Question

19.5 g of fluoro acetic acid (molar mass = 78 g mol\(^{-1}\)) is dissolved in 500 g of water at 298 K. The depression in the freezing point of water was \(1^\circ C\). What is \(K_a\) of fluoro acetic acid? (For water, \(K_f = 1.86\, K\,kg\,mol^{-1}\)). Assume molarity and molality to have same values.

• A. \(10^{-6}\)
• B. \(4\times10^{-4}\)
• C. \(3\times10^{-5}\)
• D. \(3\times10^{-3}\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Depression in freezing point is a colligative property that depends on the total number of particles in solution. For a weak acid like fluoroacetic acid, we must use the van't Hoff factor (\(i\)) to account for dissociation and then calculate the acid dissociation constant (\(K_a\)).
Step 2: Key Formula or Approach:
1. \(\Delta T_f = i \cdot K_f \cdot m\)
2. \(i = 1 + \alpha\) (for monoprotic acid)
3. \(K_a = \frac{C \alpha^2}{1-\alpha} \approx C \alpha^2\)
Step 3: Detailed Explanation:
1. Calculate molality (\(m\)):
\[ m = \frac{\text{Mass of acid / Molar mass}}{\text{Mass of solvent in kg}} = \frac{19.5 / 78}{0.5} = \frac{0.25}{0.5} = 0.5 \text{ mol/kg} \]
2. Find van't Hoff factor (\(i\)):
\[ 1 = i \times 1.86 \times 0.5 \implies 1 = i \times 0.93 \]
\[ i = \frac{1}{0.93} \approx 1.075 \]
3. Find degree of dissociation (\(\alpha\)):
\[ \alpha = i - 1 = 1.075 - 1 = 0.075 \]
4. Calculate \(K_a\) (given \(C \approx m = 0.5\)):
\[ K_a = \frac{C \alpha^2}{1-\alpha} = \frac{0.5 \times (0.075)^2}{1 - 0.075} = \frac{0.5 \times 0.005625}{0.925} \]
\[ K_a \approx 0.00304 \approx 3 \times 10^{-3} \]
Step 4: Final Answer:
The \(K_a\) of fluoroacetic acid is \(3 \times 10^{-3}\).