Question
5 g of fluoro acetic acid (molar mass $= 78\text{ }g{\text{ }mol}^{- 1}$ ) 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, $K_{f} = 1.86\text{ }K\text{ }kg{\text{ }mol}^{- 1}$ ). Assume molarity and molality to have same values.
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Solution
Sol. [Key Concept: $\Delta\ T_{f} = {iK}_{f}m$ gives the van't Hoff factor $i$; then $\alpha$ and $K_{a} = C\alpha^{2}/(1 - \alpha)$.]
$$\Delta T_{f} = iK_{r}m $$no. of moles $= \frac{19.5}{78} = \frac{1}{4}$ mole
$$1 = i \times 1.86 \times \frac{\frac{1}{4}}{\frac{1}{2}} $$
$$i = \frac{2}{1.86} $$
$$i = 1 + (n - 1)\alpha $$
$$i = 1 + \alpha $$
$$\alpha = \frac{2}{1.86} - 1 = 0.075 $$
$${K_{a} = \frac{C^{2}}{1 - \alpha} = \frac{1}{2} \times \frac{(0.075)^{2}}{(1 - 0.075)} }{= 3 \times 10^{- 3} }$$
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