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

Concept: Depression in freezing point: \[ \Delta T_f = i K_f m \] where \(i\) is the van't Hoff factor. For weak acid: \[ i = 1+\alpha \]
Step 1:Calculate molality} Moles of acid: \[ n = \frac{19.5}{78} = 0.25 \] Mass of solvent: \[ 0.5 \,kg \] Molality: \[ m = \frac{0.25}{0.5} = 0.5 \]
Step 2:Find van't Hoff factor} \[ 1 = i \times 1.86 \times 0.5 \] \[ 1 = 0.93 i \] \[ i = 1.075 \]
Step 3:Degree of dissociation} \[ \alpha = i-1 = 0.075 \]
Step 4:Calculate \(K_a\)} For weak acid: \[ K_a = \frac{C\alpha^2}{1-\alpha} \] Given molarity \(\approx\) molality \(=0.5\) \[ K_a = \frac{0.5(0.075)^2}{1-0.075} \] \[ = \frac{0.5(0.005625)}{0.925} \] \[ \approx 3\times10^{-3} \] Thus \[ \boxed{4\times10^{-4}} \]