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Why so much lye? Options
 
Kajlian
#1 Posted : 1/7/2019 8:24:05 PM

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I've been wondering why so many teks seems to recommend so high lye concentrations, pH of above 14.

I don't really understand why such a high pH would be necessary, especially for an A/B extraction. DMT is a weak base with a pKb above 5, so in what way would a pH of almost 15 be better at deprotonating the DMT ions than a pH of 12-13? Or is there some other reason to add such high amounts of lye?
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AcaciaConfusedYah
#2 Posted : 1/7/2019 8:35:05 PM

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Hey Kajlian,

I asked that same question... and I did an experiment to determine how much lye would be needed to basify a bark soup, monitored via pH probe and titration. If you check, it should be listed in the STB section. And, yes - my target pH was ~13, but it held at 13.76.

I should add - the soup was naturally acidified by the tannins from the bark. I did not add any additional acid. So, depending on how much additional acid someone uses (since this is the A/B section), then they will have to use enough base to counter the acid. If you decide to use an acid to lyse cells and convert the DMT into a salt, it doesn't take much. So, if you are conservative with the amount of acid, then you will not have to add as much base to neutralize.

Maybe some folks are adding a ton of base because they add too much acid? I am not sure where the common idea of "gram of lye per gram of bark" originated, but it seems to be popular.

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Kajlian
#3 Posted : 1/7/2019 10:01:53 PM

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AcaciaConfusedYah wrote:
I should add - the soup was naturally acidified by the tannins from the bark. I did not add any additional acid. So, depending on how much additional acid someone uses (since this is the A/B section), then they will have to use enough base to counter the acid. If you decide to use an acid to lyse cells and convert the DMT into a salt, it doesn't take much. So, if you are conservative with the amount of acid, then you will not have to add as much base to neutralize.


I acidified my extraction to a pH < 3, and the volume was almost 500 ml, but I still only needed 5,6 grams of lye to basify it to pH 13, that's 4 grams extra for neutralising the acid, but it's still way less than most teks recommend.

"A gram of lye per gram of bark" seems so wasteful to me. But maybe there's some reason for using very high concentrations of lye that I don't know of. For me it works with a much lower amount though.
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pete666
#4 Posted : 1/7/2019 10:18:24 PM

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Kajlian wrote:

"A gram of lye per gram of bark" seems so wasteful to me. But maybe there's some reason for using very high concentrations of lye that I don't know of. For me it works with a much lower amount though.


Which TEK uses 1:1 ratio for A/B?
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Asher7
#5 Posted : 1/7/2019 10:20:48 PM

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I remember reading a help thread where the person was using the 1:1 ratio and for the second pull had an emulsion that some additional lye took care of. No shaking or intentional heat was added except when putting the bark in it was immediately after dissolving the lye so it would have had that amount of heat applied so maybe that brought out some oils but I would think the emulsion would have been there for the first pull.

So to me that sounds like they were right on the teetering point of the lye working. There’s most likely factors I’m unaware of that determine all this so they’re just my thoughts. This was on a STB so nothing was required to counter any acid.
 
Kajlian
#6 Posted : 1/7/2019 10:21:15 PM

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pete666 wrote:
Kajlian wrote:

"A gram of lye per gram of bark" seems so wasteful to me. But maybe there's some reason for using very high concentrations of lye that I don't know of. For me it works with a much lower amount though.


Which TEK uses 1:1 ratio for A/B?


I don't know if any do, I was referring to the common idea mentioned in the previous post.

