Question

The molar conductivity of a weak monobasic acid, HA at 298 K is \(70\ \text{S cm}^2\text{ mol}^{-1}\). What is the percentage ionisation of HA at 298 K?
At infinite dilution \(\lambda^\circ_{H^+} = 340\ \text{S cm}^2\text{ mol}^{-1}\) and \(\lambda^\circ_{A^-} = 80\ \text{S cm}^2\text{ mol}^{-1}\)

• A. $8.35\ \%$
• B. $16.7\ \%$
• C. $20\ \%$
• D. $32.5\ \%$
• E. $15.3\ \%$

Hint:

For weak electrolytes: - $\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}$ - Add ionic conductivities to get $\Lambda^\circ_m$

Answer 1

The Correct Option is B

Concept:
• Molar conductivity at infinite dilution: \[ \Lambda^\circ_m = \lambda^\circ_{H^+} + \lambda^\circ_{A^-} \]
• Degree of ionisation: \[ \alpha = \frac{\Lambda_m}{\Lambda^\circ_m} \]
• Percentage ionisation: \[ \% = \alpha \times 100 \]

Step 1: Calculate $\Lambda^\circ_m$.
\[ \Lambda^\circ_m = 340 + 80 = 420\ \text{S cm}^2\text{ mol}^{-1} \]

Step 2: Find degree of ionisation.
\[ \alpha = \frac{70}{420} = \frac{1}{6} \approx 0.167 \]

Step 3: Calculate percentage ionisation.
\[ \% = 0.167 \times 100 = 16.7\% \]