Question

0.18 M HQ solution has molar conductivity $\frac{1}{30}$ times the molar conductivity of 0.02 M HZ solution. Find the value of pK$_a$(HQ) − pK$_a$(HZ).
[Given that $\alpha$ is very less than 1]

Hint:

Lower molar conductivity implies lower degree of dissociation and hence weaker acid strength.

Answer 1

Correct Answer: 2

Step 1: Relation between dissociation constant and degree of dissociation.
For weak electrolytes:
\[ K_a = C \alpha^2 \]
Step 2: Relation between degree of dissociation and molar conductivity.
\[ \alpha = \frac{\lambda_m}{\lambda_m^0} \]
Given that $\lambda_m^0$(Q$^+$) = $\lambda_m^0$(Z$^+$),
Step 3: Ratio of dissociation constants.
\[ \frac{K_a(HQ)}{K_a(HZ)} = \frac{C_1}{C_2} \left( \frac{\lambda_m(HQ)}{\lambda_m(HZ)} \right)^2 \]
Substituting values:
\[ = \frac{0.18}{0.02} \left( \frac{1}{30} \right)^2 = 9 \times \frac{1}{900} = \frac{1}{100} \]
Step 4: Conversion to pK values.
\[ pK_a(HQ) - pK_a(HZ) = \log 100 = 2 \]