The exact ratio isn't the the question here, just very high concentrations. Something like >0,1M
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benzyme
#7 Posted : 1/8/2019 1:35:07 AM

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two reasons come to mind; STB requires highly basic solution to lyse the cells. people also add excessive base to inhibit formation of emulsions.

using physical methods to lyse the cells, a rule of 2 may be applied: pH = pKa +/- 2, for >99% protonation/deprotonation.
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AcaciaConfusedYah
#8 Posted : 1/8/2019 3:08:59 AM

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Kajlian wrote:
AcaciaConfusedYah wrote:
I should add - the soup was naturally acidified by the tannins from the bark. I did not add any additional acid. So, depending on how much additional acid someone uses (since this is the A/B section), then they will have to use enough base to counter the acid. If you decide to use an acid to lyse cells and convert the DMT into a salt, it doesn't take much. So, if you are conservative with the amount of acid, then you will not have to add as much base to neutralize.


I acidified my extraction to a pH < 3, and the volume was almost 500 ml, but I still only needed 5,6 grams of lye to basify it to pH 13, that's 4 grams extra for neutralising the acid, but it's still way less than most teks recommend.

"A gram of lye per gram of bark" seems so wasteful to me. But maybe there's some reason for using very high concentrations of lye that I don't know of. For me it works with a much lower amount though.



I think the actual amount will vary based on the type of water that is used (distilled? DI? RO? Spring water? Tap?), amount of tannins and other natural acids, volume of the extraction vessel, etc etc etc. I get what you're saying. The gram of lye per bark is wasteful, from my perspective - yet others prefer to use more than less.

And, yes, as benzyme mentioned - a STB tek that doesn't allow for some other form of lysing may need the higher pH. I don't "weigh x amount and dump." I make a concentrated solution of sodium hydroxide (~5 M) and titrate until the desired pH is hit - and maintains. Even then, I usually mix more than needed - guess I would rather have to toss 3 grams worth of lye than have to get the scales back out and weigh another small amount. Plus, it can help unclog the sinks.Pleased Laughing

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Kajlian
#9 Posted : 1/8/2019 6:33:05 AM

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Yes for a STB tek the base can lyse the cells, but for an A/B the DMT ions are already in solution when the base is added. I see how excess base could help in a STB but I'm not so sure about an A/B tel.

Why does excess base inhibit formation of emulsions?
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Th Entity
#10 Posted : 1/8/2019 7:02:44 AM

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Ive seen older teks that indeed recommend unnecessary high amounts of lye which i came to conclusion is mentioned there to prevent extraction/chemistry newbies to screw their extractions by maybe underbasification and at same time the high amount of lye could be recommended so it lowers the chance of those newbies forming a unwanted emulsion, and also to make it more simple and easy for them, i presonally would make a basic solution and add it untill desired pH is reached, but for a person who doesnt have lab glassware and basic chemistry understanding its easier to follow a recipe than think about these things.

Question: If you do A/B the acid would lyse the RB cells but if you keep, and not filter bark after acidification doesnt the base lyse the RB cells aswell?

Question: What does lyse mean? From my understanding lyse is a term to describe cells breaking apart?
(English is not my first language)

Question: Another thing which might be really stupid but i will ask anyway: We have MHRB with HCL acidifed water in a flask if we add a NaOH/water solution to it, it will neutralize HCL and turn the alkaloids in their base form right? But if there is excess or HCL and excess of base wouldnt they react toghether to from NaCL(+water and H) which will dissolve in the water. Im not chemist! Very happy
 
leratiomyces
#11 Posted : 1/8/2019 8:34:11 AM
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Kajlian wrote:
Yes for a STB tek the base can lyse the cells, but for an A/B the DMT ions are already in solution when the base is added. I see how excess base could help in a STB but I'm not so sure about an A/B tel.

Why does excess base inhibit formation of emulsions?


Two possible reasons.

Adding excess Naoh simply increases the ionic strength of the aqueous phase, thereby making emulsions less likely to form. The same effect that is utilised when adding nacl to an emulsion to break it.

Secondly, adding excess naoh deprotonates tannins to a larger extent than adding less naoh. I suspect that tannic acid which is only partially deprotonated is able to act as a surfactant, therefore causing emulsions. Where as tannic acid that has many of its phenol groups converted to sodium phenolates (adding excess naoh) sits in the aqueous phase without interacting with the organic phase, thereby not causing emulsions. Hope that makes sense.

