JEE Main 2023 — Electrochemistry Question with Solution

Given that the specific conductance, $k$, is $5\times 10^{-5}Sc{m}^{-1}$ and the concentration, $C$, is $0.0025M$, we can find the molar conductivity, ${\lambda }_m$, as follows:

$${\lambda }_m=\frac{k}{C}\times 1000=\frac{5\times 10^{-5}\times 10^3}{0.0025}=\frac{5\times 10^{-2}}{2.5\times 10^{-3}}=20Sc{m}^{2}mo{l}^{-1}$$

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.