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

44 PYQs, one 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

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

$\Lambda = \dfrac{1000\,\kappa}{c}$ , κ in S cm⁻¹, c in mol L⁻¹; unit $\text{S cm}^2\,\text{mol}^{-1}$ . 0.02 M AgNO₃, κ = 0.00216 → $\Lambda = 108$ .
Reverse: $\kappa = \dfrac{\Lambda \cdot c}{1000}$ . 0.02 M KCl, Λ = 410 → $\kappa = 8.2 \times 10^{-3}$ . And $c = \dfrac{1000\kappa}{\Lambda}$ .
On dilution: $\kappa$ DECREASES, $\Lambda$ 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

\Lambda = \frac{1000\,\kappa}{c},\qquad \kappa = \frac{\Lambda\,c}{1000}

Worked example

The conductivity of 0.05 M NaCl is

5.5 \times 10^{-3}\ \text{S cm}^{-1}

. Find its molar conductivity.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 2 · Q83Easy

Example 1 · Electrochemistry · Molar Conductivity, Kohlrausch's Law and Degree of Dissociation

The conductivity of 0.02 M solution of

\text{AgNO}_3

0.00216\,\Omega^{-1}\text{cm}^{-1}

at 298 K. What is its molar conductivity?

Dropping the 1000

\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

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

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

Kohlrausch's law

\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⁻).

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q78Moderate

Example 2 · Electrochemistry · Molar Conductivity, Kohlrausch's Law and Degree of Dissociation

The limiting molar conductivities

\Lambda^0

for NaCl, KBr and KCl are 126, 152 and 150 S cm² mol⁻¹ respectively. What is the

\Lambda^0

of NaBr?

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

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

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

Degree of dissociation

\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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q55Easy

Example 3 · Electrochemistry · Molar Conductivity, Kohlrausch's Law and Degree of Dissociation

The molar conductivity of 0.01 M acetic acid at

25^\circ C

16.5\,\Omega^{-1}\,\text{cm}^2\,\text{mol}^{-1}

and its molar conductivity at zero concentration is

390.7\,\Omega^{-1}\,\text{cm}^2\,\text{mol}^{-1}

. What is its degree of dissociation?

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

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Molar Conductivity: Λ = 1000κ/c and Back
Molar conductivity
$\Lambda = \frac{1000\,\kappa}{c},\qquad \kappa = \frac{\Lambda\,c}{1000}$
Kohlrausch's Law: Λ₀ From the Ions
Kohlrausch's law
$\Lambda_0 = \nu_+\,\lambda_+^0 + \nu_-\,\lambda_-^0$
Degree of Dissociation: α = Λ/Λ₀
Degree of dissociation
$\alpha = \frac{\Lambda_c}{\Lambda_0}$

Watch out for (3)

Dropping the 1000
→ Molar Conductivity: Λ = 1000κ/c and Back
Adding all three values
→ Kohlrausch's Law: Λ₀ From the Ions
Inverting the ratio
→ Degree of Dissociation: α = Λ/Λ₀

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Related notes

Ionic Equilibria — Ostwald's dilution law and Ka →