Question

The conductivity of a weak acid HA of concentration 0.001 mol L$^{-1}$ is $2.0 \times 10^{-5}$ S cm$^{-1}$. If $\Lambda_m^\circ(HA) = 190$ S cm$^2$ mol$^{-1}$, the ionization constant (K$_a$) of HA is equal to _________ $\times 10^{-6}$. (Round off to the Nearest Integer)

Hint:

In Ostwald’s dilution law, \[ K_a = \frac{C\alpha^2}{1-\alpha} \] If $\alpha>0.05$, the approximation $(1-\alpha)\approx1$ should NOT be used.

Answer 1

Correct Answer: 12

Given: \[ \kappa = 2.0 \times 10^{-5}\ \text{S cm}^{-1} \] \[ C = 0.001\ \text{mol L}^{-1} \] \[ \Lambda_m^\circ = 190\ \text{S cm}^2\ \text{mol}^{-1} \] Step 1: Calculate molar conductivity $\Lambda_m$ \[ \Lambda_m = \frac{1000 \kappa}{C} \] \[ \Lambda_m = \frac{1000 \times 2.0 \times 10^{-5}}{10^{-3}} = 20\ \text{S cm}^2\ \text{mol}^{-1} \] Step 2: Calculate degree of dissociation $\alpha$ \[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{20}{190} = \frac{2}{19} \approx 0.105 \] Step 3: Apply Ostwald’s dilution law For a weak acid: \[ K_a = \frac{C\alpha^2}{1-\alpha} \] Substitute values: \[ K_a = \frac{10^{-3} \times \left(\frac{2}{19}\right)^2}{1 - \frac{2}{19}} \] \[ K_a = \frac{10^{-3} \times \frac{4}{361}}{\frac{17}{19}} \] \[ K_a = \frac{76}{6137} \times 10^{-3} \approx 1.238 \times 10^{-5} \] Step 4: Express in required format \[ K_a = 12.38 \times 10^{-6} \] Rounding to the nearest integer: \[ \boxed{12} \]