[Durlag]

Hi 

I'm pretty confused about the properties of caffeine. I've got a lot of interrogations, I hope it's not too confusing.

1) Regarding salts of amines

First of all, I'm wondering about amines in general. What salts do protonated amines form in pure water?
Another way of phrasing: what happens to a protonated amine if there are no anions to form salts? Does it precipitate? Does it bind to water molecules by H bonds?
For example, is caffeine soluble in pure water? Since its pKa is 14 at 25°C, it should be protonated in pure water (pH 7), so does it stays in solution as a "free cation"?

2) Regarding pH of water, solubility and pKa of caffeine

Next question (this is where I get really confused), I read that "Caffeine is soluble in water (approx. 16 mg/ml at room temperature, 200 mg/ml at 80°C, or 666 mg/ml in boiling water)". (http://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma-Aldrich/Product_Information_Sheet/c0750pis.pdf)

Why? Is it only because of that property of heated water to hold a greater quantity of a solute?
Is it linked to the pKa of caffeine that is 14 at 25°C but 10,4 at 40°C?
If yes, how is it linked? (I got the pKa value at 25°C from the previous link, and the one at 40°C from this link: http://www.drugbank.ca/drugs/DB00201).
Is it linked to the change in pH of water in function of the heat? I've read that this change in pH doesn't change the acidity / basicity (concentration of H3O+ / OH-), the water is still neutral.
So the pKa of caffeine at 40°C should be compared to the pH of water at 40°C, which is 6,77? All the pH scale is shifted? I'm pretty confused.
For example in order to calculate the % of dissociation of caffeine at 40°C in water, should you use the pKa of 10,4 and consider you're at a pH of 6,77?
If so, that means everything needs to be recalculated everytime you change temperature?

3) Regarding Ksp and precipitation

If I know the solubility of caffeine in water, I should be able to find the Ksp, and then to calculate the pH of a saturated solution, as well as the minimum pH required for precipitation?
But for this, I need a dissociation equation. That would need a salt of caffeine as the left term?
However, if we're in pure water, there a no salts possible? Also, I've read that these kind of salts are really soluble and shouldn't precipitate?
Or maybe calculating a Ksp in this case is not correct, because there is no salt involved?
If you're in pure water, how can it precipitate? Protonated caffeine molecules just fall out of solution when you reach a certain concentration? They link between themselves through H bonding? What about small amines that don't have O-H bonds?
Also, I've read that unprotonated alkaloids are not soluble in water, so they precipitate. What is the nature of the bonds bewteen the unprotonated caffeine molecules, that make them hold as crystals?

[Borek]

You are not confused about properties of caffeine, you are confused about chemistry in general - all your questions are based on misunderstandings of the basics.

1. No acid, no salt. Some protonation is possible, but you always need a counterion to keep the solution electrically neutral. In pure water such a counterion will be OH^-.

2. Solubility in this case has nothing to do with pH nor pKa. In general when the dissolution is endothermic solubility grows with temperature, apparently that's the case here.

3. We don't use Ksp to describe behavior of non dissociating molecules (it is possible though, just doesn't make much sense).

[Durlag]

Well, pardon my noobery 

1. Ok about the counterion being OH-, that's like a solution of NH3 that we call "NH4OH" although that compound has never been proven to exist. The ions present in solution are NH4+ and OH-.

Then I guess that in order to determine the % dissociation you would have to know the concentration of caffeine and do an ICE table. If you add caffeine to water it will raise the pH of the solution. Or you could measure the pH after addition and plug the pH and pKa into the Henderson-Hasselbalch equation to get the % dissociation.

Also, in pure water, the protonated amines are soluble because of the polarity of the water molecule.  And there will be no salts. NH4+ OH- is not a salt. However, upon the addition of an acid, you would get salts (e.g. NH4+ Cl- from the addition of HCl, which would yield NH4Cl upon evaporation). Am I right?