Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 1/30 of the molar conductivity of another weak acid HZ with concentration of 0.02M. If $$\lambda^{\circ}{}_{Q}-$$ happened to be equal with $$\lambda^{\circ}{}_{Z}-$$, then the difference of the $$pK_{a}$$ values of the two weak acids $$(pK_{a}(HQ) - pK_{a}(HZ))$$ is ___ (Nearest integer).
[Given: degree of dissociation ($$\alpha$$) << 1 for both weak acids, $$\lambda^{\circ}$$ : limiting molar conductivity of ions]

We need to find the difference between the $$pK_a$$ values of the two weak acids, $$pK_a(\text{HQ}) - pK_a(\text{HZ})$$.

1. Degree of Dissociation ($$\alpha$$) Relationship:

• For a weak electrolyte, the degree of dissociation is given by:
  $$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$$
• Since the limiting molar conductivities of the respective ions are equal ($$\lambda^\circ_{\text{Q}^-} = \lambda^\circ_{\text{Z}^-}$$), their total limiting molar conductivities are equal:
  $$\Lambda_m^\circ(\text{HQ}) = \Lambda_m^\circ(\text{HZ})$$
• Given that the molar conductivity of $$\text{HQ}$$ is $$\frac{1}{30}$$ of $$\text{HZ}$$ ($$\Lambda_m(\text{HQ}) = \frac{1}{30}\Lambda_m(\text{HZ})$$):
  $$\frac{\alpha_{\text{HQ}}}{\alpha_{\text{HZ}}} = \frac{\Lambda_m(\text{HQ})}{\Lambda_m(\text{HZ})} = \frac{1}{30}$$

2. Acid Dissociation Constant ($K_a$) and $$pK_a$$ Calculation:

• Since $$\alpha \ll 1$$, the dissociation constant simplifies to:
  $$K_a = C\alpha^2$$
• Taking the ratio of the dissociation constants for both acids:
  $$\frac{K_a(\text{HQ})}{K_a(\text{HZ})} = \frac{C_{\text{HQ}}}{C_{\text{HZ}}} \times \left(\frac{\alpha_{\text{HQ}}}{\alpha_{\text{HZ}}}\right)^2$$
• Substitute the given concentrations ($$C_{\text{HQ}} = 0.18\text{ M}$$, $$C_{\text{HZ}} = 0.02\text{ M}$$) and the derived $$\alpha$$ ratio:
  $$\frac{K_a(\text{HQ})}{K_a(\text{HZ})} = \frac{0.18}{0.02} \times \left(\frac{1}{30}\right)^2 = 9 \times \frac{1}{900} = \frac{1}{100} = 10^{-2}$$
• Taking the negative logarithm ($$-\log$$) on both sides:
  $$-\log\left(\frac{K_a(\text{HQ})}{K_a(\text{HZ})}\right) = -\log(10^{-2})$$
  $$pK_a(\text{HQ}) - pK_a(\text{HZ}) = 2$$

Conclusion:

The difference between the $$pK_a$$ values of the two weak acids is exactly 2.

Answer: 2