Question

Calculate \( \Lambda_m^0 \) for acetic acid and its degree of dissociation (\( \alpha \)) if its molar conductivity is 48.1 \( \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \). 
Given that
\( \Lambda_m^0 (\text{HC}) = 426 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \), 
\( \Lambda_m^0 (\text{NaCl}) = 126 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \), 
\( \Lambda_m^0 (\text{CH}_3\text{COONa}) = 91 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \). 
 

Hint:

The degree of dissociation \( \alpha \) indicates the fraction of the total molecules that dissociate into ions in a solution.

Answer 1

The problem requires the calculation of acetic acid's limiting molar conductivity (\( \Lambda_m^0 \)) and its degree of dissociation (\( \alpha \)). Given data includes acetic acid's molar conductivity (\( \Lambda_m \)) as 48.1 \( \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \), and the limiting molar conductivities for HCl (\( \Lambda_m^0 (\text{HCl}) = 426 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)), NaCl (\( \Lambda_m^0 (\text{NaCl}) = 126 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)), and CH3COONa (\( \Lambda_m^0 (\text{CH}_3\text{COONa}) = 91 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)).

1. Determination of \( \Lambda_m^0 \) for Acetic Acid:
Acetic acid (\( \text{CH}_3\text{COOH} \)) dissociates into \( \text{H}^+ \) and \( \text{CH}_3\text{COO}^- \). Kohlrausch's law states \( \Lambda_m^0 (\text{CH}_3\text{COOH}) = \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) \). The individual ionic conductivities are derived from the provided data:
- \( \Lambda_m^0 (\text{HCl}) = \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{Cl}^-) = 426 \),
- \( \Lambda_m^0 (\text{NaCl}) = \lambda_m^0 (\text{Na}^+) + \lambda_m^0 (\text{Cl}^-) = 126 \),
- \( \Lambda_m^0 (\text{CH}_3\text{COONa}) = \lambda_m^0 (\text{Na}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) = 91 \).
Subtracting the second equation from the first yields: \( \lambda_m^0 (\text{H}^+) - \lambda_m^0 (\text{Na}^+) = 426 - 126 = 300 \).
Using the third equation, \( \lambda_m^0 (\text{CH}_3\text{COO}^-) = 91 - \lambda_m^0 (\text{Na}^+) \).
To find \( \Lambda_m^0 (\text{CH}_3\text{COOH}) \), we add \( \lambda_m^0 (\text{H}^+) - \lambda_m^0 (\text{Na}^+) = 300 \) and \( \lambda_m^0 (\text{Na}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) = 91 \). This results in \( \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) = 300 + 91 = 391 \).
Therefore, \( \Lambda_m^0 (\text{CH}_3\text{COOH}) = 391 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).

2. Calculation of the Degree of Dissociation (\( \alpha \)):
The degree of dissociation is calculated using the formula \( \alpha = \frac{\Lambda_m}{\Lambda_m^0} \).

\( \alpha = \frac{48.1}{391} \approx 0.123 \).

Final Results:
The limiting molar conductivity for acetic acid (\( \Lambda_m^0 \)) is \( 391 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \), and its degree of dissociation (\( \alpha \)) is approximately \( 0.123 \).