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MHT-CET Chemistry · Electrochemistry

Molar Conductivity, Kohlrausch's Law and Degree of Dissociation

Molar conductivity Λ = 1000κ/c is the conductivity of one mole of electrolyte; it rises on dilution to a limit Λ₀ that Kohlrausch's law builds from the ions, and the ratio Λ/Λ₀ is the degree of dissociation of a weak electrolyte.

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Why this matters

24 PYQs, none HARD. Half are the conversion between κ and Λ in either direction (the 1000 is where marks are lost), a quarter are Kohlrausch — add and subtract Λ₀ values, or count ions with their coefficients — and the rest are α = Λ/Λ₀ as a fraction or a percentage. Two formulas and one factor of 1000.

Concept 1 of 3

Molar Conductivity: Λ = 1000κ/c and Back

Intuition

κ counts the ions in one cm³; Λ counts the conductance one whole mole would give. Dividing κ by the concentration in mol per cm³ (c/1000 for c in mol L⁻¹) does it. Diluting a solution lowers κ (fewer ions per cm³) but raises Λ (each mole is more completely dissociated and less crowded).

Definition

  • Λ=1000 κc\Lambda = \dfrac{1000\,\kappa}{c}, κ in S cm⁻¹, c in mol L⁻¹; unit S cm2 mol−1\text{S cm}^2\,\text{mol}^{-1}. 0.02 M AgNO₃, κ = 0.00216 → Λ=108\Lambda = 108.
  • Reverse: κ=Λ⋅c1000\kappa = \dfrac{\Lambda \cdot c}{1000}. 0.02 M KCl, Λ = 410 → κ=8.2×10−3\kappa = 8.2 \times 10^{-3}. And c=1000κΛc = \dfrac{1000\kappa}{\Lambda}.
  • On dilution: κ\kappa DECREASES, Λ\Lambda INCREASES. The most dilute solution has the highest Λ (0.001 M beats 0.005 M).
  • Strong electrolytes: Λ rises slowly and linearly in √c to Λ₀. Weak electrolytes: Λ shoots up at high dilution, so Λ₀ cannot be read off a graph — it comes from Kohlrausch.

Molar conductivity

Λ=1000 κc,κ=Λ c1000\Lambda = \frac{1000\,\kappa}{c},\qquad \kappa = \frac{\Lambda\,c}{1000}

Worked example

The conductivity of 0.05 M NaCl is 5.5×10−3 S cm−15.5 \times 10^{-3}\ \text{S cm}^{-1}. Find its molar conductivity.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1ElectrochemistryEASY
The conductivity of 0.02 M solution of AgNO3\text{AgNO}_3 is 0.00216 Ω−1cm−10.00216\,\Omega^{-1}\text{cm}^{-1} at 298 K. What is its molar conductivity?

[Q83 · 15th May Shift 2 · 2023]

Dropping the 1000

κ/c\kappa/c with c in mol L⁻¹ is a thousand times too small. The factor converts litres to cm³ because κ is per cm. Options are set one factor of 1000 apart to catch exactly this.
Drill 12 more on molar conductivity: λ = 1000κ/c and back

Concept 2 of 3

Kohlrausch's Law: Λ₀ From the Ions

Intuition

At infinite dilution every ion moves independently, so Λ₀ of an electrolyte is the sum of its ions' limiting conductivities, each weighted by how many of that ion the formula releases. It also lets you build Λ₀ of a weak acid from three strong electrolytes: add the ones that carry the ions you want, subtract the one that carries the ions you don't.

