Calculate the depression in the freezing point of water when 10 g of $CH_3CH_2CHClCOOH$ is added to 250 g of water. $K_a = 1.4 \times 10^{-3}$, $K_f = 1.86$ K kg mol$^{-1}$.

Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water

Exercises · 1.32

CH_3CH_2CHClCOOH

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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 depression in freezing point depends on the total number of particles after dissociation. Since the acid is weak, we first find its degree of dissociation using $K_a$ , then compute the van’t Hoff factor $i$ , and finally apply $\Delta T_f = i \, K_f \, m$ . The result is $\boxed{0.65\ \text{K}}$ .

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 weak acid like $CH_3CH_2CHClCOOH$ , the solute partially dissociates into ions. So the effective number of particles is greater than the number of formula units dissolved.

We need the van’t Hoff factor $i$ , which is the ratio of actual particles after dissociation to the number of formula units dissolved. For a weak acid HA dissociating as

HA \rightleftharpoons H^+ + A^-

if the degree of dissociation is $\alpha$ , then

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

So $i = 1 + \alpha$ .

Once we find $\alpha$ from the given $K_a$ , we can compute the effective molality and then $\Delta T_f$ .

Step-by-step solution

1. Find the molar mass of the acid

The formula is $CH_3CH_2CHClCOOH$ (4 carbons, 7 hydrogens, 2 oxygens, 1 chlorine).

Atomic masses: C = 12, H = 1, O = 16, Cl = 35.5

M = 4(12) + 7(1) + 2(16) + 35.5 = 48 + 7 + 32 + 35.5 = 122.5\ \text{g mol}^{-1}

2. Calculate the molality of the solution (if no dissociation)

Mass of solute = 10 g

Moles of solute = $\frac{10}{122.5} = 0.08163\ \text{mol}$

Mass of solvent (water) = 250 g = 0.250 kg

Molality (apparent, assuming no dissociation)

m = \frac{0.08163}{0.250} = 0.3265\ \text{mol kg}^{-1}

3. Find the degree of dissociation $\alpha$ from $K_a$

The acid is weak, so we use the relation for a weak acid in water. Let the initial concentration be $c$ mol/L. But here we have molality — for dilute aqueous solutions, molality ≈ molarity, so we can use $c \approx 0.3265\ \text{M}$ .

For $HA \rightleftharpoons H^+ + A^-$ ,

K_a = \frac{c \alpha^2}{1 - \alpha}

Given $K_a = 1.4 \times 10^{-3}$ , $c = 0.3265$ .

Since $K_a$ is small, $\alpha$ will be small, so $1 - \alpha \approx 1$ . Then

$\alpha \approx \sqrt{\frac{K_a}{c}} = \sqrt{\frac{1.4 \times 10^{-3}}{0.3265}} = \sqrt{4.287 \times 10^{-3}} = 0.0655$ …

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Method: Van’t Hoff Factor Approach for Weak Acid (Association/Dissociation)

This problem involves a weak acid that partially dissociates in water. The depression in freezing point depends on the total number of particles in solution, so we must account for the degree of dissociation using the Van’t Hoff factor ( $i$ ).

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

Molar mass of $CH_3CH_2CHClCOOH$ :

$C_4H_7ClO_2 = (4 \times 12) + (7 \times 1) + 35.5 + (2 \times 16) = 48 + 7 + 35.5 + 32 = 122.5$ g/mol
Moles of acid = $\frac{10}{122.5} = 0.0816$ mol
Mass of water = 250 g = 0.250 kg
Molality ( $m$ ) = $\frac{0.0816}{0.250} = 0.3264$ mol/kg

Step 2: Find the degree of dissociation ( $\alpha$ ) using $K_a$

The acid dissociates as:

$HA \rightleftharpoons H^+ + A^-$

Initial: $c$ 0 0

At equilibrium: $c(1-\alpha)$ $c\alpha$ $c\alpha$

K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(c\alpha)^2}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha}

Given $K_a = 1.4 \times 10^{-3}$ and $c = 0.3264$ M (since molality ≈ molarity for dilute solutions):

1.4 \times 10^{-3} = \frac{0.3264 \times \alpha^2}{1-\alpha}

Since $K_a$ is small, assume $\alpha \ll 1$ , so $1-\alpha \approx 1$ :

\alpha^2 \approx \frac{1.4 \times 10^{-3}}{0.3264} = 4.29 \times 10^{-3}

\alpha \approx 0.0655

Check: $1-\alpha = 0.9345 \approx 1$ , so the approximation is valid.

--- …

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Let’s break down the common mistakes and how to avoid each.

✗ Mistake 1: Assuming no dissociation (treating it as a non-electrolyte)

What students do:

They calculate molality using molar mass of $CH_3CH_2CHClCOOH$ and directly plug into

$\Delta T_f = K_f \cdot m$

Why it’s wrong:

The acid dissociates partially in water, increasing the number of particles. Colligative properties depend on total particle concentration, not just the formula unit concentration.

✓ How to avoid:

Always check if the solute is an electrolyte (strong or weak). For weak acids, you must:

Find the van’t Hoff factor $i$ using the degree of dissociation $\alpha$ .
Use $\Delta T_f = i \cdot K_f \cdot m$

✗ Mistake 2: Forgetting to calculate the degree of dissociation ( $\alpha$ ) correctly

What students do:

They either skip $\alpha$ or assume $\alpha = 1$ (complete dissociation).

Why it’s wrong:

$K_a = 1.4 \times 10^{-3}$ is small — the acid is weak. $\alpha$ is much less than 1.

✓ How to avoid:

For a weak monoprotic acid HA:

K_a = \frac{C \alpha^2}{1 - \alpha}

Since $K_a$ is small, you can approximate $1 - \alpha \approx 1$ :

\alpha \approx \sqrt{\frac{K_a}{C}}

Where $C$ is the molar concentration of the acid in mol/L (not molality).

Then find $i = 1 + \alpha$ (since HA ⇌ H⁺ + A⁻ gives 2 particles from 1).

✗ Mistake 3: Using molality when $K_a$ needs molarity

What students do:

They calculate molality $m$ and use it in the $K_a$ expression.

Why it’s wrong:

$K_a$ is defined in terms of molar concentration (mol/L), not molality.

✓ How to avoid:

Calculate moles of acid from mass and molar mass.
Find volume of water (assume density = 1 g/mL, so 250 g ≈ 0.250 L).
Get molarity $C = \frac{\text{moles}}{\text{volume in L}}$ .
Use this $C$ to find $\alpha$ and $i$ .

✗ Mistake 4: Using mass of solution instead of solvent

What students do:

They take total mass (acid + water) as the solvent mass.

Why it’s wrong: …

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Q.Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOHCH_3CH_2CHClCOOHCH3​CH2​CHClCOOH is added to 250 g of water. Ka=1.4×10−3K_a = 1.4 \times 10^{-3}Ka​=1.4×10−3, Kf=1.86K_f = 1.86Kf​=1.86 K kg mol−1^{-1}−1.

Why this approach works

Step-by-step solution

1. Find the molar mass of the acid

2. Calculate the molality of the solution (if no dissociation)

3. Find the degree of dissociation α\alphaα from KaK_aKa​

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Q.Calculate the depression in the freezing point of water when 10 g of $CH_3CH_2CHClCOOH$ is added to 250 g of water. $K_a = 1.4 \times 10^{-3}$ , $K_f = 1.86$ K kg mol $^{-1}$ .

3. Find the degree of dissociation $\alpha$ from $K_a$

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