κ = 1/ρ, Λm = κ/c. Kohlrausch's law for weak electrolytes.

Conductivity and Molar Conductivity

Question

The molar conductivity of 0.025 mol/L methanoic acid (HCOOH) is 46.1 S cm² mol⁻¹. Its limiting molar conductivity $\Lambda_m^\circ$ is 404.5 S cm² mol⁻¹. Calculate the degree of dissociation $\alpha$ and the dissociation constant $K_a$ of methanoic acid at this concentration.

(JEE Main 2023 pattern — weak electrolyte Kohlrausch application)

Solution — Step by Step

The degree of dissociation is the fraction of molecules that have actually ionised. We get it directly from the ratio of measured to limiting molar conductivity:

\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{46.1}{404.5} = 0.114

So about 11.4% of HCOOH molecules are dissociated at this concentration. The rest remain as neutral molecules.

\text{HCOOH} \rightleftharpoons \text{HCOO}^- + \text{H}^+

At equilibrium, if initial concentration is $c$ and degree of dissociation is $\alpha$ :

Species	Initial	Equilibrium
HCOOH	$c$	$c(1-\alpha)$
HCOO⁻	0	$c\alpha$
H⁺	0	$c\alpha$

K_a = \frac{[\text{HCOO}^-][\text{H}^+]}{[\text{HCOOH}]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha}

This is the standard weak acid equilibrium expression — same one used for acetic acid, carbonic acid, all weak acids.

c = 0.025 \text{ mol/L}, \quad \alpha = 0.114

K_a = \frac{0.025 \times (0.114)^2}{1 - 0.114} = \frac{0.025 \times 0.01300}{0.886}

K_a = \frac{3.25 \times 10^{-4}}{0.886} = \boxed{3.67 \times 10^{-4} \text{ mol/L}}

Why This Works

Kohlrausch’s law tells us that at infinite dilution, every ion contributes independently to conductivity — there are no interionic attractions to slow things down. At any finite concentration $c$ , the measured $\Lambda_m$ is lower because only the fraction $\alpha$ of molecules have actually ionised AND the ions that do exist are slowed by each other.

The ratio $\Lambda_m / \Lambda_m^\circ$ captures both effects in one clean number. For weak electrolytes like HCOOH, the dominant reason $\Lambda_m$ is low is incomplete dissociation (small $\alpha$ ), not just ion-ion interactions. This is why the approximation $\alpha \approx \Lambda_m / \Lambda_m^\circ$ works well for weak electrolytes but is less meaningful for strong electrolytes like KCl (where $\alpha = 1$ always, conductivity drops only due to interactions).

The $K_a$ we calculate here is a thermodynamic constant — it shouldn’t change with concentration (at the same temperature). If you calculate $K_a$ at different concentrations of HCOOH and get roughly the same value each time, that confirms the equilibrium model is valid.

Alternative Method — Using κ directly

If the problem gives conductivity $\kappa$ instead of $\Lambda_m$ , convert first:

\Lambda_m = \frac{\kappa \times 1000}{c}

where $\kappa$ is in S/cm, $c$ in mol/L, and $\Lambda_m$ comes out in S cm² mol⁻¹.

Say the problem gives $\kappa = 1.1525 \times 10^{-3}$ S/cm for 0.025 mol/L HCOOH:

\Lambda_m = \frac{1.1525 \times 10^{-3} \times 1000}{0.025} = 46.1 \text{ S cm}^2 \text{ mol}^{-1}

Same value — then proceed as above. Many JEE problems give $\kappa$ and expect you to do this conversion as the first step. Students who skip the ×1000 factor lose the mark.

Common Mistake

Forgetting the 1000 factor in the κ → Λm conversion.

The formula $\Lambda_m = \kappa / c$ only works if $\kappa$ is in S/m and $c$ in mol/m³. When you use the more common units (S/cm and mol/L), you must multiply by 1000:

\Lambda_m \text{ (S cm}^2\text{ mol}^{-1}) = \frac{\kappa \text{ (S/cm)} \times 1000}{c \text{ (mol/L)}}

In JEE Main 2023, this unit mismatch was the #1 source of wrong answers in conductivity numericals. Always check your units before plugging in.

For weak acids with small $\alpha$ (less than 0.1), you can approximate $1 - \alpha \approx 1$ , giving $K_a \approx c\alpha^2$ . Here $\alpha = 0.114$ is just above the threshold, so using the full expression gives a slightly more accurate answer — worth the extra step in JEE.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using κ directly

Common Mistake

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