Question

Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be $\dfrac{1{30}$ of the molar conductivity of another weak acid HZ with concentration 0.02 M. If $\alpha_Q$ happened to be equal with $\alpha_Z$, 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 ($\alpha \ll 1$ for both weak acids, $\lambda^\circ$ : limiting molar conductivity of ions)

Hint:

For weak acids with same degree of dissociation, $K_a \propto$ concentration.

Answer 1

Correct Answer: 1

Step 1: Relation between molar conductivity and dissociation.
For weak electrolytes:
\[ \Lambda_m = \alpha \Lambda_m^\circ \]
Step 2: Given ratio of molar conductivities.
\[ \frac{\Lambda_{HQ}}{\Lambda_{HZ}} = \frac{1}{30} \]
Since $\alpha_Q = \alpha_Z$:
\[ \frac{\alpha \Lambda^\circ_{HQ}}{\alpha \Lambda^\circ_{HZ}} = \frac{1}{30} \Rightarrow \frac{\Lambda^\circ_{HQ}}{\Lambda^\circ_{HZ}} = \frac{1}{30} \]
Step 3: Using Ostwald’s dilution law.
\[ K_a = \frac{C \alpha^2}{1-\alpha} \approx C \alpha^2 \]
Step 4: Ratio of dissociation constants.
\[ \frac{K_{a,HQ}}{K_{a,HZ}} = \frac{C_{HQ}}{C_{HZ}} = \frac{0.18}{0.02} = 9 \]
Step 5: Difference in pK$_a$.
\[ \Delta \text{p}K_a = \log 9 \approx 1 \]