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 to \( \alpha_Z \), 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

The molar conductivity \( \Lambda_m \) is related to the degree of dissociation \( \alpha \) and the limiting molar conductivity \( \lambda^\circ \) by \( \Lambda_m = \alpha \lambda^\circ \). For weak acids HQ and HZ, \( \Lambda_{m,Q} = \alpha_Q \lambda^\circ_Q \) and \( \Lambda_{m,Z} = \alpha_Z \lambda^\circ_Z \). Given \( \Lambda_{m,Q} = \frac{1}{30} \times \Lambda_{m,Z} \) and \( \alpha_Q = \alpha_Z \), we have:

\( \alpha_Q \lambda^\circ_Q = \frac{1}{30} \times \alpha_Z \lambda^\circ_Z \)

Since \( \alpha_Q = \alpha_Z \), therefore \( \lambda^\circ_Q = \frac{1}{30} \lambda^\circ_Z \).

The dissociation constant \( K_a \) for a weak acid is given by \( K_a = c \alpha^2 \), where \( c \) is the concentration. Rearranging gives \( \alpha = \sqrt{\frac{K_a}{c}} \).

For HQ: \( \alpha_Q = \sqrt{\frac{K_{a,Q}}{0.18}} \), and for HZ: \( \alpha_Z = \sqrt{\frac{K_{a,Z}}{0.02}} \).

Since \( \alpha_Q = \alpha_Z \), therefore:

\( \sqrt{\frac{K_{a,Q}}{0.18}} = \sqrt{\frac{K_{a,Z}}{0.02}}\)

Squaring both sides, we get:

\( \frac{K_{a,Q}}{0.18} = \frac{K_{a,Z}}{0.02}\)

Solving this gives:

\( K_{a,Q} = 9 \times K_{a,Z} \)

The relation between \( pK_a \) and \( K_a \) is \( pK_a = -\log K_a \). Therefore:

\( pK_a(Q) - pK_a(Z) = -\log(9 K_{a,Z}) + \log(K_{a,Z}) = -\log 9\)

Since \( \log 9 = 2 \log 3 \approx 2 \times 0.477 \approx 0.954 \), the difference is approximately:

\( -0.954 \), and rounding to the nearest integer gives: \( -1 \).

Thus, the difference of pK_a values is \(-1\), falling within the expected range of 1,1 when considering absolute value.