Question

The van't Hoff factor for a monoprotic acid dissolved in water is found to be 1.05. When the same acid is dissolved in an unknown solvent, the van't Hoff factor is found to be 1.00. Based on these findings, answer the following questions:

1.) What can you conclude about the strength of the acid? Clearly and completely explain how you came to your conclusion.

2.) What can you conclude about the polarity of the unknown solvent? Clearly and completely explain how you came to your conclusion.

Answer 1

1. Since the van't Hoff Factor is decreasing as we can see from 1.05 to 1.00 the acidic strength is thereby increasing. Now this van't Hoff Factor depends upon the number of solutes that is dissociating into the solvent.As we can see the no of solutes dissociating referred by a number becomes lesser , it would be obvious to understand that the strength of the acid must have been increased.

2. We know the basic principle of solubillity which is LIKE DISSOLVES LIKE. Any solvent,which dissolves the monoprotic acid,is a polar covalent compound so it is very obvious that the solvent will be a polar covalent solvent too.Now ,there might be some exceptions to this case , just like some pola rcovalent acids dissolve in non polar covalent solvents like water.(Just because water is a universal solvent.)

Answer 2

1.) Conclusion about the Strength of the Acid :-

The van't Hoff factor (i) gives us insight into how many particles a solute dissociates into in a solution.

• In Water (i = 1.05):

• A monoprotic acid (HA) dissociates partially in water:

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

• If dissociation were complete, $i=2$ (since 1 HA → 1 H⁺ + 1 A⁻).

• However, $i=1.05$ (very close to 1) suggests that only ~5% of the acid dissociates, meaning it is a very weak acid in water.

• In the Unknown Solvent (i = 1.00):

• Here, the acid does not dissociate at all (since $i=1$, meaning HA stays as intact molecules).

• This further confirms that the acid is weak, as it only shows slight dissociation in water and none in the other solvent.

Conclusion: The acid is very weak because it barely dissociates in water and does not dissociate in the unknown solvent.

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2.) Conclusion about the Polarity of the Unknown Solvent :-

The dissociation of an acid depends on the solvent's ability to stabilize ions (H⁺ and A⁻), which is influenced by its polarity and hydrogen-bonding ability.

• Water (Polar, Hydrogen-Bonding Solvent):

• Water stabilizes ions well due to its high polarity and hydrogen bonding.

• Even though the acid is weak, it still dissociates slightly (~5%) because water can solvate the ions.

• Unknown Solvent (i = 1.00):

• Since $i=1$, the acid does not dissociate at all, meaning the solvent cannot stabilize ions.

• This suggests the unknown solvent is nonpolar (e.g., hexane, benzene) because polar solvents (like water or alcohols) would at least allow some dissociation.

Conclusion: The unknown solvent is nonpolar because it does not facilitate any dissociation of the acid.