Analysis of the Problem:

We are given the molar conductivities of two weak acids, HQ and HZ, at specific concentrations. We know that the limiting molar conductivity of their conjugate bases ($Q^-$ and $Z^-$) are equal. The degree of dissociation ($\alpha$) is much less than 1 for both acids. We need to find the difference between their $\text{pK}_a$ values.

Relating Molar Conductivity and Dissociation

For a weak electrolyte, the molar conductivity ($\Lambda_m$) is related to the degree of dissociation ($\alpha$) and the limiting molar conductivity ($\Lambda_m^\circ$) by:

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

Using Ostwald's dilution law, for a weak acid with dissociation constant $K_a$ and concentration $C$, the degree of dissociation is approximately:

$ \alpha \approx \sqrt{\frac{K_a}{C}} \quad (\text{since } \alpha \ll 1) $

Substituting this into the molar conductivity equation:

$ \Lambda_m \approx \sqrt{\frac{K_a}{C}} \Lambda_m^\circ $

Comparing the Two Acids

Let $C_1 = 0.18 \text{ M}$ for HQ and $C_2 = 0.02 \text{ M}$ for HZ.

The limiting molar conductivity is $\Lambda_m^\circ = \lambda^\circ_{\text{H}^+} + \lambda^\circ_{\text{A}^-}$.

Given $\lambda^\circ_{Q^-} = \lambda^\circ_{Z^-}$, this implies:

$ \Lambda_m^\circ(\text{HQ}) = \lambda^\circ_{\text{H}^+} + \lambda^\circ_{Q^-} $

$ \Lambda_m^\circ(\text{HZ}) = \lambda^\circ_{\text{H}^+} + \lambda^\circ_{Z^-} $

Therefore, $\Lambda_m^\circ(\text{HQ}) = \Lambda_m^\circ(\text{HZ})$. Let this common value be $\Lambda^\circ$.

We are given $\Lambda_m(\text{HQ}) = \frac{1}{30} \Lambda_m(\text{HZ})$.

Using the approximation $\Lambda_m \approx \alpha \Lambda_m^\circ$:

$ \alpha_1 \Lambda^\circ = \frac{1}{30} \alpha_2 \Lambda^\circ $

$ \alpha_1 = \frac{1}{30} \alpha_2 $

Calculating the $K_a$ Ratio

Substitute $\alpha \approx \sqrt{K_a/C}$:

$ \sqrt{\frac{K_{a,HQ}}{C_1}} = \frac{1}{30} \sqrt{\frac{K_{a,HZ}}{C_2}} $

Square both sides:

$ \frac{K_{a,HQ}}{C_1} = \frac{1}{900} \frac{K_{a,HZ}}{C_2} $

Rearrange to find the ratio of $K_a$ values:

$ \frac{K_{a,HQ}}{K_{a,HZ}} = \frac{1}{900} \frac{C_1}{C_2} $

Substitute the concentrations ($C_1 = 0.18 \text{ M}$, $C_2 = 0.02 \text{ M}$):

$ \frac{K_{a,HQ}}{K_{a,HZ}} = \frac{1}{900} \frac{0.18}{0.02} = \frac{1}{900} \times 9 = \frac{9}{900} = \frac{1}{100} $

So, $K_{a,HQ} = \frac{1}{100} K_{a,HZ}$.

Calculating the $\text{pK}_a$ Difference

The $\text{pK}_a$ is defined as $\text{pK}_a = -\log_{10}(K_a)$.

We need to find $\text{pK}_a(\text{HQ}) - \text{pK}_a(\text{HZ})$:

$ \text{pK}_a(\text{HQ}) - \text{pK}_a(\text{HZ}) = -\log_{10}(K_{a,HQ}) - (-\log_{10}(K_{a,HZ})) $

$ = \log_{10}(K_{a,HZ}) - \log_{10}(K_{a,HQ}) $

$ = \log_{10}\left(\frac{K_{a,HZ}}{K_{a,HQ}}\right) $

Substitute the ratio $K_{a,HQ}/K_{a,HZ} = 1/100$:

$ \text{pK}_a(\text{HQ}) - \text{pK}_a(\text{HZ}) = \log_{10}\left(\frac{K_{a,HZ}}{K_{a,HZ}/100}\right) $

$ = \log_{10}(100) $

$ = 2 $

The difference in $\text{pK}_a$ values is 2. Since the correct value is stated to be between 2 and 2, the nearest integer is 2.