Animated Solution for Chemistry - Solutions: In a $0.2$ molal aqueous solution of a weak acid $\text{HX}$, the degree of ionisation is $0.3$. Taking $K_f$ for water as $1.85$, the freezing point of the solution will be nearest to

Have you ever wondered why we throw salt on icy roads in the winter? It’s not to melt the ice with heat, but to fundamentally change the physical properties of the water itself! This phenomenon is governed by what we call colligative properties.

Colligative properties are unique because they don't care about the identity of the solute—whether it's sugar, salt, or a weak acid like in our problem. They only care about the number of particles dissolved in the solvent. The more particles you have, the more they disrupt the solvent's ability to freeze, leading to a phenomenon known as freezing point depression.

In a perfect world, one molecule of solute would yield one particle in solution. But chemistry is rarely that simple! When we dissolve an ionic compound or an acid, it can break apart (dissociate) into multiple ions.

This is where the van't Hoff factor ($i$) comes to the rescue. It acts as a multiplier to account for the actual number of particles present in the solution after dissociation. For a weak acid $\text{HX}$ that only partially dissociates, we can't just assume $i=2$. We have to calculate it based on its degree of ionization ($\alpha$).

Let's look at the equilibrium of our weak acid:

If we start with $1$ mole of $\text{HX}$, and a fraction $\alpha$ dissociates, we are left with $(1-\alpha )$ moles of intact $\text{HX}$. At the same time, we produce $\alpha$ moles of ${\text{H}}^{+}$ and $\alpha$ moles of ${\text{X}}^{-}$.

The total number of moles at equilibrium is the sum of all these species:

Since we started with $1$ mole, our van't Hoff factor $i$ is simply $1+\alpha$. Given that the degree of ionization $\alpha$ is $0.3$, we find:

This means for every mole of $\text{HX}$ we dissolve, we effectively get $1.3$ moles of particles in the solution!

Now we bring in our master equation for freezing point depression:

Here, $K_f$ is the molal freezing point depression constant for water ($1.85{\text{ K kg mol}}^{-1}$), and $m$ is the molality of the solution ($0.2\text{ molal}$).

Let's plug in our values:

First, multiplying the molality and the constant:

Next, we multiply by our van't Hoff factor to account for the extra particles:

This value, $0.481^{\circ }\text{C}$, represents the drop in the freezing point, not the freezing point itself!

To find the actual freezing point of the solution ($T_f$), we must subtract the depression from the freezing point of the pure solvent ($T_f^{\circ }$). For water, the pure freezing point is exactly $0^{\circ }\text{C}$.

And there we have it! The presence of the partially dissociated weak acid has forced the water to freeze at a colder temperature of $-0.481^{\circ }\text{C}$. This perfectly matches option (a).

Always remember: when dealing with electrolytes, never forget to invite the van't Hoff factor to the party!