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The vapor pressure of acetic acid is 0.033 atm at 25°C. If 5.00g of $\ce{LiF}$ is added to 100g of acetic acid, what would be the vapor pressure of the solution at 25°C?

The lowering in vapor pressure (VP) can be calculated with this equation:

${∆P = X_{solute} * P°_{solvent} * i}$

${∆P}$ is the change (lowering) in VP.

${X_{solute}}$ is the mole fraction of the solute.

${P°_{solvent}}$ is the VP of the pure solvent.

${i}$ is the van't Hoff factor of the solute.

What is the "correct" van't Hoff for this problem, from the perspective of a general chemistry student? I know that $\ce{LiF}$ is not very soluble in water, and I know there are papers on lithium fluoride and acetic acid, but that's all beyond the scope of a general chemistry class.

Therefore, I think it's 2, because $\ce{LiF}$ is expected to be a strong electrolyte. It's ionic, and the solvent is similarly polar (acetic acid has hydrogen bonds). The solvent even resembles water in that both have hydrogen-bonding. So I would expect $\ce{LiF}$ to dissolve fully in acetic acid.

Remember, this is a general chemistry class, and the students aren't expected to know much beyond "like dissolves like." $\ce{LiF}$is polar, and so is acetic acid. So to them, $\ce{LiF}$ should dissolve in acetic acid.

Problem is that I'm in trouble with my supervisor since I told someone that i = 2. My supervisor is telling me that the van't Hoff factor should be 1. I looked over his work to see if he accounted for the dissolving of $\ce{LiF}$when calculating the mole fraction of the solute but he didn't - he just found the mole fraction of $\ce{LiF}$, not the combined mole fractions of $\ce{Li+}$ and $\ce{F-}$.

So, what's the deal with this problem? Is there some weird exception I'm not aware of regarding $\ce{LiF}$ and acetic acid? The van't Hoff factor should be 2, correct?

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According to Acid-Base Equilibria in Glacial Acetic Acid. III. Acidity Scale. Potentiometric Determination of Dissociation Constants of Acids, Bases and Salts J. Am. Chem. Soc., 1956, 78 (13), pp 2974–2979:

It is generally agreed [references 3 and 4] that acids, bases and salts are only slightly dissociated in glacial acetic acid.

While the article does not specifically address lithium fluoride, it has quantitative data for lithium chloride.

Lithium Chloride dissociates with a pK of 7.08 +/- 0.02 in acetic acid.

So considering the high concentration of LiF in the problem (5 in 100 grams), if LiF is similar to LiCl and the other salts studied, 1 rather than 2 is correct.

This knowledge is not part of a standard general chemistry curriculum, so unless a specific book or lecture provided relevant information, I wouldn't expect students to know this.

Furthermore, the solubility of LiF in acetic acid is only 0.84 grams per kilogram according to Glacial Acetic Acid as a Non-aqueous Solvent for Metal Fluorides, so the problem is not realistic. (Only 0.084g of the 5.00 g would dissolve). Also, HF is a strong acid in acetic acid according to that paper, so LiF might even be an exception to the general observation that salts do not dissociate in acetic acid.

Also, keep in mind that in the liquid phase, acetic acid exists as mostly cyclic dimers, which gives the solvent some non-polar characteristics.

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