13 replies to this topic

[greenskeeper]

Here's my problem:

If I estimate X mg of mescaline freebase is in my solvent, I should use Y drops of acid to salt all of the freebase with minimum excess acid.

 

In order to work this out, I'd like to know:

For each drop of 35% hydrochloric acid, how many mg of mescaline hydrochloride should result?

What about for 20% HCl?

 

HCl - 36.46 g/mol

mescaline - 211.26 g/mol

1 drop = 0.05ml

 

?

Edited by greenskeeper, 06 May 2021 - 09:53 PM.

[Ringo]

I don't mean to stick my nose in where I don't know what I'm saying but are you the same person from that other thread that posted that Klean Muriatic Acid?

 

If so, that is not 35%. It is 20%. The BLUE label stuff is 35-37%.The stuff in the picture in the other thread was GREEN. 20%

 

I see you have both listed above.

 

If I am out of place, please excuse me and I'll delete the post.

 

Peace

 

Edit: I just went back to review that other thread. Please excuse me. You know what you have. I just reported this post for a mod to please delete this post.

 

Sorry 'bout that.

Edited by Ringo, 06 May 2021 - 10:53 PM.

[greenskeeper]

The common concentrations for sale seem be approx. 35% or 20% so the answers for those should be useful in most cases.

I'm glad people are interested :)

[Phineas_Carmichael]

Good catch, Ringo! And a great reminder to always check the SDS documents for whatever chemicals you're using in your Clandestine Chemistry Lab!

A Thought Experiment
Ultimately a Pointless Mathematical Exercise
©Phineas_Carmichael 2021

You need equimolar amounts of freebase(X) and acid(Y).

1 mole is 6.02214076x10²³ or 602,214,076,000,000,000,000,000 atoms or molecules.

602.2 SEPTILLION molecules!!!

First, convert your estimated mescaline content (X) into moles. For the sake of round numbers, let's say you're expecting to get 1g of mescaline from your extraction.

1g ÷ 211.26 g/mol = 0.0047 moles mescaline.

That's the easy part...

The tricky part is calculating molarity (Moles/Liter) of your acid solution (Y) from the % by weight given in the SDS document. For the sake of round numbers let's use a 20% HCl solution. 20% by weight is 200g HCl in 1000g solution.

We need the density of the solution to convert the weight to volume . Our SDS gives the density as 9.09Lb/Gal because of course it does... Let's convert that to a more usable unit!

9.09Lb/Gal ÷ 3.794L/Gal = 2.40Lb/L
2.40Lb/L × 454g/Lb = 1,089.6g/L
1,089.6g/L ÷ 1,000mL/L = 1.09g/mL

So...
1mL of solution weighs 1.09g and 20% of that is HCl:
1.09g × 0.2 = 0.218g
Each mL of 20% HCl contains 0.218g of HCl
0.218g ÷ 36.46g/mol = 0.006mol
Each mL of 20% HCL contains 0.006mol of HCl

Remember, we need equimolar amounts and looking back in our notes we see that we estimated 0.0047 moles of mescaline...

0.0047mol Mesc ÷ 0.006 mol/mL HCl = 0.78mL HCl

We need 0.78mL 20% HCl to convert our 1g of mescaline freebase into mescaline HCl. Conventional wisdom is that there are 20 drops in 1mL

20 drops/mL × 0.78mL = 15.6 drops

=========

But wait! All of that math up there assumes that the equation is:
Mescaline Freebase + HCl --> Mescaline HCl

In reality, this is Equilibrium Chemistry, so the equation is actually:
Mescaline Freebase + HCl <--> Mescaline HCl

The reaction runs both ways at all times! Molecules are crashing into each other and bouncing around in solution making all that math up there completely useless because we assumed it was a one-way reaction!

Sorry for the math-troll lol

=========

In equilibrium chemistry we want to force the reaction to favor the products. We do this by adding reactants or subtracting products.

Adding to the left side of the equation pushes the equilibrium to the right and subtracting from the right side pulls it to the right.

In some reactions water (or an alcohol or something) is formed so we distill it off to force the reactants to keep crashing into each other. In this case we can't do that so we have to add an excess of 1 reactant.

Since our estimated Mescaline Freebase is fixed at 1g we have to add extra HCl. Mescaline's pKa is 9.56, so when exposed to pH 5.56 HCl solution 99.99% of the target molecule is in salt form in the aqueous.

We do multiple pulls because water and xylene don't like to interact with each other so we need to smash them together several times to make sure all the molecules we want get salted out.

Edited by Phineas_Carmichael, 07 May 2021 - 03:50 AM.

[greenskeeper]

Thank so much for this answer, math-troll and all. This is a great learning step, a lot to consider.

 

If we have 1g of freebase and add 15.6 drops - what happens? What is the pH of the water? How much water am I using? How much is freebase? How much is salt? Does that depend on the solubility of the freebase in the xylene and the solubility of the salt in water? How much xylene am I using?

