19.5 g of $CH_2FCOOH$ is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0$^\circ$C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1

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CH_2FCOOH

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Concept understanding — Colligative Properties Association

Colligative Properties: The Intuition First

Imagine you're at a party. The room is full of people dancing — that's your solvent molecules, moving freely. Now, someone brings in a few heavy, slow-moving guests who just stand in one spot — those are your solute particles. They don't dance, they don't interact much, they just take up space.

What happens? The dancers now have less room to move. They bump into the standing guests more often. The whole atmosphere changes — the dancers can't move as freely, they can't escape the room as easily, and the overall "energy" of the party shifts.

That's the core idea of colligative properties. When you add a non-volatile solute (like salt) to a solvent (like water), the solute particles don't do anything special — they just exist in the solution. But their mere presence changes four measurable properties of the solvent:

Vapour pressure decreases
Boiling point increases
Freezing point decreases
Osmotic pressure increases

The key insight: these changes depend only on the number of solute particles, not on what kind of particles they are. One molecule of sugar and one ion of salt (if they don't dissociate) affect these properties identically — provided they're the same number of particles.

Note

This is why "colligative" comes from the Latin colligatus meaning "bound together" — the properties are bound to the quantity of solute, not its identity.

The Precise Statement

Colligative properties are properties of a solution that depend solely on the ratio of the number of solute particles to the number of solvent molecules in a given solution, and not on the chemical nature of the solute.

Mathematically, for a dilute solution of a non-volatile, non-electrolyte solute:

\Delta P = P^0 \cdot x_2

Where:

$\Delta P$ = lowering of vapour pressure
$P^0$ = vapour pressure of pure solvent
$x_2$ = mole fraction of solute
$\Delta T_b$ = elevation in boiling point
$K_b$ = ebullioscopic constant (depends only on solvent)
$m$ = molality of solution
$\Delta T_f$ = depression in freezing point
$K_f$ = cryoscopic constant (depends only on solvent)
$\Pi$ = osmotic pressure
$i$ = van't Hoff factor (accounts for dissociation/association)
$M$ = molarity
$R$ = gas constant
$T$ = absolute temperature

The Crucial Distinction: Association vs. Dissociation

Now, here's where the association part comes in — and it's the twist that catches most students.

The formulas above assume the solute particles remain as individual, independent particles. But in reality:

Dissociation: Some solutes break apart into smaller particles (e.g., NaCl → Na⁺ + Cl⁻). This increases the number of particles, so the colligative effect is larger than expected.
Association: Some solutes clump together into larger particles (e.g., acetic acid in benzene forms dimers: 2 CH₃COOH → (CH₃COOH)₂). This decreases the number of particles, so the colligative effect is smaller than expected.

Watch out

A common mistake: students think "association" means the solute interacts with the solvent. No — association means solute particles bind to each other, reducing the effective particle count. Solvent-solute interactions affect non-colligative properties like solubility.

The van't Hoff Factor $i$

To account for these real-world effects, we introduce the van't Hoff factor:

i = \frac{\text{Actual number of particles in solution}}{\text{Number of formula units dissolved}}

For a non-electrolyte that doesn't associate or dissociate: $i = 1$

For dissociation (e.g., NaCl): $i > 1$ (ideally 2 for NaCl)

For association (e.g., acetic acid dimerizing): $i < 1$

The corrected formulas become:

\Delta T_b = i \cdot K_b \cdot m

\Delta T_f = i \cdot K_f \cdot m

\Pi = i \cdot MRT

A Concrete Example

Consider acetic acid (CH₃COOH) dissolved in benzene. In benzene, acetic acid molecules form hydrogen-bonded dimers:

2 \text{CH}_3\text{COOH} \rightleftharpoons (\text{CH}_3\text{COOH})_2

If you dissolve 1 mole of acetic acid, you might end up with only 0.6 moles of particles (0.4 moles of dimers + 0.2 moles of monomers). So $i = 0.6$ . …

Why this formula?

Colligative Properties & Association: Why the Formula Holds

Let's build this from first principles — understanding why association changes colligative properties, not just memorizing the formula.

The Core Idea: What Are Colligative Properties?

Colligative properties depend only on the number of solute particles in solution, not on their chemical identity. The four key ones are:

Vapor pressure lowering
Boiling point elevation
Freezing point depression
Osmotic pressure

When a solute associates (e.g., two molecules dimerize), the effective number of particles decreases. This is the entire reason the formula changes.

The van't Hoff Factor: The Bridge

We define the van't Hoff factor $i$ as:

i = \frac{\text{actual number of particles in solution}}{\text{number of formula units dissolved}}

For a non-electrolyte that does not associate, $i = 1$ .

For association, $i < 1$ .

Example: Dimerization of Benzoic Acid in Benzene

Benzoic acid ( $C_6H_5COOH$ ) forms dimers in benzene:

2 \, C_6H_5COOH \rightleftharpoons (C_6H_5COOH)_2

If we dissolve $n$ moles of monomer, but only $n/2$ moles of dimer exist, then:

i = \frac{n/2}{n} = 0.5

Deriving the Modified Formula

Step 1: Start with the Normal Colligative Formula

For freezing point depression (the most common exam case):

\Delta T_f = K_f \cdot m

where $m$ is the molality of the solute (moles per kg solvent).

