QJEEM23Chb9e4: JEE Main 2023 Chemistry - Class 12 JEEM23Chb9e4 Solution: The specific conductance of acetic acid is at a certain…

Given that the specific conductance, $k$ , is $5 \times 10^{-5} ~S~cm^{-1}$ and the concentration, $C$ , is $0.0025~M$ , we can find the molar conductivity, $\lambda_m$ , as follows:

$\lambda_m = \frac{k}{C} \times 1000 = \frac{5 \times 10^{-5} \times 10^3}{0.0025} = \frac{5 \times 10^{-2}}{2.5 \times 10^{-3}} = 20 ~S~cm^2~mol^{-1}$

Next, we find the degree of dissociation, $\alpha$ , by dividing $\lambda_m$ by the limiting molar conductivity, $\lambda_m^o$ :

$\alpha = \frac{20}{400} = \frac{1}{20}$

Finally, we use the formula for the dissociation constant of a weak acid, $K_a$ :

$K_a = \frac{C \alpha^{2}}{1-\alpha} = \frac{0.0025 \times \frac{1}{20} \times \frac{1}{20}}{\frac{19}{20}} = \frac{0.0025}{19 \times 20} = 6.6 \times 10^{-6} = 66 \times 10^{-7}$

So, the correct answer is 66.