mcat question of the day

[deleted647690]

How would a decrease in the pH of a solution affect the volatility of decanoic acid?


The volatility continues to increase as the pH decreases.


Why would being protonated make something more volatile? Because it is nonpolar? Why does being nonpolar increase volatility? I see online that polar compounds have increased intermolecular forces, therefore, they can hold together better.

My line of reasoning was that being protonated would increase molecular weight, meaning it is less likely to evaporate.....haha does that even make sense? Would the additional weight of a few protons not make much of a difference?

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[aldol16]

Protonation would eliminate any charge on the molecule, thus making it uncharged. Uncharged molecules generally evaporate better than charged ones.

The added molecular weight of a single proton is negligible because you have to think about why volatility generally goes down as molecular weight increases. It's because increasing molecular weight provides greater van der Waals interaction area and so greater intermolecular interactions. But a single proton is tiny and negligible in something as relatively large as decanoic acid.

 

[deleted647690]

Protonation would eliminate any charge on the molecule, thus making it uncharged. Uncharged molecules generally evaporate better than charged ones.

The added molecular weight of a single proton is negligible because you have to think about why volatility generally goes down as molecular weight increases. It's because increasing molecular weight provides greater van der Waals interaction area and so greater intermolecular interactions. But a single proton is tiny and negligible in something as relatively large as decanoic acid.

Because uncharged molecules only experience Van der waals forces, which are weaker than the polar forces that charged molecules have? Thank you. So it follows that volatility follows this trend? Van der waals forces<polar bonds<Hydrogen bonds? Molecules that experience each of these types of bonds would have increasingly higher volatility, correct? Since each increasing interaction offers stronger intermolecular forces.

 

[aldol16]

Because uncharged molecules only experience Van der waals forces, which are weaker than the polar forces that charged molecules have? Thank you. So it follows that volatility follows this trend? Van der waals forces<polar bonds<Hydrogen bonds? Molecules that experience each of these types of bonds would have increasingly higher volatility, correct? Since each increasing interaction offers stronger intermolecular forces.

I think you meant to say lower volatility, or higher boiling point. In other words, increasing interactions also means harder to evaporate.

 

[betterfuture]

This relates to vapor pressure right? Because say you have hexane and alcohol. Due to their stronger solute-solute and solvent-solvent interactions, they are less likely to mix and form new solute-solvent bonds. This increases the vapor pressure above the solution.