Question

Dissociation constant and molar conductance of acetic acid are $1.78\times10^{-5}~mol~L^{-1}$ and $48.15~S~cm^{2}~mol^{-1}$. Conductivity is:

• A. $4.9\times10^{-2}~S~cm^{-1}$
• B. $4.9\times10^{2}~S~cm^{-1}$
• C. $4.9\times10^{-5}~S~cm^{-1}$
• D. $4.9\times10^{-5}~S~cm^{-1}$

Hint:

Ostwald's Dilution Law: $K_a = C\alpha^2$ is key for weak electrolytes.

Answer 1

The Correct Option is D

Step 1: Concept
Conductivity $\kappa = \frac{\Lambda_m \times C}{1000}$. First, find concentration $C$ using $K_a = C\alpha^2$.
Step 2: Analysis
$\alpha = \Lambda_m / \Lambda_m^\infty = 48.15 / 390.5 \approx 0.123$.
$C = K_a / \alpha^2 = 1.78\times10^{-5} / (0.123)^2 \approx 1.17\times10^{-3}~M$.
$\kappa = (\Lambda_m \times C) / 1000 \approx 4.9\times10^{-5}~S~cm^{-1}$.
Step 3: Conclusion
The calculated conductivity is $4.9\times10^{-5}~S~cm^{-1}$. Final Answer:(D)