0.18 M HQ solution has molar conductivity 1/30 times the molar conductivity of 0.02 M HZ solution. Find the value of pKa(HQ) – pKa(Hz). [Given that α <<< 1]

Assume that λ_m^∞Q = λ_m^∞Z

The correct answer is 2.

Let us solve the problem step-by-step.
We are given two weak acid solutions:
1. HQ solution with concentration $C_1=0.18\text{M}$ and molar conductivity ${\Lambda }_{m,1}$.
2. HZ solution with concentration $C_2=0.02\text{M}$ and molar conductivity ${\Lambda }_{m,2}$.

The relationship between the molar conductivities is:
${\Lambda }_{m,1}=\frac{1}{30}{\Lambda }_{m,2}$

We are also given that the limiting molar conductivities of the anions are equal:
${\lambda }_m^{\infty }\left(Q^{-}\right)={\lambda }_m^{\infty }\left(Z^{-}\right)$
Since both acids release $H^{+}$, their limiting molar conductivities are equal:
${\Lambda }_{m,1}^{\infty }={\Lambda }_{m,2}^{\infty }={\Lambda }^{\infty }$

The degree of dissociation ($\alpha$) for a weak electrolyte is given by:
$\alpha =\frac{{\Lambda }_m}{{\Lambda }^{\infty }}$

Therefore, the ratio of their degrees of dissociation is:
$\frac{{\alpha }_1}{{\alpha }_2}=\frac{{\Lambda }_{m,1}}{{\Lambda }_{m,2}}=\frac{1}{30}$

For a weak acid, the acid dissociation constant ($K_a$) is related to $C$ and $\alpha$ by:
$K_a=\frac{C{\alpha }^2}{1-\alpha }$
Given that $\alpha \ll 1$, we can approximate this as:
$K_a\approx C{\alpha }^2$

Let us find the ratio of $K_a\left(\text{HQ}\right)$ to $K_a\left(\text{HZ}\right)$:
$\frac{K_a\left(\text{HQ}\right)}{K_a\left(\text{HZ}\right)}=\frac{C_1{\alpha }_1^2}{C_2{\alpha }_2^2}=\frac{C_1}{C_2}\cdot {\left(\frac{{\alpha }_1}{{\alpha }_2}\right)}^2$

Substitute the given values into the equation:
$\frac{K_a\left(\text{HQ}\right)}{K_a\left(\text{HZ}\right)}=\frac{0.18}{0.02}\cdot {\left(\frac{1}{30}\right)}^2=9\cdot \frac{1}{900}=\frac{1}{100}={10}^{-2}$

Taking the negative logarithm on both sides:
$-{\log }_{10}\left(\frac{K_a\left(\text{HQ}\right)}{K_a\left(\text{HZ}\right)}\right)=-{\log }_{10}\left({10}^{-2}\right)$
$-{\log }_{10}{K}_a\left(\text{HQ}\right)-(-{\log }_{10}{K}_a(\text{HZ}\left)\right)=2$
${\text{pKa}}_{\text{(HQ)}}-{\text{pKa}}_{\text{(HZ)}}=2$