 

Holy shit. I have to read some things...

[greenskeeper]

Regarding the multiple pulls because of xylene and water's mutual dislike, I have a question:

 

If my salting-out water reaches pH 5 and on evap'ing yields for example 100mg of salt, approximately how much do you think might remain to be pulled from the xylene in the next round? (assuming I shake the mixture for a minute after adding drops of acid).

[pharmer]

I applaud you're desire to get to the finicky fine details of proper chemistry.

 

Some of us don't have that kind of  discipline for math (god bless those who do)  and just add a drop of concentrate and watch what happens.  You'll be able to see the reaction in the liquid and then the crystals drop out. Some of us do this until nothing precipitates out of solution. Usually what happens is you get brown crystals once you've evap'd away the over-acidified water.

 

It's far from perfect but gets the job done.

 

Knowing the math, and why it's important, is master class level stuff. Knowing what the reactions are and why they happen at each step of the process is where the meat and potatoes of home chem projects live.

[greenskeeper]

I understand. This is my current obsession. I learned the method and was so amazed at what I'd done I want to understand what's going on, what to consider in this beginner's home chemistry.

 

Although I have an idea what pKa is and decided to go 4 (pHs?) higher than the pKa of mescaline to push the equilibrium far over to the freebase side in my basified cactus solution, I don't know why I didn't think to apply that to salting out.

 

I think if I can find out what relationships are involved and what is the scale and the nature of the curves (i.e. linear, log, exponential, etc) I can figure out how to estimate some things.

 

Thank you again for the help.

[Phineas_Carmichael]

Quick note of pKa as it applies to extraction chemistry while I'm on break from work. In a solution where pH=pKa 50% of the target molecule is freebase and 50% is a salt at any given time. The curve is logarithmic from there.

At pH=pKa +1 then 90% is freebase.
At pH=pKa +2 then 99% is freebase.
At pH=pKa +3 then 99.9% is freebase.
At pH=pKa +4 then 99.99% is freebase.

It goes in the other direction towards the salt as you decrease pH of course.

I'll try to answer most of your other questions after work when I'll have access to my textbooks. You're really putting me through my paces with some of this stuff!

[greenskeeper]

Please don't feel you have to put time and effort into this. It's not like I'm working towards a particular thing.
On the other hand, if you enjoy it then please tell me everything...

[Phineas_Carmichael]

Oh, no, you're not getting off that easily. You asked the questions so now you're getting a lecture

I use bits and pieces of this stuff in a practical application at work every day; it's fun for me to go back and consider it from the textbook side every once in a while and try to explain it. Makes me a better employee and a better writer.

Thank so much for this answer, math-troll and all. This is a great learning step, a lot to consider.

If we have 1g of freebase and add 15.6 drops - what happens? What is the pH of the water? How much water am I using? How much is freebase? How much is salt? Does that depend on the solubility of the freebase in the xylene and the solubility of the salt in water? How much xylene am I using?

Holy shit. I have to read some things...

Today's lesson is a bunch of me blathering about the pH equation: pH = -log₁₀[H⁺] where [H⁺] is the concentration of acid in solution expressed in Moles/L. This is easy to do with HCl because it breaks apart completely into H⁺ (Hydronium ions aka protons) and Cl^- (chloride ions) in water; it's more difficult to do with other acids, like acetic, that don't completely break apart.

pH is a dimensionless quantity, and as such there is no unit. The equation simply describes the amount of free hydrogen (H⁺ ions) in solution from a range of 0 (lots of free hydrogen) to 14 (Very little free hydrogen) Because this is a logarithmic function, each whole number indicates a 10x jump in the amount of H⁺ ions. So a pH 6 solution has 10x more than pure water (pH 7), pH 5 solution has 100x more than pure water, ph 4 solution has 1000x more, etc

Looking back at our notes from last night we can see that 1mL of our 20% solution contains 0.006mol of HCl. Let's convert that to moles/L quick so we can calculate the pH of our 20% solution.

0.006mol x 1,000mL/L = 6mol/L
-log₁₀[6] = -0.79

Yes, pH can be negative, even though the scale goes from 0-14, but you only ever really see it when calculating the pH of concentrated acids.

If you have 1L of xylene that contains 1g of mescaline and you add 15.6 drops of 20% HCl you'll be able to see the mescaline HCl crystals crashing out with each drop you add, but in the end you'll have 1L of xylene with 0.78mL of water at the bottom. The salting reaction happens only at the interface between the xylene and the water, and with such a tiny amount of water you could shake that 1L container for a thousand years and that 0.78mL of water would only touch a fraction of the xylene to react with a fraction of the mescaline. The pH of the water would remain relatively unchanged at around -0.79 (I think... I hope pharmer can chime in a bit to clarify the Theory vs Practice on this for me)

If I remember correctly from your extraction thread you run 100mL of strong HCl solution through your xylene and check the pH, adding more acid dropwise until the pH is 7 or lower. This is excellent Practical Chemistry, as a pH of 7 or less indicates that all of the mescaline in the xylene has reacted with the H⁺ ions in the aqueous solution. The brown impurities we see in "over-acidified extractions" are probably other alkaloids that have a much lower pKa than mescaline that come over in the overly-strong HCl solution.

