JEE Chemistry 2026 Question with Solution

Given: Molar conductivity ratio $\dfrac{\Lambda_{HQ}}{\Lambda_{HZ}} = \dfrac{1}{30}$, concentrations are $C_{HQ} = 0.18 \, \text{M}$ and $C_{HZ} = 0.02 \, \text{M}$, and $\alpha_Q = \alpha_Z$ with $\alpha \ll 1$.

Find: $\text{p}K_a(\text{HQ}) - \text{p}K_a(\text{HZ})$.

For weak electrolytes,

$$\Lambda_m = \alpha \Lambda_m^\circ$$

Given,

$$\frac{\Lambda_{HQ}}{\Lambda_{HZ}} = \frac{1}{30}$$

Since $\alpha_Q = \alpha_Z$,

$$\frac{\alpha \Lambda^\circ_{HQ}}{\alpha \Lambda^\circ_{HZ}} = \frac{1}{30}$$

So,

$$\frac{\Lambda^\circ_{HQ}}{\Lambda^\circ_{HZ}} = \frac{1}{30}$$

Using Ostwald’s dilution law for weak acids,

$$K_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2$$

Because both acids have the same degree of dissociation, the ratio of dissociation constants becomes

$$\frac{K_{a,HQ}}{K_{a,HZ}} = \frac{C_{HQ}}{C_{HZ}} = \frac{0.18}{0.02} = 9$$

Now,

$$\text{p}K_a = -\log K_a$$

Therefore,

$$\text{p}K_a(\text{HQ}) - \text{p}K_a(\text{HZ}) = -\log\left(\frac{K_{a,HQ}}{K_{a,HZ}}\right) = -\log 9$$

Its magnitude is approximately

$$\log 9 \approx 1$$

Hence, the nearest integer asked in the solution is $1$.

Given: $\alpha_Q = \alpha_Z$ and $\alpha \ll 1$ for both weak acids.

Find: The difference in their $\text{p}K_a$ values.

From Ostwald’s dilution law,

$$K_a \approx C\alpha^2$$

If the two weak acids have the same $\alpha$, then

$$K_a \propto C$$

Thus,

$$\frac{K_{a,HQ}}{K_{a,HZ}} = \frac{0.18}{0.02} = 9$$

Now convert to $\text{p}K_a$ scale:

$$\Delta \text{p}K_a = \log 9 \approx 1$$

So the required nearest integer is $1$.