Question

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If $K_f$ for water is $1.86{}^{o}C/m$, the freezing point of the solution will be:

• A. $-0.18{}^{o}C$
• B. $-0.54{}^{o}C$
• C. $-0.36{}^{o}C$
• D. $-0.24{}^{o}C$

Answer 1

Answer: (D)

$-0.24{}^{o}C$

The relationship between the depression in the freezing point and the molality of the solution is as given below.
$\Delta T_f=i{K}_{f}m$......(1)

Let $\alpha$ b the degree of dissociation of the weak acid.

The dissociation equilibrium of the weak acid is as represented below.
$HA⟶H^{+}+A^{-}$
$1-\alpha \alpha \alpha$
$1-0.30.30.3$

The Van't Hoff factor i is the total number of ions/molecules given by the dissociation of 1 molecule of solute.
$i=1-0.3+0.3+0.3$
$i=1.3$

Substitute values in the expression (1).
$\Delta T_f=1.3\times 1.86\times 0.1=0.2418$
$T_f=0-0.2418=-{0.2418}^{0}C$

Hence, the freezing point of the solution will be $-{0.24}^{0}C$.