Step 2: Replace $m$ with Effective Molality

Because only the number of particles matters, we replace $m$ with $i \cdot m$ :

\Delta T_f = K_f \cdot (i \cdot m)

This is the general formula for any colligative property when association or dissociation occurs.

Step 3: Express $i$ in Terms of Degree of Association

Let:

$\alpha$ = degree of association (fraction of molecules that associate)
$n$ = number of molecules that combine to form one associated particle (e.g., $n=2$ for dimerization)

For a dimerization ( $n=2$ ):

Initially: 1 mole of monomer
After association: $(1 - \alpha)$ moles remain as monomer, and $\alpha/2$ moles of dimer form
Total particles = $(1 - \alpha) + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$

Thus:

i = \frac{1 - \frac{\alpha}{2}}{1} = 1 - \frac{\alpha}{2}

General formula for association of $n$ molecules:

i = 1 - \alpha + \frac{\alpha}{n} = 1 - \alpha\left(1 - \frac{1}{n}\right)

Why This Makes Physical Sense …

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The van't Hoff factor $i$ is found by comparing the observed freezing-point depression to the expected value for a non-electrolyte, giving $i \approx 1.075$ . Using this $i$ and the initial concentration, the dissociation constant $K_a$ of fluoroacetic acid is calculated to be approximately $3.07 \times 10^{-3}$ .

This is a problem that connects colligative properties with chemical equilibrium. The freezing-point depression tells us how many particles are actually present in solution. For a weak acid like fluoroacetic acid ( $CH_2FCOOH$ ), the observed depression is less than what you'd expect if it fully dissociated, but more than if it didn't dissociate at all. The van't Hoff factor $i$ captures this "effective number of particles per formula unit." Once we have $i$ , we can work backwards to find the degree of dissociation $\alpha$ , and from there the acid dissociation constant $K_a$ .

Let's go step by step.

Calculate the expected (theoretical) freezing-point depression for a non-electrolyte.

First, find the molality of the solution. Molar mass of $CH_2FCOOH$ :

C: 12 \times 2 = 24,\quad H: 3 \times 1 = 3,\quad F: 19,\quad O: 16 \times 2 = 32

Total = $24 + 3 + 19 + 32 = 78\ \text{g/mol}$ .

Moles of acid:

n = \frac{19.5}{78} = 0.25\ \text{mol}

Mass of water = 500 g = 0.5 kg. So molality:

m = \frac{0.25}{0.5} = 0.5\ \text{mol/kg}

For water, the cryoscopic constant $K_f = 1.86\ \text{K kg mol}^{-1}$ . If the acid did not dissociate at all, the depression would be:

\Delta T_f(\text{expected}) = K_f \cdot m = 1.86 \times 0.5 = 0.93^\circ\text{C}

But the observed depression is $1.0^\circ\text{C}$ , which is larger than 0.93°C. This tells us the acid is dissociating — more particles are present than just the undissociated molecules.

Calculate the van't Hoff factor $i$ .

The van't Hoff factor is defined as:

i = \frac{\text{observed } \Delta T_f}{\text{expected } \Delta T_f \text{ (if no dissociation)}}

So:

i = \frac{1.0}{0.93} \approx 1.0753

Watch out

In $\Delta T_f = i\,K_f\,m$ , the molality $m$ is that of the acid as prepared (0.5 mol kg $^{-1}$ ) — the factor $i$ alone accounts for the extra particles produced by dissociation. Don't also inflate $m$ by the ion count, or you double-count the dissociation.

So $i \approx 1.075$ .

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

For a weak acid $HA$ that dissociates as:

HA \rightleftharpoons H^+ + A^-

If we start with 1 mole of $HA$ and let $\alpha$ be the fraction dissociated, then:

Moles of $HA$ remaining: $1 - \alpha$
Moles of $H^+$ produced: $\alpha$
Moles of $A^-$ produced: $\alpha$
Total moles after dissociation: $(1 - \alpha) + \alpha + \alpha = 1 + \alpha$

The van't Hoff factor is the ratio of total particles after dissociation to initial particles:

i = \frac{1 + \alpha}{1} = 1 + \alpha

Therefore:

\alpha = i - 1 = 1.0753 - 1 = 0.0753

Tip

This relation $i = 1 + \alpha$ holds only for a binary electrolyte (one that gives two ions). For a salt like $CaCl_2$ that gives three ions, the formula would be $i = 1 + 2\alpha$ . Always check the stoichiometry of dissociation.

