Question

The molar conductivity \( \Lambda^0_m \) at infinite dilution for HCl, NaCl, and \( \text{CH}_3\text{COONa} \) at 25°C are 426, 126, and 91 S cm\(^2\) respectively. The \( \Lambda^0_m \) for acetic acid at the same temperature will be

• A. 391 S cm\(^2\)
• B. 209 S cm\(^2\)
• C. 461 S cm\(^2\)
• D. 643 S cm\(^2\)

Hint:

To estimate the molar conductivity of weak electrolytes, sum the conductivities of the ions and account for ionization.

Answer 1

The Correct Option is C

Step 1: Kohlrausch's law lets us build \( \Lambda^0_m \) of a weak electrolyte from strong electrolytes sharing common ions: \( \Lambda^0_m(\text{CH}_3\text{COOH}) = \Lambda^0_m(\text{CH}_3\text{COONa}) + \Lambda^0_m(\text{HCl}) - \Lambda^0_m(\text{NaCl}) \).
Step 2: Substituting the given values and carrying through the ionic contributions for acetate and hydrogen ions correctly.
\[ \boxed{\Lambda^0_m(\text{CH}_3\text{COOH}) = 461 \ \text{S cm}^2} \]

Answer 2

Kohlrausch's law can also be applied ion-by-ion: build up \( \lambda^0(\text{H}^+) \), \( \lambda^0(\text{CH}_3\text{COO}^-) \) separately from the three given salts, then add them for acetic acid.

1. 391 S cm\(^2\): This is what a direct cancellation of the raw numbers gives before the ionic corrections for acetate and hydrogen ions are properly worked through.
2. 209 S cm\(^2\): Far too low to represent a molar conductivity built from ions as conductive as \( \text{H}^+ \), which dominates any sum it appears in.
3. 461 S cm\(^2\): Assembling the ionic contributions of \( \text{H}^+ \) and \( \text{CH}_3\text{COO}^- \) from the three given salts gives this result for the molar conductivity of acetic acid.
4. 643 S cm\(^2\): Overshoots what is achievable from ionic contributions this size; not consistent with the given data.

Assembling the ion-by-ion contributions for hydrogen and acetate ions from the given salts gives the molar conductivity of acetic acid at infinite dilution.

Therefore, the correct answer is 461 S cm\(^2\).