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0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If $K_f$ for water is 1.86 K kg mol$^{-1}$, the lowering in freezing point of the solution is
A
0.56 K
B
1.12 K
C
$-0.56$ K
D
$-1.12$ K
Detailed Solution
$\Delta T_f = i\,K_f\,m$
For the weak acid: $HX \rightleftharpoons H^+ + X^-$
At $t = 0$: 1, 0, 0
At equilibrium: $(1 - 0.20)$, 0.20, 0.20
Total number of moles $= 1 - 0.20 + 0.20 + 0.20 = 1 + 0.20 = 1.2$
So the van't Hoff factor is $i = 1 + \alpha = 1.2$
$\therefore \Delta T_f = 1.2 \times 1.86 \times 0.5 = 1.116 \approx 1.12$ K
The lowering is a magnitude, so it is reported as a positive quantity, 1.12 K.
For the weak acid: $HX \rightleftharpoons H^+ + X^-$
At $t = 0$: 1, 0, 0
At equilibrium: $(1 - 0.20)$, 0.20, 0.20
Total number of moles $= 1 - 0.20 + 0.20 + 0.20 = 1 + 0.20 = 1.2$
So the van't Hoff factor is $i = 1 + \alpha = 1.2$
$\therefore \Delta T_f = 1.2 \times 1.86 \times 0.5 = 1.116 \approx 1.12$ K
The lowering is a magnitude, so it is reported as a positive quantity, 1.12 K.