If we have estimated our xylene to contain 1g (0.0047mol) of mescaline freebase then we need to introduce it to 0.0047mol of HCl in a solution that has a pH that isn't so low that all the icky stuff comes over with it. Let's just see what happens if we put our 15.6 drops (0.0047mol) of 20% HCl in 1L of pure water.

-log₁₀[0.0047] = 2.33

So we have created a pH 2.33 solution that if we run it through the xylene in 5 or so pulls with vigorous mixing each time should theoretically convert 1g of mescaline freebase (211.26 g/mol) into 1.16g mescaline HCl(247.72g/mol)

I'm kind of getting into the weeds here, so I'm going to stop and encourage further questions and maybe some input from more experienced extractors...

Theory and Practice are VERY different animals, and these numbers are not quite lining up with established extraction protocols as far as I recall; it's been a hot minute since my buddy Freddy was doing this kind of stuff regularly and textbooks only go so far...

Regarding the multiple pulls because of xylene and water's mutual dislike, I have a question:

If my salting-out water reaches pH 5 and on evap'ing yields for example 100mg of salt, approximately how much do you think might remain to be pulled from the xylene in the next round? (assuming I shake the mixture for a minute after adding drops of acid).

Multiple pulls are done because the reaction only happens at the places where the xylene and the water touch. Each time they touch they decrease the probability of the reaction happening again so we shake them up to increase the number of places they touch.

Forgive me, but I don't quite know your salting process well enough to speak to the numbers. I know it's a "titration salting" but can you run me through it again please?
Perhaps with example numbers from previous extractions? (I know that's a big ask, and if you don't have actual data you can just make up numbers)

I applaud you're desire to get to the finicky fine details of proper chemistry.

Some of us don't have that kind of discipline for math (god bless those who do) and just add a drop of concentrate and watch what happens. You'll be able to see the reaction in the liquid and then the crystals drop out. Some of us do this until nothing precipitates out of solution. Usually what happens is you get brown crystals once you've evap'd away the over-acidified water.

It's far from perfect but gets the job done.

Knowing the math, and why it's important, is master class level stuff. Knowing what the reactions are and why they happen at each step of the process is where the meat and potatoes of home chem projects live.

That's the beauty of Chemistry! You can go at it in several different ways. None of the famous Counterculture chemists (Nicholas Sand, Owlsley Stanley, William Leonard Pickard etc) had chemistry degrees; they were simply exceptionally intelligent individuals with the means and drive to produce these exciting molecules...

"Far from perfect but gets the job done" is just as good as "I got perfect results."
Either way the job is done!

Can you tell us more about this adding concentrated acid dropwise into xylene method please? I've never read anything about it & I'm mighty intrigued by watching anything precipitate out of anything else. That's some meat and potatoes I want to know more about...

Edited by Phineas_Carmichael, 08 May 2021 - 01:28 AM.

[greenskeeper]

Now I don't think my salting out method is right - I was only doing one acid pull on the xylene (before returning it to the base mixture) because I was thinking in terms of a one-way reaction.

I will try three and see how that works out.

 

I use 35% HCl

I start with 1 drop per 100mg of estimated yield into the xylene - I came up with this number from experience. (I only put the acid conc. directly into the xylene because it looks cool - some crystals can be seen briefly sometimes)

100ml water goes in next before mixing it up.

I take drops from the aqueous layer using my modified baster and check the pH with indicator paper.

If it's basic I add another drop or two, mix it, and check again.

It often ends up quite acidic, around pH 3, especially on salting out from the last base/xylene pulls. Not too much problem, I stay upwind during evap :)

 

When I started out I was using more like 300ml water but decided to use 100ml instead to make the evap quicker.

Should I use more water to avoid the low pH and increase contact surface in the mixing?

[greenskeeper]

Does concentration gradient have any affect on the freebase/salt equilibrium? Is the salt more preferred if there is more room in the water, considering it is insoluble in xylene?

[pharmer]

Please don't feel you have to put time and effort into this. It's not like I'm working towards a particular thing.
On the other hand, if you enjoy it then please tell me everything...

Lots of people will see this thread. Everybody gets to learn.

 

Some of us are reminded of stuff we've forgotten.

 

It's alllllllllllllllllllll gooooooood!

 

Think about the very tiny percentage of people who actually know, or ever will know, how to purify cactus. It's an elite club