Q.The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

The depression in freezing point is a colligative property, directly proportional to the effective number of solute particles in the solution.

1. Concept: Colligative Properties (Freezing Point Depression) and Acid Dissociation.
2. Reasoning:
   • The depression in freezing point ($\Delta T_f$) is given by $\Delta T_f = i K_f m$, where $i$ is the van't Hoff factor, $K_f$ is the cryoscopic constant, and $m$ is the molality. For the same amount of acid and solvent, $\Delta T_f$ is directly proportional to $i$.
   • The van't Hoff factor $i$ for a weak acid is $1 + \alpha$, where $\alpha$ is the degree of dissociation. A higher $\alpha$ means more particles in solution.
   • The acid strength increases in the order: Acetic acid < Trichloroacetic acid < Trifluoroacetic acid. This is because the electron-withdrawing inductive effect of the substituents stabilizes the conjugate base, enhancing acid strength. The $-\text{CF}_3$ group is more electron-withdrawing than $-\text{CCl}_3$, which is more electron-withdrawing than $-\text{CH}_3$. …

The depression in freezing point is a colligative property, meaning it depends solely on the number of solute particles present in a given amount of solvent, not on the chemical nature of those particles. When we dissolve an acid in water, it dissociates to some extent, producing more than one particle per molecule initially added. The greater the extent of dissociation, the more particles are formed, and consequently, the greater the observed freezing point depression.

Let's break down the explanation:

1. Understanding Freezing Point Depression:

   The depression in freezing point ($\Delta T_f$) is given by the formula:

   $\Delta T_f = i K_f m$

   Here,

   • $\Delta T_f$ is the depression in freezing point.
   • $K_f$ is the cryoscopic constant of the solvent (for water, $K_f = 1.86 \text{ K kg mol}^{-1}$).
   • $m$ is the molality of the solute.
   • $i$ is the van't Hoff factor, which accounts for the number of particles a solute produces in solution.

2. Identifying Constant Factors:

   In this problem, we are comparing the depression for the "same amount" of each acid in water. This implies that the molality ($m$) would be the same if the acids did not dissociate. Also, the solvent is water, so $K_f$ is constant. Therefore, the variation in $\Delta T_f$ must be due to differences in the van't Hoff factor ($i$).

3. The van't Hoff Factor ($i$) and Dissociation:

   For a weak acid (HA) that dissociates in water:

   $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$

   If $\alpha$ is the degree of dissociation (the fraction of acid molecules that dissociate), then for every 1 mole of HA initially dissolved, we will have:

   • $(1 - \alpha)$ moles of undissociated HA
   • $\alpha$ moles of $\text{H}^+$ ions
   • $\alpha$ moles of $\text{A}^-$ ions The total number of moles of particles in solution will be $(1 - \alpha) + \alpha + \alpha = 1 + \alpha$. Thus, the van't Hoff factor $i = 1 + \alpha$. A higher degree of dissociation ($\alpha$) leads to a higher van't Hoff factor ($i$).

4. Relating Dissociation to Acid Strength:

   The degree of dissociation ($\alpha$) is directly related to the strength of the acid. A stronger acid dissociates to a greater extent in water, meaning it has a higher $\alpha$. Consequently, a stronger acid will have a higher $i$ value.

5. Comparing Acid Strengths:

   Now, let's compare the strengths of the three acids:

   • Acetic acid ($\text{CH}_3\text{COOH}$): The methyl ($\text{CH}_3$) group is slightly electron-donating. This destabilizes the conjugate base ($\text{CH}_3\text{COO}^-$) by increasing electron density on the carboxylate oxygen, making it less stable. Thus, acetic acid is a relatively weak acid. …

Method: Van't Hoff Factor and Degree of Dissociation (Weak-Acid Trend)

This problem is explained using the van't Hoff factor ($i$), which accounts for the dissociation of each weak acid into ions -- not association.

Step-by-step reasoning

Step 1 - Recall the freezing point depression formula

where $i$ = van't Hoff factor, $K_f$ = cryoscopic constant (solvent property), $m$ = molality.

