JEE Mains Chemistry · Electrochemistry

Molar Conductivity, Dilution and Kohlrausch's Law

How molar conductivity changes with dilution for strong and weak electrolytes, how Kohlrausch's law builds a limiting value from ion values, and how that gives the degree of dissociation, Ka and solubility.

All Electrochemistry notes JEE Mains Chemistry strategy

Why this matters

Twenty-one PYQs, twelve of them multiple choice, and four from 2026. Nine ask how Λm behaves with dilution, for strong and weak electrolytes or in a conductometric titration. Five build a limiting molar conductivity from other salts, and seven turn Λm into a degree of dissociation, a Ka or a solubility. Three ideas cover the page.

Concept 1 of 3: Strong and weak electrolytes on dilution

A strong electrolyte is fully ionised already. Dilution only lets its ions move a little more freely, so

\Lambda_m

rises slowly and in a straight line against

\sqrt c

. A weak electrolyte ionises more as it is diluted, so its

\Lambda_m

stays low and then shoots up near zero concentration.

Definition

Strong (KCl, NaCl, HCl): $\Lambda_m=\Lambda_m^\circ-A\sqrt c$ . A plot of $\Lambda_m$ against $\sqrt c$ is a straight line with slope $-A$ and intercept $\Lambda_m^\circ$ .
The unit of $A$ is $\mathrm{S\,cm^2\,mol^{-1}\,(mol\,L^{-1})^{-1/2}}$ , that is $\mathrm{S\,cm^2\,mol^{-3/2}\,L^{1/2}}$ .
Weak ( $\mathrm{CH_3COOH}$ , $\mathrm{NH_4OH}$ , $\mathrm{H_2CO_3}$ ): the plot is not a line. Its $\Lambda_m^\circ$ cannot be found by extrapolation, only by Kohlrausch's law.
A weak acid's $\Lambda_m^\circ$ can still be LARGE: $\mathrm{CH_3COOH}$ is 390.5 against KCl's 149.8, because it contains $\mathrm{H^+}$ .
Conductometric titration with NaOH: a strong acid falls sharply ( $\mathrm{H^+}$ replaced by the slower $\mathrm{Na^+}$ ) then rises (excess $\mathrm{OH^-}$ ). A weak acid dips slightly, rises gently as the salt forms, then rises steeply after the end point. A mixture of the two shows the fall, the gentle rise and the steep rise in turn.

Debye–Hückel–Onsager (strong electrolytes)

\Lambda_m=\Lambda_m^\circ-A\sqrt c

Worked example

A strong electrolyte has

\Lambda_m=141.0

S cm² mol⁻¹ at 0.01 M and 138.0 at 0.04 M. Find

A

and

\Lambda_m^\circ

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q48Moderate

Example 1 · Electrochemistry · Molar Conductivity, Dilution and Kohlrausch's Law

For a strong electrolyte,

\Lambda_{m}

increases slowly with dilution and can be represented by the equation

\Lambda_{m} = \Lambda_{m}^{\circ} - Ac^{1/2}

. Molar conductivity values of the solutions of a strong electrolyte AB at

18^{\circ}C

are given below :

$c$ ( $mol\ L^{-1}$ )	0.04	0.09	0.16	0.25
$\Lambda_{m}$ ( $S\ cm^{2}\ mol^{-1}$ )	96.1	95.7	95.3	94.9

The value of the constant A based on the above data [in

S\ cm^{2}\ mol^{-1}/(mol/L)^{1/2}

unit] is

\_\_\_\_

Extrapolating a weak electrolyte

Only a strong electrolyte gives a straight line to read

\Lambda_m^\circ

from. For a weak one the curve is nearly vertical near zero, so there is no intercept to read.

Plotting against c instead of √c

The straight line is against

\sqrt c

. Taking the slope from

c

values gives the wrong

A

Drill 8 more on strong and weak electrolytes on dilution

Concept 2 of 3: Kohlrausch's law of independent migration

At infinite dilution each ion moves on its own, so the limiting molar conductivity is just the sum of the ions' contributions. When the ion values are not given, add and subtract whole salts until the unwanted ions cancel.

Definition

$\Lambda_m^\circ=\nu_+\lambda_+^\circ+\nu_-\lambda_-^\circ$ , where $\nu$ is the number of each ion in the formula.
$\Lambda_m^\circ(\mathrm{CH_3COOH})=\Lambda_m^\circ(\mathrm{CH_3COONa})+\Lambda_m^\circ(\mathrm{HCl})-\Lambda_m^\circ(\mathrm{NaCl})$ .
$\Lambda_m^\circ(\mathrm{BaSO_4})=\Lambda_m^\circ(\mathrm{BaCl_2})+\Lambda_m^\circ(\mathrm{H_2SO_4})-2\Lambda_m^\circ(\mathrm{HCl})$ .
$\Lambda_m^\circ(\mathrm{AgI})=\Lambda_m^\circ(\mathrm{NaI})+\Lambda_m^\circ(\mathrm{AgNO_3})-\Lambda_m^\circ(\mathrm{NaNO_3})$ .
A divalent salt MX (such as $\mathrm{MgSO_4}$ ) has one of each ion: $\Lambda_m^\circ=\lambda_+^\circ+\lambda_-^\circ$ .
Mohr's salt, $\mathrm{FeSO_4\cdot(NH_4)_2SO_4\cdot6H_2O}$ : $\lambda^\circ(\mathrm{Fe^{2+}})+2\lambda^\circ(\mathrm{NH_4^+})+2\lambda^\circ(\mathrm{SO_4^{2-}})$ .