Calculate the dissociation constant $K_a$ . …

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

Method name: Colligative property → van’t Hoff factor → dissociation constant (Ostwald’s dilution law)

Step 1 — Calculate the expected (theoretical) freezing point depression

Given:

Mass of solute $CH_2FCOOH$ = 19.5 g
Molar mass of $CH_2FCOOH$ = $12 + 2 + 19 + 12 + 16 + 16 + 1 = 78 \text{ g/mol}$
Mass of solvent (water) = 500 g = 0.5 kg
$K_f$ for water = $1.86 \text{ K kg mol}^{-1}$

Moles of solute

n = \frac{19.5}{78} = 0.25 \text{ mol}

Molality

m = \frac{0.25}{0.5} = 0.5 \text{ mol/kg}

Expected depression (if no dissociation)

\Delta T_f(\text{theoretical}) = K_f \cdot m = 1.86 \times 0.5 = 0.93^\circ\text{C}

Step 2 — Calculate van’t Hoff factor $i$

Observed depression = $1.0^\circ\text{C}$

i = \frac{\Delta T_f(\text{observed})}{\Delta T_f(\text{theoretical})} = \frac{1.0}{0.93} \approx 1.0753

Van’t Hoff factor

\boxed{i \approx 1.075}

Step 3 — Relate $i$ to degree of dissociation $\alpha$

For fluoroacetic acid ( $CH_2FCOOH$ ), dissociation is:

CH_2FCOOH \rightleftharpoons CH_2FCOO^- + H^+

If initial moles = 1, at equilibrium:

Undissociated = $1 - \alpha$
Ions produced = $\alpha + \alpha = 2\alpha$
Total particles = $1 - \alpha + 2\alpha = 1 + \alpha$

Thus:

i = 1 + \alpha \quad \Rightarrow \quad \alpha = i - 1 = 1.0753 - 1 = 0.0753

Degree of dissociation

$\boxed{\alpha \approx 0.0753 \ (\text{or } 7.53\%)}$ …

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Here are the most common mistakes students make when solving this problem, along with the conceptual corrections to avoid them.

Mistake 1: Forgetting to Convert Mass to Moles Correctly

The Error: Students often use the molar mass of the acid incorrectly or forget to convert the mass of the solvent (water) from grams to kilograms for the molality formula.

Why it happens: The formula for depression in freezing point is $\Delta T_f = i \cdot K_f \cdot m$ , where $m$ is molality (moles of solute per kg of solvent). Using grams for the solvent mass directly gives a wrong molality.

How to Avoid:

Always calculate moles of solute first: Molar mass of $CH_2FCOOH$ = $2(12) + 3(1) + 19 + 2(16) = 78 \text{ g/mol}$ Moles of acid = $\frac{19.5}{78} = 0.25 \text{ mol}$
Convert solvent mass to kg: $500 \text{ g} = 0.5 \text{ kg}$
Then calculate observed molality from the given $\Delta T_f$ : $m_{\text{obs}} = \frac{\Delta T_f}{K_f} = \frac{1.0}{1.86} \approx 0.5376 \text{ mol/kg}$
Theoretical molality (if no dissociation) = $\frac{0.25}{0.5} = 0.5 \text{ mol/kg}$

Mistake 2: Confusing the van't Hoff Factor Formula

The Error: Students incorrectly use $i = \frac{\text{observed molar mass}}{\text{theoretical molar mass}}$ without realizing that for freezing point depression, the correct relation is $i = \frac{\text{observed colligative property}}{\text{theoretical colligative property}}$ .

Why it happens: Many textbooks give multiple forms of the van't Hoff factor, and students mix them up.

How to Avoid:

For freezing point depression, use:

i = \frac{\Delta T_f (\text{observed})}{\Delta T_f (\text{theoretical})}

Since $\Delta T_f \propto m$ , you can also write:

i = \frac{m_{\text{observed}}}{m_{\text{theoretical}}}

Here: $i = \frac{0.5376}{0.5} = 1.0752 \approx 1.075$

Mistake 3: Assuming Complete Dissociation

The Error: Students assume the acid dissociates completely (like a strong acid) and set $i = 2$ for a monoprotic acid, then get confused when the calculated $i$ is close to 1.

Why it happens: Fluoroacetic acid is a weak acid — it only partially dissociates. The observed $i$ (1.075) is between 1 and 2, indicating partial dissociation.

How to Avoid:

Recognize that $i$ for a weak acid is given by:

i = 1 + \alpha

where $\alpha$ is the degree of dissociation.

Solve for $\alpha$ : $\alpha = i - 1 = 1.0753 - 1 = 0.0753$ (about 7.5% dissociation)

Mistake 4: Overcomplicating the Concentration Used in $K_a$

The Error: Students know $K_a$ is defined with molarity (mol/L) while the freezing-point data gave molality (mol/kg), and launch into a density-based conversion mid-problem — ending up with a final answer that disagrees with the rest of their own working.

Why it happens: Mixing two different concentration conventions inside one solution.

How to Avoid:

For a dilute aqueous solution, molality ≈ molarity, so take $C = 0.5 \text{ mol L}^{-1}$ — the standard assumption for this problem.
Then use:

K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.5 \times (0.0753)^2}{1 - 0.0753}

Calculate: $K_a = \frac{0.5 \times 0.00567}{0.9247} \approx \frac{0.002835}{0.9247} \approx 3.07 \times 10^{-3}$ …

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Q.19.5 g of CH2FCOOHCH_2FCOOHCH2​FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0∘^\circ∘C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

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Q.19.5 g of $CH_2FCOOH$ is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0 $^\circ$ C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

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