Q.0.6 mL of acetic acid ($CH_3COOH$), having density 1.06 g mL$^{-1}$, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205$^\circ$C. Calculate the van't Hoff factor and the dissociation constant of acid.

Concept: Colligative Properties – Association/Dissociation

The observed depression is greater than expected for a non-electrolyte, indicating partial dissociation of acetic acid.

Step 1 – Molality of solution

Mass of acetic acid = $0.6 \times 1.06 = 0.636$ g.

Molar mass = 60 g/mol -> moles = $0.636 / 60 = 0.0106$ mol.

Mass of water = 1 kg -> molality $m = 0.0106$ m.

Step 2 – Expected depression (no dissociation)

$K_f$ for water = 1.86 K kg mol$^{-1}$.

$\Delta T_f(\text{theoretical}) = K_f \cdot m = 1.86 \times 0.0106 = 0.0197^\circ$C.

Step 3 – van't Hoff factor

$i = \dfrac{\Delta T_f(\text{observed})}{\Delta T_f(\text{theoretical})} = \dfrac{0.0205}{0.0197} \approx 1.041$ (NCERT's own printed value for this example).

Step 4 – Dissociation constant

For weak acid $HA \rightleftharpoons H^+ + A^-$, degree of dissociation $\alpha = i - 1 = 0.041$. …

Why this approach works

Freezing point depression is a colligative property — it depends only on the number of solute particles in solution, not on their identity. For a non-electrolyte like sugar, one mole of solute gives one mole of particles. But acetic acid is a weak electrolyte: it partially dissociates into ions:

So the actual number of particles in solution is greater than the number of acid molecules dissolved. The van’t Hoff factor $i$ captures this:

Once we know $i$, we can relate it to the degree of dissociation $\alpha$, and from $\alpha$ and the initial concentration, calculate the dissociation constant $K_a$.

Step-by-step solution

1. Find the molality of acetic acid

We have 0.6 mL of acetic acid, density 1.06 g/mL.

Mass of acetic acid = volume × density = $0.6 \times 1.06 = 0.636$ g.

Molar mass of $CH_3COOH$ = $12 + 3 + 12 + 16 + 16 + 1 = 60$ g/mol.

Moles of acetic acid = $\frac{0.636}{60} = 0.0106$ mol.

This is dissolved in 1 litre of water. For dilute aqueous solutions, 1 L water ≈ 1 kg water (since density of water is 1 g/mL). So the molality $m$ is:

2. Calculate the expected freezing point depression (no dissociation)

The cryoscopic constant for water is $K_f = 1.86\ ^\circ\text{C kg mol}^{-1}$.

If acetic acid did not dissociate at all, the depression would be:

3. Find the van’t Hoff factor $i$

The observed depression is $\Delta T_f(\text{obs}) = 0.0205\ ^\circ\text{C}$.

By definition:

The NCERT textbook's own printed value for this worked example carries one more significant figure:

4. Relate $i$ to degree of dissociation $\alpha$

For a weak acid that dissociates as:

If we start with 1 mole of acid and $\alpha$ is the fraction dissociated, then:

• Moles of undissociated acid = $1 - \alpha$
• Moles of acetate ions = $\alpha$
• Moles of hydrogen ions = $\alpha$ …

Method: Van't Hoff Factor & Dissociation Constant from Freezing Point Depression

Method name: Colligative property method using freezing point depression to find van't Hoff factor ($i$) and then dissociation constant ($K_a$).

Step 1: Find the molality of acetic acid (assuming no dissociation)

Given:

• Volume of acetic acid = $0.6 \, \text{mL}$
• Density = $1.06 \, \text{g mL}^{-1}$
• Mass of acetic acid = $0.6 \times 1.06 = 0.636 \, \text{g}$
• Molar mass of $CH_3COOH$ = $60 \, \text{g mol}^{-1}$

Moles of acetic acid:

Water = 1 L ≈ 1 kg (since density of water ≈ 1 g/mL)

So molality (assuming no dissociation):

Step 2: Calculate observed van't Hoff factor ($i$)

Freezing point depression formula:

For water, $K_f = 1.86 \, \text{K kg mol}^{-1}$

Given $\Delta T_f = 0.0205 \, ^\circ\text{C}$

So:

Calculate:

Van't Hoff factor, $i = 1.041$ (keep this third decimal — over-rounding to 1.04 propagates into a wrong $K_a$ later).

Step 3: Relate $i$ to degree of dissociation ($\alpha$)

For acetic acid dissociation:

If $\alpha$ = degree of dissociation:

• Initial moles = 1 …

🧠 The Core Concept

This problem connects colligative properties (freezing point depression) with ionic equilibrium (weak acid dissociation).

You are given:

• Volume of acetic acid = 0.6 mL
• Density = 1.06 g/mL
• Solvent = 1 L water
• Observed $\Delta T_f = 0.0205^\circ\text{C}$
• Need: van't Hoff factor $i$ and dissociation constant $K_a$

✗ Common Mistake #1: Wrong mass of acetic acid

The error:

Students directly use volume (0.6 mL) as mass, or forget to multiply by density.

Why it happens:

They see "0.6 mL" and treat it like grams.

How to avoid:

Always use:

So here:

✓ Write this step explicitly.

✗ Common Mistake #2: Wrong molar mass or moles calculation

The error:

Using molar mass of acetic acid as 60 g/mol (correct) but then miscalculating moles.

How to avoid:

Double-check:

$0.636 \div 60 = 0.0106$ exactly.

✓ Always show the division clearly.

✗ Common Mistake #3: Confusing molality with molarity

The error:

Using 1 L water as 1 kg (correct for dilute aqueous solutions), but then writing molality formula wrong.

Why it happens:

Students mix up $m$ (molality) and $M$ (molarity).

How to avoid:

For freezing point depression:

where $m = \frac{\text{moles of solute}}{\text{kg of solvent}}$

Here, 1 L water ≈ 1 kg, so:

✓ State: "Since water density ≈ 1 g/mL, 1 L = 1 kg"

✗ Common Mistake #4: Using wrong $K_f$ value

The error:

Using $K_f$ for water = 1.86 K kg mol⁻¹ (correct), but forgetting units or using 0.52 (which is $K_b$).

How to avoid:

Memorise:

• $K_f(\text{water}) = 1.86 \, \text{K kg mol}^{-1}$
• $K_b(\text{water}) = 0.52 \, \text{K kg mol}^{-1}$

✓ Write the value with units at the start.

✗ Common Mistake #5: Over-rounding the van't Hoff factor

The error:

Computing $i$ and rounding it to two decimal places (1.04) before using it further.

How to avoid:

First compute denominator:

$1.86 \times 0.0106 = 0.019716$

Then:

✓ Keep 3–4 significant figures: $i = 1.041$, as NCERT does. Rounding to 1.04 here silently corrupts $\alpha$ and then $K_a$ in the next steps.

✗ Common Mistake #6: Confusing $i$ with degree of dissociation $\alpha$

The error:

Thinking $i = \alpha$ directly.

Why it happens:

For weak acids, $i = 1 + \alpha$ (since each molecule gives 2 ions when dissociated).

How to avoid:

For acetic acid (CH₃COOH ⇌ CH₃COO⁻ + H⁺):

So:

✓ Always write the dissociation equation first — and keep $\alpha = 0.041$, not 0.04.

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