For the same amount (same molality $m$) of each acid in the same solvent (water), $\Delta T_f$ depends only on $i$.

Step 2 - Understand where $i$ comes from for a weak acid

Each of these is a carboxylic acid that partially ionises (dissociates) in water:

If $\alpha$ is the degree of dissociation, then for every 1 mole of HA dissolved, the total particle count becomes $(1-\alpha) + \alpha + \alpha = 1+\alpha$, so:

A stronger acid dissociates more (larger $\alpha$), giving a larger $i$ -- this is the opposite of association, which would give $i<1$. Nothing in this problem involves molecules combining into dimers; the effect here is purely ionisation.

Step 3 - Compare the three acids' strength

The order given is: acetic acid < trichloroacetic acid < trifluoroacetic acid (increasing $\Delta T_f$).

This is explained by the electron-withdrawing inductive effect of the halogen substituents:

• $-\text{CH}_3$ (acetic acid) is weakly electron-donating -> destabilises the conjugate base -> weakest acid, smallest $\alpha$, smallest $i$.
• $-\text{CCl}_3$ (trichloroacetic acid): the three electronegative Cl atoms withdraw electron density, stabilising the conjugate base ($\text{CCl}_3\text{COO}^-$) -> stronger acid, larger $\alpha$, larger $i$. …

Here’s a breakdown of the common mistakes students make on this specific colligative properties question, and how to avoid each.

The Core Concept (Why this happens)

The depression in freezing point ($\Delta T_f$) is a colligative property, meaning it depends on the number of solute particles in solution, not their identity.

• $i$ = van't Hoff factor (number of particles per formula unit)
• $K_f$ = cryoscopic constant (for water, it's fixed)
• $m$ = molality (same amount of acid, so same $m$)

Since $K_f$ and $m$ are constant here, $\Delta T_f$ depends only on $i$. The order of $\Delta T_f$ is:

Acetic acid < Trichloroacetic acid < Trifluoroacetic acid

This happens because the acid strength increases as more electronegative chlorine/fluorine atoms are added. A stronger acid dissociates more in water, giving more particles ($i$ is larger).

Common Mistake #1: Forgetting the "Same Amount" Condition

The Mistake: Students compare the molar masses of the acids and conclude that the heaviest acid (trifluoroacetic) has the lowest $\Delta T_f$ because it has fewer moles for the same mass.

Why it's wrong: The question explicitly says "same amount" — this means same number of moles (or same mass, but the key is equimolar). Since $\Delta T_f$ depends on molality ($m$), if the number of moles is the same, $m$ is the same. The difference comes from $i$, not $m$.

How to avoid: Always check the wording. If the problem says "same amount" or "equimolar", molality is constant. Focus on the van't Hoff factor ($i$).

Common Mistake #2: Ignoring the Role of Acid Strength

The Mistake: Students treat all three acids as equally strong (or equally weak) and assume $i = 1$ for all, leading to the wrong conclusion that $\Delta T_f$ is the same.

Why it's wrong: Acetic acid is a weak acid ($K_a \approx 1.8 \times 10^{-5}$). Trichloroacetic acid is much stronger ($K_a \approx 0.23$). Trifluoroacetic acid is even stronger ($K_a \approx 0.59$). The degree of dissociation ($\alpha$) increases in this order.

So $i$ increases: Acetic < Trichloroacetic < Trifluoroacetic.

How to avoid: Memorize the trend: Electronegative substituents (Cl, F) increase acid strength by inductive effect. More dissociation → more particles → higher $\Delta T_f$.

Common Mistake #3: Confusing "Depression in Freezing Point" with "Freezing Point"

The Mistake: Students say "trifluoroacetic acid has the highest freezing point" or mix up the direction of change.

Why it's wrong: Depression means the freezing point lowers. A higher $\Delta T_f$ means a lower freezing point. So trifluoroacetic acid gives the lowest freezing point (most depression). …