Question

(i) The depression in freezing point of water observed for the same molar concentration of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order as stated above. Explain.
(ii) Calculate the depression in freezing point of water when 20.0 g of $C{H}_{3}C{H}_{2}CHClCOOH$ is added to 500g of water.

Answer 1

(i) The degree of ionisation acetic acid, trichloroacetic acid and trifluoroacetic acids will decrease in the following order.

Trifluoroacetic acid $>$ Trichloroacetic acid $>$ Acetic acid.

It depends on the strength of the acid. Trifluoroacetic acid is more acidic than trichloroacetic acid which is further more acidic than acetic acid. Now greater the degree of ionisation, greater will be the depression of freezing point. Hence the order will be

$C{H}_{3}COOH<CC{l}_{3}COOH<C{F}_{3}COOH$

(ii) Molar mass of $C{H}_{3}C{H}_{2}CHClCOOH=122.5g/mol$

$\therefore$ Moles of $C{H}_{3}C{H}_{2}CHClCOOH=20g/122.5=0.163mol$

$\therefore$ Molality of the solution= $\frac{0.163\times 1000}{500}=0.326molk{g}^{-1}$

Now, $C{H}_{3}C{H}_{2}CHClCOOH\leftrightarrow C{H}_{3}C{H}_{2}CHClCO{O}^{-}+H^{+}$

Initial $C$ $0$ $0$

At eqn $C(1-\alpha )$ $C\alpha$ $C\alpha$

So, $K_a=\frac{C{\alpha }^2}{1-\alpha }$

$K_a=C{\alpha }^2$ as $1-\alpha \simeq 1$

$\Rightarrow \alpha =\sqrt{K_a/C}$

$\therefore \alpha =\sqrt{1.4\times {10}^{-3}/0.326}$

$=0.0655$

Now at equilibrium, Van't Hoff factor

$\Rightarrow i=1-\alpha +\alpha +\alpha$

$=1+0.0655=1.0655$

Hence the depression in the freezing point

$\Delta T_f=i{K}_{f}m$

$=1.0655\times 1.86\times 0.326$

$={0.647}^o$