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

To solve the problem, we need to calculate the limiting molar conductivity (\( \Lambda_m^0 \)) of acetic acid and its degree of dissociation (\( \alpha \)), given its molar conductivity (\( \Lambda_m \)) as 48.1 \( \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \), and the limiting molar conductivities: \( \Lambda_m^0 (\text{HCl}) = 426 \), \( \Lambda_m^0 (\text{NaCl}) = 126 \), and \( \Lambda_m^0 (\text{CH}_3\text{COONa}) = 91 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).

1. Calculate \( \Lambda_m^0 \) for Acetic Acid:
Acetic acid (\( \text{CH}_3\text{COOH} \)) dissociates into \( \text{H}^+ \) and \( \text{CH}_3\text{COO}^- \). We use Kohlrausch’s law to find \( \Lambda_m^0 (\text{CH}_3\text{COOH}) \): \( \Lambda_m^0 (\text{CH}_3\text{COOH}) = \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) \).
We can find these ionic conductivities using the given data:
- \( \Lambda_m^0 (\text{HCl}) = \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{Cl}^-) = 426 \),
- \( \Lambda_m^0 (\text{NaCl}) = \lambda_m^0 (\text{Na}^+) + \lambda_m^0 (\text{Cl}^-) = 126 \),
- \( \Lambda_m^0 (\text{CH}_3\text{COONa}) = \lambda_m^0 (\text{Na}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) = 91 \).
Subtract the second equation from the first to eliminate \( \lambda_m^0 (\text{Cl}^-) \):
\( \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{Cl}^-) - (\lambda_m^0 (\text{Na}^+) + \lambda_m^0 (\text{Cl}^-)) = 426 - 126 \),
\( \lambda_m^0 (\text{H}^+) - \lambda_m^0 (\text{Na}^+) = 300 \).
Now use the third equation with \( \lambda_m^0 (\text{Na}^+) \):
From the third equation, \( \lambda_m^0 (\text{CH}_3\text{COO}^-) = 91 - \lambda_m^0 (\text{Na}^+) \).
We need \( \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) \):
Add \( \lambda_m^0 (\text{H}^+) - \lambda_m^0 (\text{Na}^+) = 300 \) to \( \lambda_m^0 (\text{Na}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) = 91 \):
\( \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) = 300 + 91 = 391 \).
Thus, \( \Lambda_m^0 (\text{CH}_3\text{COOH}) = 391 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).

2. Calculate the Degree of Dissociation (\( \alpha \)):
The degree of dissociation is given by \( \alpha = \frac{\Lambda_m}{\Lambda_m^0} \):

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

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