Question

The pH and conductance of a weak acid (HX) was found to be 5 and \(4\times10^{-5}\) S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1 cm\(^2\) were at a distance of 15 cm apart. The value of the limiting molar conductivity (\(\Lambda_m^\circ\)) is___ S cm\(^2\) mol\(^{-1}\) (nearest integer) (Given: degree of dissociation \(\alpha \ll 1\))
  

Hint:

The relation \(\Lambda_m^\circ = \frac{1000 \kappa}{[H^+]}\) is a useful shortcut for weak monobasic acids when \(\alpha \ll 1\), combining the dissociation equilibrium and conductivity definitions.

Answer 1

Correct Answer: 6

   Step 1: Calculate Conductivity (\(\kappa\))
Conductance \(G = 4 \times 10^{-5}\) S. Cell parameters: \(l = 15\) cm, \(A = 1\) cm\(^2\). Cell constant \(G^* = \frac{l}{A} = \frac{15}{1} = 15 \text{ cm}^{-1}\). \[ \kappa = G \times G^* = (4 \times 10^{-5}) \times 15 = 60 \times 10^{-5} = 6 \times 10^{-4} \text{ S cm}^{-1} \] Step 2: Relate \(\Lambda_m^\circ\) to Data
For a weak acid, \([H^+] = c \alpha\). Also, \(\Lambda_m = \frac{1000 \kappa}{c}\) and \(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\). Substituting \(\Lambda_m\): \[ \alpha = \frac{1000 \kappa}{c \Lambda_m^\circ} \] Since \([H^+] = c \alpha\), we can substitute \(\alpha = [H^+]/c\): \[ \frac{[H^+]}{c} = \frac{1000 \kappa}{c \Lambda_m^\circ} \] \[ [H^+] = \frac{1000 \kappa}{\Lambda_m^\circ} \implies \Lambda_m^\circ = \frac{1000 \kappa}{[H^+]} \] Step 3: Calculation
Given pH = 5 \(\implies [H^+] = 10^{-5}\) M. \[ \Lambda_m^\circ = \frac{1000 \times (6 \times 10^{-4})}{10^{-5}} \] \[ \Lambda_m^\circ = \frac{0.6}{10^{-5}} = 0.6 \times 10^5 = 60,000 \text{ S cm}^2 \text{ mol}^{-1} \] (Note: While 60,000 is physically unlikely for an aqueous ion—typical values are<500—this is the correct mathematical result derived from the specific numbers provided in the question text.) Final Answer: 60000.