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

Correct Answer: 19

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: \(x = 19\).

Answer 2

Step 1: Given data.
Dissociation constant, K_a = 6.25 × 10⁻⁵
Volume of acetic acid = 5 mL = 5 × 10⁻³ L
Density of acetic acid = 1.2 g/mL
Molar mass of acetic acid = 60 g/mol
Volume of solution = 1 L water
K_f(water) = 1.86 K·kg·mol⁻¹

Step 2: Calculate the number of moles of acetic acid.
Mass of acetic acid = 5 mL × 1.2 g/mL = 6 g
\[ \text{Moles of CH₃COOH} = \frac{6}{60} = 0.1 \, \text{mol} \]

Step 3: Calculate molality of the solution.
Since 1 L of water ≈ 1 kg water:
\[ \text{Molality (m)} = \frac{0.1}{1} = 0.1 \, \text{mol/kg} \]

Step 4: Dissociation of acetic acid.
CH₃COOH ⇌ CH₃COO⁻ + H⁺
If degree of dissociation is α,
Initial concentration = 0.1
At equilibrium: [CH₃COOH] = 0.1(1 − α), [CH₃COO⁻] = 0.1α, [H⁺] = 0.1α

Step 5: Apply dissociation constant formula.
\[ K_a = \frac{(0.1\alpha)^2}{0.1(1 − \alpha)} = 0.1\frac{\alpha^2}{1 − \alpha} \]
Since α is small, (1 − α) ≈ 1:
\[ 6.25 × 10^{-5} = 0.1 \alpha^2 \]
\[ \alpha^2 = 6.25 × 10^{-4} \quad \Rightarrow \quad \alpha = 0.025 \]

Step 6: Calculate van’t Hoff factor (i).
For weak acid dissociating into 2 particles:
\[ i = 1 + \alpha = 1.025 \]

Step 7: Depression in freezing point.
\[ \Delta T_f = i \, K_f \, m \] \[ \Delta T_f = 1.025 × 1.86 × 0.1 = 0.19065 \, \text{K} \]

Step 8: Final freezing point.
Freezing point = 0 − 0.19065 = −0.19 °C
\[ x = 19 \]

Final Answer: x = 19