Volatility

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victorias

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I though that decanoic acid is pretty large molecule (10 carbons) and at low pH, it would be protonated, which means that it would be capable of making H bonds in this stats - causing it to have a higher boiling point and lower volatility.

How should I approach this question? Can someone explain this?

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Also what do the terms protonated and deprotonated state really mean? This is my understanding:

So if I put a molecule in a solution of low pH, there are already lots of protons in the the solution and the molecule should not lose its protons - it would remain protonated. In fact if the molecule is capable if accepting protons, it will accept some from the solution.

Conversely, if I put a molecule in a solution with high pH, there are not many protons in the solution so the molecule can give off its protons and be in a deprotonated state.

The degree to which a molecule will give off or accept protons will also depend on its strength as an acid or base.

Is this correct?
 
Your first post's reasoning is correct. The protonated decanoic acid will be able to form hydrogen bonds. Yet, you need to consider both the deprotonated and protonated states.

You need to consider them in relation to the molecules pka:

pH<pKa - the molecule will be protonated
pH=pKa- molecule will be neutral (half will be protonated and half unprotonated)
pH>pka- the molecule will be mostly deprotonated

In the case for decanoic acid, when the hydroxyl group is deprotonated it will mostly bear a negative charge. The charged the molecule will be able to make stronger intermolecular bonds. Remember, as the strength of intermolecular bonds increases, the volatility decreases.

Meanwhile, if you being to lower the pH, the proportion of protonated decanoic acid increases and it eliminates the negative charge, making it neutral. You have to note that electrostatic interactions are much stronger than hydrogen bonds. As a result, intermolecular bond strength decreases and volatility increases.
 
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Where did you get that question from? It sounds very.... odd, to put it politely.

They are basically asking what the boiling point of oil is when mixed with water. Whaaaaaat? How will they go about testing that? Because I am pretty sure no matter what they do the water will boil first and then only oil is left.
 
They are basically asking what the boiling point of oil is when mixed with water. Whaaaaaat? How will they go about testing that? Because I am pretty sure no matter what they do the water will boil first and then only oil is left.

Only if you don't have an azeotropic mixture of water and oil (butyric and propionic acid come to mind) :p
 
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