Chemistry · Electrochemistry · NEET
At infinite dilution (very high dilution), every ion moves completely independently, as if the other ions are not there. So the total conducting power of an electrolyte is just the sum of the conducting powers of its separate ions. This total conducting power at infinite dilution is called limiting molar conductivity, Λ°m. That is why the law is also called the 'law of independent migration of ions'.
Λm is the molar conductivity of a solution at some concentration c. As you keep diluting, Λm rises and finally reaches a maximum value that does not increase anymore — that maximum is Λ°m, the limiting molar conductivity (at zero concentration or infinite dilution). The small circle (°) just means 'at infinite dilution'. NEET numericals almost always give or ask for Λ°m.
For an electrolyte that gives ν₊ cations and ν₋ anions: Λ°m = ν₊ λ°₊ + ν₋ λ°₋. Here λ°₊ and λ°₋ are the limiting molar conductivities of one mole of each ion. Example: Λ°m(NaCl) = λ°(Na⁺) + λ°(Cl⁻); Λ°m(BaCl₂) = λ°(Ba²⁺) + 2 λ°(Cl⁻) because there are two chloride ions.
For strong electrolytes, you plot Λm against √c and draw a straight line back to c = 0 to read Λ°m. Weak electrolytes like CH₃COOH do NOT give a straight line — Λm shoots up steeply near zero concentration, so you cannot extrapolate reliably. Kohlrausch law rescues us: we build Λ°m of the weak acid from Λ°m values of strong electrolytes instead.
Combine them so the spectator ions cancel: Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl). Check the ions: CH₃COONa gives CH₃COO⁻+Na⁺, HCl gives H⁺+Cl⁻, and we subtract NaCl (Na⁺+Cl⁻). The Na⁺ and Cl⁻ cancel, leaving λ°(CH₃COO⁻)+λ°(H⁺), which is exactly Λ°m(CH₃COOH). This is the most repeated NEET pattern.
Two big uses. First, degree of dissociation: α = Λm / Λ°m (how much of the weak acid actually splits into ions). Second, using α you get the dissociation constant Ka = cα²/(1−α) ≈ cα² by Ostwald's dilution law. NEET loves asking for α or Ka using a given Λm and Λ°m from Kohlrausch law.
The molar conductances of NaCl, HCl and CH₃COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm² mol⁻¹ respectively. The molar conductance of CH₃COOH at infinite dilution is:
If the molar conductivity (Λm) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its degree of dissociation will be [λ°₊ = 349.6 S cm² mol⁻¹, λ°₋ = 50.4 S cm² mol⁻¹]:
Given Λ°m(H₂SO₄)=x, Λ°m(K₂SO₄)=y, Λ°m(CH₃COOK)=z (all S cm² mol⁻¹). Λ°m for CH₃COOH will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. It is strictly a law of independent migration of ions at infinite dilution (zero concentration). It uses limiting molar conductivities Λ°m, not the Λm at ordinary concentrations, because only at infinite dilution do ions move fully independently.
Yes. Λ°m = ν₊ λ°₊ + ν₋ λ°₋ applies to any electrolyte. But its special usefulness is for weak electrolytes, whose Λ°m cannot be found by extrapolating the Λm vs √c graph.
(1) Finding Λ°m of weak electrolytes like CH₃COOH from strong electrolyte data. (2) Calculating the degree of dissociation α = Λm/Λ°m, and from it the dissociation constant Ka of a weak acid.
Λm is molar conductivity at a real, finite concentration. Λ°m (limiting molar conductivity) is its value when concentration approaches zero (infinite dilution) — the maximum, where ions no longer interact. The ratio Λm/Λ°m gives the degree of dissociation.
H⁺ ions conduct by a special fast 'hopping' (Grotthuss) mechanism through water instead of physically moving, so λ°(H⁺) is much larger (~349.6 S cm² mol⁻¹) than most ions. This is why acids show high conductivity.
