Why it could be counterproducing to add excess of acid in A/B

[Adivino]

I've seen a lot of people reporting that using extra vinegar (acetic acid to be more specific) doesn't necessarily increase the yield of an A/B extraction. I intuitively though that an excess of acetic acid is a sure way to neutralize a larger fraction of mescaline freebase, but now I see how this could be wrong.

Firstly, mescaline is a weak base which reacts with acetic acid to form mescaline acetate. It has a certain solubility product which is something like this: Ksp = [Mescaline+][Ac-]. Now, if there's an excess of the acetate ion in solution, which means increasing [Ac-], [Mescaline+] has to decrease because Ksp remains constant. I think that this could shift the equilibrium back to mescaline freebase, which is not what we want.

The easiest solution I see is adding small amounts of acetic acid and doing a pull, then adding more and doing another pull with polar solvents.

What do you think? Does this sound reasonable?

 

[Infundibulum]

Firstly, mescaline is a weak base which reacts with acetic acid to form mescaline acetate. It has a certain solubility product which is something like this: Ksp = [Mescaline+][Ac-]. Now, if there's an excess of the acetate ion in solution, which means increasing [Ac-], [Mescaline+] has to decrease because Ksp remains constant. I think that this could shift the equilibrium back to mescaline freebase, which is not what we want.

No, this is incorrect chemistry.

First off, mescaline acetate is a salt and it fully dissociates in solution, so:
Mescaline acetate -> [mescaline+] + [acetate-] as an irreversible reaction.

In addition, irrespective of the presence or absence of acetate, low pH would keep mescaline as a cation. It is actually the low pH that makes the mescaline a cation and not the acetate; even in the theoretical scenario where acetates were totally depleted but somehow the pH was kept acidic, mescaline would still be a cation; acidic pH means too little [OH-] and too much [H+] and mescaline as a base reacts with water to provide [OH-] to try to restore the [OH-] - [H+] equilibrium:

[Mescaline-NH2] + [H20] <-> [Mescaline-NH3+] and [OH-]

So basically using too much vinegar is not a problem, the only problem is having too much volume to evaporate plus the infernal stink of the acetic acid. In the case you describe above, nothing of that happens. First off, there is not such a thing as mescaline acetate if it is completely dissolved in water. In your example above there is [CH3COOH], [H+], [CH3COO-] and [mescaline+]; the low pH caused the formation of [mescaline+] from mescaline freebase in the NP solvent. And since mescaline freebase is a base and provides OH- in the solution, then the only thing that will happen would be the dissociation of more [CH3COOH] to form [H+] and [CH3COO-].

If, now one takes mescaline acetate and dissolves it in vinegar, then equilibria will change; mescaline acetate dissociates and provides [CH3COO-], which shift the dissociation equilibrium of the acetic acid to the left, i.e.:
add [CH3COO-], then

[CH3COOH] <- [H+] and [CH3COO-]

 

[ThirdEyeVision]

Yeah....what he said

 

[Psilocin Dreams]

Ok so you cant use too much acetic acid (vinegar right?) in a mescaline extract. Does the same go for DMT if not please help me to understand.

 

[dg]

i'm sorry, why is any acid needed?
(at least until salting step)

is the op stating that the plant naturally contains mescaline freebase that needs to be freed up? if so, that is not true, there is no mescaline freebase in cacti naturally
thats why water or other polar solovents work to extract fresh material, with or with out any acid being added

basically, the A step in cacti teks is silly, unless defatting is being done for some reason, even then, no acid is needed

 

[dg]

post question in cacti forum?

 

[gammagore]

Why is this in the DMT extraction sub-forum, or am I missing somthing?

Il move it over to Cacti.