Question

The values of limiting ionic conductance of H\textsuperscript{+ and HCOO\textsuperscript{--} ions are respectively 347 and 53 S cm\textsuperscript{2} mol\textsuperscript{--1} at 298 K. If the molar conductance of 0.025M methanoic acid at 298 K is 40 S cm\textsuperscript{2} mol\textsuperscript{--1}, the dissociation constant of methanoic acid at 298 K is}

• A. $1 \times 10^{-5}$
• B. $2 \times 10^{-5}$
• C. $1.5 \times 10^{-4}$
• D. $2.5 \times 10^{-5}$
• E. $2.5 \times 10^{-4}$

Hint:

In competitive exams, if $\alpha$ is exactly 0.1, it is usually too large to ignore in the denominator ($1-\alpha$). However, if the options are spread out, the $C\alpha^2$ approximation gets you close enough to pick the right answer!

Answer 1

The Correct Option is D

Concept: This problem requires Kohlrausch's Law to find limiting molar conductivity and Ostwald's Dilution Law to find the dissociation constant.
• Kohlrausch's Law: $\Lambda_m^\circ = \lambda^0(H^+) + \lambda^0(HCOO^-)$.
• Degree of Dissociation ($\alpha$): $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
• Dissociation Constant ($K_a$): $K_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2$ (if $\alpha$ is small).

Step 1: Calculate $\Lambda_m^\circ$ and $\alpha$. $\Lambda_m^\circ = 347 + 53 = 400 \text{ S cm}^2 \text{ mol}^{-1}$. Given $\Lambda_m = 40 \text{ S cm}^2 \text{ mol}^{-1}$: \[ \alpha = \frac{40}{400} = 0.1 \]

Step 2: Calculate the dissociation constant $K_a$. Given Concentration $C = 0.025$ M: \[ K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.025 \times (0.1)^2}{1 - 0.1} = \frac{0.025 \times 0.01}{0.9} \] \[ K_a = \frac{0.00025}{0.9} = \frac{2.5 \times 10^{-4}}{0.9} \approx 2.7 \times 10^{-5} \] Evaluating the closest option based on the provided choices: $2.5 \times 10^{-5}$.