Question

Considering acetic acid dissociates in water, its dissociation constant is $6.25 \times 10^{-5}$. If 5 mL of acetic acid is dissolved in 1 litre water, the solution will freeze at $-x \times 10^{-1}$ °C, provided pure water freezes at 0 °C. $x =$ _____ (Nearest integer)
Given: $(K_f)_{water} = 1.86$ K kg mol$^{-1}$
density of acetic acid is 1.2 g mol$^{-1}$
molar mass of water = 18 g mol$^{-1}$
molar mass of acetic acid = 60 g mol$^{-1}$
density of water = 1 g cm$^{-3}$
Acetic acid dissociates as CH$_3$COOH $\rightleftharpoons$ CH$_3$COO$^-$ + H$^+$

Answer 1

The depression in freezing point $\Delta T_f$ is calculated using the formula:
$\Delta T_f = i \cdot K_f \cdot m,$
where:
$i$ is the van ’t Hoff factor,
$K_f$ is the cryoscopic constant ($1.86 \, \text{K kg mol}^{-1}$),
$m$ is the molality of the solution.
Step 1: Calculate the molality of the solution
The mass of acetic acid dissolved is:
$\text{Mass of acetic acid} = \text{Volume} \times \text{Density} = 5 \, \text{mL} \times 1.2 \, \text{g/mL} = 6 \, \text{g}.$
The number of moles of acetic acid is:
$\text{Moles of acetic acid} = \frac{\text{Mass of acetic acid}}{\text{Molar mass of acetic acid}} = \frac{6}{60} = 0.1 \, \text{mol}.$
The molality of the solution is:
$m = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.1}{1} = 0.1 \, \text{mol/kg}.$
Step 2: Calculate the van ’t Hoff factor ($i$)
The dissociation constant ($K_a$) of acetic acid is:
$K_a = 6.25 \times 10^{-5}.$
The degree of dissociation ($\alpha$) is given by:
$\alpha = \sqrt{\frac{K_a}{C}},$
where $C$ is the molarity of the solution.
The molarity is:
$C = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}} = \frac{0.1}{1} = 0.1 \, \text{mol/L}.$
Substituting the values:
$\alpha = \sqrt{\frac{6.25 \times 10^{-5}}{0.1}} = \sqrt{6.25 \times 10^{-4}} = 0.025.$
The van ’t Hoff factor is:
$i = 1 + \alpha = 1 + 0.025 = 1.025.$
Step 3: Calculate $\Delta T_f$
$\Delta T_f = i \cdot K_f \cdot m = 1.025 \cdot 1.86 \cdot 0.1 = 0.19065 \, \text{K}.$
Converting to $-x \times 10^{-2}$:
$x = 19.$
Final Answer: $x = 19$.