The pH and conductance of a weak acid (HX) was found to be 5 and… - Solution | JEE Chemistry

NVAMediumJEE 2026Electrolytic Conductance & Kohlrausch's Law

The pH and conductance of a weak acid ( $HX$ ) was found to be $5$ and $4\times10^{-5} \, \text{S}$ , 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 $15 \, \text{cm}$ apart. The value of the limiting molar conductivity ( $\Lambda_m^\circ$ ) is _____ $\text{S cm}^2 \, \text{mol}^{-1}$ (nearest integer) (Given: degree of dissociation $\alpha \ll 1$ )

Answer

Correct answer:60000

Step-by-step solution

Standard Method

Given: weak acid $HX$ , pH = $5$ , conductance $G = 4\times10^{-5} \, \text{S}$ , plate separation $l = 15 \, \text{cm}$ , electrode area $A = 1 \, \text{cm}^2$ , and $\alpha \ll 1$ .

Find: limiting molar conductivity $\Lambda_m^\circ$ .

First calculate conductivity:

\frac{l}{A} = \frac{15}{1} = 15 \, \text{cm}^{-1}

\kappa = G \times \frac{l}{A} = \left(4\times10^{-5}\right) \times 15 = 6\times10^{-4} \, \text{S cm}^{-1}

For a weak monobasic acid:

[H^+] = c\alpha

Also,

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

and

\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}

So,

\alpha = \frac{1000\kappa}{c\Lambda_m^\circ}

Using $[H^+] = c\alpha$ ,

\frac{[H^+]}{c} = \frac{1000\kappa}{c\Lambda_m^\circ}

Hence,

[H^+] = \frac{1000\kappa}{\Lambda_m^\circ}

Therefore,

\Lambda_m^\circ = \frac{1000\kappa}{[H^+]}

From pH = $5$ ,

[H^+] = 10^{-5} \, \text{M}

Substitute the values:

\Lambda_m^\circ = \frac{1000 \times 6\times10^{-4}}{10^{-5}} = \frac{0.6}{10^{-5}} = 0.6\times10^5 = 60000 \, \text{S cm}^2 \, \text{mol}^{-1}

Therefore, the limiting molar conductivity is $60000 \, \text{S cm}^2 \, \text{mol}^{-1}$ . The numerical value answer is 60000.

The provided answer key $6$ disagrees with the extracted solution working, and the solution gives 60000.

Use the direct weak-acid relation

Given: $\alpha \ll 1$ for weak acid $HX$ .

Find: $\Lambda_m^\circ$ .

For a weak monobasic acid, combining $\Lambda_m = \frac{1000\kappa}{c}$ , $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$ , and $[H^+] = c\alpha$ gives the direct relation:

\Lambda_m^\circ = \frac{1000\kappa}{[H^+]}

Now,

\kappa = G \times \frac{l}{A} = 4\times10^{-5} \times 15 = 6\times10^{-4} \, \text{S cm}^{-1}

and from pH = $5$ ,

[H^+] = 10^{-5} \, \text{M}

Therefore,

\Lambda_m^\circ = \frac{1000\times6\times10^{-4}}{10^{-5}} = 60000 \, \text{S cm}^2 \, \text{mol}^{-1}

So, the numerical answer is 60000.

Common mistakes

Using conductance $G$ directly in place of conductivity $\kappa$ is incorrect because cell geometry matters. First convert with $\kappa = G\frac{l}{A}$ , then use conductivity formulas.
Treating pH = $5$ as $[H^+] = 5$ is wrong. pH is logarithmic, so $[H^+] = 10^{-5} \, \text{M}$ .
Using $\Lambda_m^\circ = \frac{1000\kappa}{c}$ is incorrect because that expression gives $\Lambda_m$ , not limiting molar conductivity. For this weak acid with $\alpha \ll 1$ , connect $\Lambda_m$ , $\alpha$ , and $[H^+]$ before solving for $\Lambda_m^\circ$ .

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