Calculate Dissociation Constant of Fluoroacetic Acid from Freezing Point Depression

To find the dissociation constant ($K_a$) of fluoroacetic acid, we follow these step-by-step calculations:

1. Calculate the Molality ($m$) of the Solution: The number of moles of fluoroacetic acid ($\text{CH}_2\text{FCOOH}$) is given by: $\text{Moles of solute} = \frac{\text{Mass of solute}}{\text{Molar mass}} = \frac{19.5\text{ g}}{78\text{ g mol}^{-1}} = 0.25\text{ mol}$

   The mass of the solvent (water) in kilograms is: $\text{Mass of solvent} = 500\text{ g} = 0.5\text{ kg}$

   Therefore, the molality ($m$) is: $m = \frac{0.25\text{ mol}}{0.5\text{ kg}} = 0.5\text{ mol kg}^{-1}$

2. Calculate the Van 't Hoff Factor ($i$): The formula for depression in freezing point is: $\Delta T_f = i \cdot K_f \cdot m$

   Given that $\Delta T_f = 1^\circ\text{C} = 1\text{ K}$, $K_f = 1.86\text{ K kg mol}^{-1}$, and $m = 0.5\text{ mol kg}^{-1}$: $1 = i \times 1.86 \times 0.5$ $1 = i \times 0.93$ $i = \frac{1}{0.93} \approx 1.0753$

3. Determine the Degree of Dissociation ($\alpha$): Fluoroacetic acid dissociates in water according to the reaction: $\text{CH}_2\text{FCOOH} \rightleftharpoons \text{CH}_2\text{FCOO}^- + \text{H}^+$

   The relation between the van 't Hoff factor $i$ and degree of dissociation $\alpha$ for a monobasic acid is: $i = 1 + \alpha$ $\alpha = i - 1 = 1.0753 - 1 = 0.0753$

4. Calculate the Dissociation Constant ($K_a$): Given that molarity ($C$) is equal to molality ($m = 0.5\text{ M}$), the expression for $K_a$ is: $K_a = \frac{C \alpha^2}{1 - \alpha}$

   Substituting the values of $C$ and $\alpha$: $K_a = \frac{0.5 \times (0.0753)^2}{1 - 0.0753}$ $K_a = \frac{0.5 \times 0.00567}{0.9247} \approx 3.07 \times 10^{-3} \approx 3 \times 10^{-3}$

Thus, the correct option is D.