Question

19.5 gm $FCH_2COOH$ is dissolved in 500 gm water due to which depression in freezing point is found to be $1^{\circ} C$. Calculate $K_a$ of $FCH_2COOH$. $K_f$ of water = 1.86 K-kg/mole, m = M

• A. $2.8 \times 10^{-3}$
• B. $2.8 \times 10^{-2}$
• C. $1.4 \times 10^{-3}$
• D. $5.6 \times 10^{-3}$

Answer 1

The Correct Option is A

Application of the freezing point depression formula to find the degree of ionization, and subsequently the acid dissociation constant.

LOGIC:
1. Molality calculation: Moles of Fluoracetic acid = $19.5 / 78 = 0.25$. Mass of water = $0.5 \text{ kg}$. Molality = $0.25/0.5 = 0.5 \text{ mol/kg}$.
2. Using $\Delta T_f = i K_f m$: $1 = i \times 1.86 \times 0.5$. So, $i = 1 / 0.93 = 1.075$.
3. Relation for monobasic acid dissociation: $i = 1 + \alpha$. Thus, $\alpha = i - 1 = 0.075$.
4. Calculate $K_a$: $K_a = \frac{m \alpha^2}{1 - \alpha}$. Assuming $1-\alpha \approx 1$ for dilute solution:
$K_a = 0.5 \times (0.075)^2 = 0.5 \times 0.005625 = 0.0028125$.
5. In scientific notation: $2.8 \times 10^{-3}$. Matches option (1).