The freezing point depression of a solution is given by the formula

$$\Delta T_f = i \, m \, K_f,$$

where

• $$\Delta T_f$$ is the freezing point depression,

• $$i$$ is the van’t Hoff factor (the number of particles the solute dissociates into),

• $$m$$ is the molality of the solution, and

• $$K_f$$ is the cryoscopic constant of the solvent.

For a 1 molal solution, the only variable that changes among these acids is the van’t Hoff factor. When an acid dissolves in water, it can dissociate into its ions. A higher degree of dissociation (a larger $$i$$ value) means more particles in solution and a greater freezing point depression (resulting in a lower freezing point).

Here’s what happens with the given acids:

$$\mathrm{CF_3COOH}$$ (trifluoroacetic acid)

– The $$\mathrm{CF_3}$$ group is highly electron-withdrawing, making the acid very strong. It nearly fully dissociates, giving $$i \approx 2.$$

$$\mathrm{CCl_3COOH}$$ (trichloroacetic acid)

– With the strong electron-withdrawing $$\mathrm{CCl_3}$$ group, this acid is also very strong and nearly fully dissociates (again, $$i \approx 2$$).

$$\mathrm{Cl_2CHCOOH}$$ (dichloroacetic acid)

– Also considerably stronger than acetic acid, this acid will dissociate almost completely ($$i \approx 2$$).

$$\mathrm{CH_3COOH}$$ (acetic acid)

– The methyl group ($$\mathrm{CH_3}$$) is electron-donating, making acetic acid much weaker (pKa ≈ 4.76). It ionizes only partially in water, so its effective van’t Hoff factor is closer to 1.

Since the freezing point depression is directly proportional to the van’t Hoff factor, a solution with a lower $$i$$ will have a smaller depression and hence a higher freezing point. In this case, acetic acid (Option D) dissociates the least in water, meaning its 1 molal solution will have the highest freezing point compared to the nearly fully dissociated acids.

Thus, the compound with the highest freezing point of its 1 molal aqueous solution is:

Option D: $$\mathrm{CH_3COOH}$$.

Liquid Solution

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