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

This problem involves colligative properties (depression in freezing point) and ionic equilibrium (acid dissociation constant).
1. Calculate molality (m):
Molar mass of $FCH_2COOH = 19 + 12 + 2 + 12 + 16 + 16 + 1 = 78 \text{ g/mol}$.
Moles of acid = $\frac{19.5}{78} = 0.25 \text{ moles}$.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.25}{0.5} = 0.5 \text{ m}$.

2. Find Van't Hoff factor (i):
$\Delta T_f = i \cdot K_f \cdot m$
$1 = i \cdot 1.86 \cdot 0.5$
$1 = i \cdot 0.93 \Rightarrow i = \frac{1}{0.93} \approx 1.0753$

3. Find degree of dissociation ($\alpha$):
$FCH_2COOH \rightleftharpoons FCH_2COO^- + H^+$
The number of particles produced, $n = 2$.
$i = 1 + (n-1)\alpha \Rightarrow 1.0753 = 1 + \alpha$
$\alpha = 0.0753$

4. Calculate $K_a$:
$K_a = \frac{C\alpha^2}{1-\alpha}$
Since $\alpha$ is small relative to 1, we can approximate $K_a \approx C\alpha^2$. Here $C = \text{molality} = 0.5$.
$K_a = 0.5 \cdot (0.0753)^2 \approx 0.5 \cdot 0.00567 \approx 0.00283$
$K_a \approx 2.8 \times 10^{-3}$

Thus, option (1) is correct.