Kohlrausch's law

\Lambda_m^\circ=\nu_+\lambda_+^\circ+\nu_-\lambda_-^\circ

Worked example

\Lambda_m^\circ

values in S cm² mol⁻¹:

\mathrm{CH_3COONa}

91.0, HCl 425.9, NaCl 126.4. Find

\Lambda_m^\circ(\mathrm{CH_3COOH})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q124Moderate

Example 2 · Electrochemistry · Molar Conductivity, Dilution and Kohlrausch's Law

The molar conductance of an infinitely dilute solution of ammonium chloride was found to be

185\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 1}

and the ionic conductance of hydroxyl and chloride ions are 170 and

70\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 1}

, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is

85.5\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 1}

, its degree of dissociation is given by

x \times 10^{- 1}

. The value of

x

\_\_\_\_

. (Nearest integer)

Doubling a divalent salt

\mathrm{MgSO_4}

has one

\mathrm{Mg^{2+}}

and one

\mathrm{SO_4^{2-}}

. Its

\Lambda_m^\circ

\lambda_+^\circ+\lambda_-^\circ

, not twice that.

Leaving an ion uncancelled

Write the ions of every salt you add and subtract. The ions left over must be exactly those of the target, with the right counts.

Drill 4 more on kohlrausch's law of independent migration

Concept 3 of 3: Degree of dissociation, Ka and solubility

A weak electrolyte conducts only through the fraction that has ionised. So the ratio of what it does conduct to what it would conduct fully ionised is that fraction,

\alpha

. For a sparingly soluble salt the solution is so dilute that

\Lambda_m

\Lambda_m^\circ

, so conductivity gives the solubility.

Definition

$\alpha=\frac{\Lambda_m}{\Lambda_m^\circ}$ .
$K_a=\frac{c\alpha^2}{1-\alpha}\approx c\alpha^2$ when $\alpha\ll1$ .
From pH: $[\mathrm{H^+}]=c\alpha$ , so $\Lambda_m^\circ=\frac{1000\,\kappa}{c\alpha}=\frac{1000\,\kappa}{[\mathrm{H^+}]}$ .
Sparingly soluble salt: solubility $s=\frac{1000\,\kappa}{\Lambda_m^\circ}$ mol L⁻¹. Then $K_{sp}=s^2$ for a 1:1 salt, and $108s^5$ for $\mathrm{A_2X_3}$ .
A mixture of ions: $\kappa=\sum\lambda_i^\circ c_i$ , with $c$ in mol m⁻³ for SI units.

Degree of dissociation

\alpha=\frac{\Lambda_m}{\Lambda_m^\circ},\qquad K_a=\frac{c\alpha^2}{1-\alpha}

Worked example

0.01 M acetic acid has

\kappa=1.95\times10^{-4}

S cm⁻¹.

\Lambda_m^\circ=390

S cm² mol⁻¹. Find

\alpha

and

K_a

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q47Moderate

Example 3 · Electrochemistry · Molar Conductivity, Dilution and Kohlrausch's Law

The pH and conductance of a weak acid (HX) was found to be 5 and

4 \times 10^{- 5}\text{ }S

, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of

1{\text{ }cm}^{2}

were at a distance of 15 cm apart. The value of the limiting molar conductivity is

\_\_\_\_

Sm^{2}{\text{ }mol}^{- 1}

. (nearest integer) (Given: degree of dissociation of the weak acid (a) << 1)

Dropping the 1000

With

\kappa

in S cm⁻¹ and

c

in mol L⁻¹, the 1000 converts litres to cm³. Leave it out and every answer is off by a thousand.

The wrong Ksp expression

K_{sp}=s^2

only for a 1:1 salt. For

\mathrm{A_2X_3}

K_{sp}=(2s)^2(3s)^3=108s^5

Drill 6 more on degree of dissociation, ka and solubility

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)

Strong and weak electrolytes on dilution
Debye–Hückel–Onsager (strong electrolytes)
$\Lambda_m=\Lambda_m^\circ-A\sqrt c$
Kohlrausch's law of independent migration
Kohlrausch's law
$\Lambda_m^\circ=\nu_+\lambda_+^\circ+\nu_-\lambda_-^\circ$
Degree of dissociation, Ka and solubility
Degree of dissociation
$\alpha=\frac{\Lambda_m}{\Lambda_m^\circ},\qquad K_a=\frac{c\alpha^2}{1-\alpha}$

Watch out for (6)

Extrapolating a weak electrolyte
→ Strong and weak electrolytes on dilution
Plotting against c instead of √c
→ Strong and weak electrolytes on dilution
Doubling a divalent salt
→ Kohlrausch's law of independent migration
Leaving an ion uncancelled
→ Kohlrausch's law of independent migration
Dropping the 1000
→ Degree of dissociation, Ka and solubility
The wrong Ksp expression
→ Degree of dissociation, Ka and solubility

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