Animated Solution for Chemistry - Solutions: The freezing point of benzene decreases by $0.45^{\circ }\text{C}$ when $0.2\text{ g}$ of acetic acid is added to $20\text{ g}$ of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be ($K_f$ for benzene = $5.12{\text{ K kg mol}}^{-1}$)

Imagine you are conducting an experiment in a chemistry lab. You take a beaker filled with exactly $20\text{ g}$ of benzene, a classic non-polar solvent. Then, you carefully add $0.2\text{ g}$ of acetic acid (${\text{CH}}_3\text{COOH}$) to it. You might expect the acetic acid molecules to simply swim around independently. But there is a catch! Because benzene is non-polar, the acetic acid molecules feel lonely and start forming hydrogen bonds with each other. They pair up to form dimers. This means two molecules of acetic acid act as a single particle in the solution.

This phenomenon is called association, and it fundamentally changes the colligative properties of the solution, such as the freezing point depression.

To account for this molecular pairing, we use the van't Hoff factor, denoted by $i$. Let's set up the equilibrium equation for the dimerization process:

Let the initial moles of acetic acid be $1$. If the degree of association is $\alpha$, then at equilibrium, $\alpha$ moles of the monomer have associated. Because it takes two monomers to make one dimer, this forms $\frac{\alpha }{2}$ moles of the dimer. The remaining unassociated monomer is $1-\alpha$.

The total number of moles at equilibrium is the sum of the unassociated monomers and the newly formed dimers:

Since the van't Hoff factor $i$ is the ratio of the total moles at equilibrium to the initial moles, we get:

Freezing Point Depression

Now, let's look at the physical effect. The freezing point of the benzene drops by $0.45^{\circ }\text{C}$. The formula linking this depression to the concentration is:

Here, $K_f$ is the molal depression constant for benzene ($5.12{\text{ K kg mol}}^{-1}$), and $m$ is the molality of the solution. Molality is defined as the moles of solute per kilogram of solvent. The molar mass of acetic acid is $60{\text{ g mol}}^{-1}$. Let's calculate the raw molality:

Substituting everything into our master equation, we get:

This is where we need to be careful with our algebra. Let's isolate the term containing $\alpha$:

Simplifying the right side:

Now, solving for $\alpha$:

The degree of association is $0.946$. To find the percentage association, we simply multiply by $100$.

This tells us that a whopping $94.6\%$ of the acetic acid molecules have paired up in the benzene solvent! It is a beautiful demonstration of how microscopic molecular interactions dictate macroscopic physical properties.