$0.5$ molal aqueous solution of a weak acid ($ ext{HX}$) is $20\%$ ionised. If $K_f$ for water is $1.86 ext

Question

$0.5$ molal aqueous solution of a weak acid ($	ext{HX}$) is $20\%$ ionised. If $K_f$ for water is $1.86 	ext{ K kg mol}^{-1}$, the lowering in freezing point of the solution is:

Accepted Answer

For the weak acid $	ext{HX}$, the ionization reaction is:
$	ext{HX} ightleftharpoons 	ext{H}^+ + 	ext{X}^-$
The number of ions produced per molecule upon dissociation, $n = 2$.
Given, the degree of ionization $\alpha = 20\% = 0.2$.

The van't Hoff factor ($i$) is calculated as:
$i = 1 + \alpha(n - 1) = 1 + 0.2(2 - 1) = 1.2$

The depression (lowering) in freezing point is given by the formula:
$\Delta T_f = i 	imes K_f 	imes m$

Given values:
Molality, $m = 0.5 	ext{ mol kg}^{-1}$
Molal depression constant, $K_f = 1.86 	ext{ K kg mol}^{-1}$

Substituting the values:
$\Delta T_f = 1.2 	imes 1.86 	imes 0.5 = 1.116 	ext{ K} \approx 1.12 	ext{ K}$.

Therefore, the lowering in freezing point of the solution is $1.12 	ext{ K}$.

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