear plot extrapolates cleanly to c = 0. A weak electrolyte’s Λm depends on its degree of dissociation α, which itself rises steeply as c → 0, producing a curve that shoots upward with no linear region to extrapolate.

√c → (concentration decreases this way) Λm Λm° (strong) — clean linear extrapolation Λm (weak) — rises steeply, never linear ? Unreadable from the graph — needs Kohlrausch’s Law
Strong electrolyte (HCl, NaCl, KCl) Weak electrolyte (CH₃COOH, NH₄OH)

6 Limiting Ionic Molar Conductivities (25°C, aqueous)

Cationλ°+Anionλ°
H⁺349.6OH⁻199.1
K⁺73.5Cl⁻76.3
NH₄⁺73.5Br⁻78.1
Ag⁺61.9NO₃⁻71.4
Na⁺50.1CH₃COO⁻40.9
Li⁺38.7HCO₃⁻44.5
½ Ca²⁺119.0½ SO₄²⁻80.0
½ Mg²⁺106.1½ CO₃²⁻69.3
Divalent ions are conventionally tabulated per unit charge (equivalent conductivity) since Λmo itself must be built using the stoichiometric ν± from Section 1 — e.g. Λmo(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻), where λ°(Ca²⁺) = 2 × 119.0 = 238.0.

7 Dissociation Deepens on Dilution

For a weak electrolyte, α = Λmmo is literally the fraction of dissolved molecules that exist as free ions rather than associated pairs. Diluting the solution shifts the equilibrium toward dissociation (Le Chatelier), which is exactly why Λm climbs toward Λmo as c → 0.

Concentrated — low α
H
A
H
A
H
A
H
A
Most HA molecules stay paired; only a small fraction separates into free H⁺ and A⁻.
Dilute — high α
H
A
H
A
H
A
Fewer molecules overall (dilution), but most of those present have separated — α approaches 1 as c → 0.

8 The Payoff — Ostwald’s Dilution Law

Combining α = Λmmo with the equilibrium expression for a weak monoprotic acid HA ⇌ H⁺ + A⁻ gives:

Ka = cα² / (1−α) c = analytical concentration · α obtained from conductivity via Kohlrausch’s Λm°
This is the single biggest practical use of Kohlrausch’s Law: it converts a conductivity measurement — something purely electrical — into a thermodynamic equilibrium constant, without ever needing to isolate or titrate the species involved.

9 Key Takeaways

The law itselfAt infinite dilution, each ion contributes independently and additively to Λmo.
Why H⁺/OH⁻ are outliersThey travel by Grotthuss proton relay, not by dragging a hydration shell like ordinary ions.
Why it’s neededWeak electrolytes rise too steeply near c → 0 for any linear extrapolation to work.
What it unlocksα and, via Ostwald’s dilution law, Ka or Kb — from conductivity data alone.

CogitaVerse · Physical Chemistry Series

Download Complete Notes for Kohlrausch’s Law Below

(4) Calculation of the Degree of Dissociation or Conductance Ratio

λH+ = 349.8 mho cm-1

λ OH– = 198.5 mho cm-1

λH2O = λH+ + λ OH

349.8 + 198.5 = 548.3 mho cm-1

Since one water molecule gives one H+ ion and one OH ion

So

H2O = H+ + OH

Assuming that conductance and ionic concentration are proportionate, we have

[H+] = [OH] = (is 5.54 × 10-8) / 548.3= 1.01 × 10-7 g ion litre-1

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