Definition

  • Λ0=ν+λ+0+ν−λ−0\Lambda_0 = \nu_+\lambda_+^0 + \nu_-\lambda_-^0. AB₃: λA3+0+3λB−0\lambda^0_{A^{3+}} + 3\lambda^0_{B^-}. A₂B₃: 2λA3+0+3λB2−02\lambda^0_{A^{3+}} + 3\lambda^0_{B^{2-}}. Al₂(SO₄)₃: 2(189)+3(50.1)=528.32(189) + 3(50.1) = 528.3.
  • Weak acid from salts: Λ0(CH2ClCOOH)=Λ0(HCl)+Λ0(CH2ClCOOK)−Λ0(KCl)\Lambda_0(\text{CH}_2\text{ClCOOH}) = \Lambda_0(\text{HCl}) + \Lambda_0(\text{CH}_2\text{ClCOOK}) - \Lambda_0(\text{KCl}) = 4.2 + 1.1 − 1.5 = 3.8.
  • Same trick for any salt: Λ0(NaBr)=Λ0(NaCl)+Λ0(KBr)−Λ0(KCl)\Lambda_0(\text{NaBr}) = \Lambda_0(\text{NaCl}) + \Lambda_0(\text{KBr}) - \Lambda_0(\text{KCl}) = 126 + 152 − 150 = 128.
  • Halve a 2:1 salt when you need one ion of it: Λ0(NH4OH)=Λ0(NH4Cl)+12Λ0(Ba(OH)2)−12Λ0(BaCl2)\Lambda_0(\text{NH}_4\text{OH}) = \Lambda_0(\text{NH}_4\text{Cl}) + \tfrac{1}{2}\Lambda_0(\text{Ba(OH)}_2) - \tfrac{1}{2}\Lambda_0(\text{BaCl}_2) = 129 + 260 − 140 = 249.

Kohlrausch's law

Λ0=ν+ λ+0+ν− λ−0\Lambda_0 = \nu_+\,\lambda_+^0 + \nu_-\,\lambda_-^0

Worked example

Λ₀ of NaOH, HCl and NaCl are 248, 426 and 126 S cm² mol⁻¹. Find Λ₀ of water (H⁺ + OH⁻).
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2ElectrochemistryMODERATE
The limiting molar conductivities Λ0\Lambda^0 for NaCl, KBr and KCl are 126, 152 and 150 S cm² mol⁻¹ respectively. What is the Λ0\Lambda^0 of NaBr?

[Q78 · 3rd May Shift 2 · 2023]

Adding all three values

The third electrolyte supplies the ions you must REMOVE, so it is subtracted. 4.2 + 1.1 + 1.5 = 6.8 is nowhere near an option; 4.2 + 1.1 − 1.5 = 3.8 is the key.
Drill 6 more on kohlrausch's law: λ₀ from the ions

Concept 3 of 3

Degree of Dissociation: α = Λ/Λ₀

Intuition

A weak electrolyte's molar conductivity at concentration c is low because only a fraction α of its molecules are ions. If all were ions it would show Λ₀. So α = Λ_c/Λ₀ — a ratio of two numbers the paper hands you, then sometimes asks as a percentage.

Definition

  • α=ΛcΛ0\alpha = \dfrac{\Lambda_c}{\Lambda_0}. 0.01 M acetic acid: 16.5/390.7=0.042216.5/390.7 = 0.0422.
  • Percentage dissociation =100α= 100\alpha: 3.3/132=0.025=2.5%3.3/132 = 0.025 = 2.5\%.
  • Dissociation constant follows: Ka=cα21−α≈cα2K_a = \dfrac{c\alpha^2}{1 - \alpha} \approx c\alpha^2 (Ostwald's dilution law).
  • α rises with dilution — which is why Λ_c climbs towards Λ₀.

Degree of dissociation

α=ΛcΛ0\alpha = \frac{\Lambda_c}{\Lambda_0}

Worked example

Λ of 0.1 M HCOOH is 5.2 and Λ₀ is 404.5 S cm² mol⁻¹. Find α and Ka.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3ElectrochemistryEASY
The molar conductivity of 0.01 M acetic acid at 25∘C25^\circ C is 16.5 Ω−1 cm2 mol−116.5\,\Omega^{-1}\,\text{cm}^2\,\text{mol}^{-1} and its molar conductivity at zero concentration is 390.7 Ω−1 cm2 mol−1390.7\,\Omega^{-1}\,\text{cm}^2\,\text{mol}^{-1}. What is its degree of dissociation?

[Q55 · 10th May Shift 2 · 2023]

Inverting the ratio

α is the SMALL number over the big one and must come out below 1. Λ₀/Λ_c gives 23.7 for acetic acid — no option, but a sign the fraction is upside down.
Drill 3 more on degree of dissociation: α = λ/λ₀

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