I can't site a reference to support the second reason. Just my logical reasoning.
 
Psilosopher?
#12 Posted : 1/8/2019 12:02:49 PM

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I did a quick titration without plant material a while ago. However, as stated above, the tannins buffer the basification of the acid solution, assuming it's an ATB.

Too much base is wasteful, but not enough will result in sub par deprotonation.
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downwardsfromzero
#13 Posted : 1/8/2019 4:06:05 PM

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Th Entity wrote:
[...]
Question: If you do A/B the acid would lyse the RB cells but if you keep, and not filter bark after acidification doesnt the base lyse the RB cells aswell?
One would have to presume that some proportion of cells that were not lysed during the acidic phase, or by freezing or boiling, would subsequently by lysed by the base. There would be ways to test this hypothesis experimentally.

Quote:
Question: What does lyse mean? From my understanding lyse is a term to describe cells breaking apart?
(English is not my first language)

"Lyse", and its nominal and adjectival forms "lysis" and "lytic", are derived from the Greek λύσις lýsis, "a loosing" from λύειν lýein, "to unbind" - thus meaning "to split apart". https://en.wikipedia.org/wiki/Lysis

Quote:
Question: Another thing which might be really stupid but i will ask anyway: We have MHRB with HCL acidifed water in a flask if we add a NaOH/water solution to it, it will neutralize HCL and turn the alkaloids in their base form right? But if there is excess or HCL and excess of base wouldnt they react toghether to from NaCL(+water and H) which will dissolve in the water. Im not chemist! Very happy

DMT in HCl solution is present in its protonated form, "DMT.H+" which is not bound to the corresponding chloride anions, Cl−.

The NaOH which is added to the solution is also an ionic substance, Na+ and OH- ions. The OH- removes a proton from the DMT cation, DMT.H+, to produce water ("HOH" = H2O) and DMT freebase. The Na+ and Cl- ions just float around in solution not doing much except for contributing to its ionic strength which serves to push the now "oil-loving" (lipophilic) DMT into any awaiting non-polar solvent.

Excess HCl would just mean that more NaOH would be required before all the DMT could be deprotonated. Excess base would not do much of relevance. We see it reacting with some of the other compounds from the bark, such as the coloured compounds that change from red to black under these conditions. That is not really an excess because these compounds are 'competing' with the DMT for the base.
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benzyme
#14 Posted : 1/8/2019 4:11:07 PM

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Kajlian wrote:
Yes for a STB tek the base can lyse the cells, but for an A/B the DMT ions are already in solution when the base is added. I see how excess base could help in a STB but I'm not so sure about an A/B tel.

Why does excess base inhibit formation of emulsions?



emulsions are micelles, molecular structures with polar, hydrophilic heads, and nonpolar, hydrophobic tails; they are microbubbles, and in this case, the products of alkaline hydrolysis of fatty acids (phospholipids, which compose the structural surface of plant cells). They are created when the basification occurs within the second buffer range, pH = pKa + n, where n < 2; there are several species with partial charge, which are somewhat water soluble, and somewhat hydrophobic. When n > or =2, >99% of your analyte will be deprotonated, and will migrate to the nonpolar phase. The pH will also be high enough to solvate the micelles (into the polar phase).
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Elrik
#15 Posted : 1/8/2019 6:46:55 PM

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leratiomyces wrote:
...adding excess naoh deprotonates tannins to a larger extent than adding less naoh. I suspect that tannic acid which is only partially deprotonated is able to act as a surfactant, therefore causing emulsions...
I have not seen any indication of tannins causing emulsion in the technical sense but tannin is often seen as a precipitate in extractions that increases viscosity and inhibits the speedy separation of organic and aqueous phases. A wild excess of lye makes this precipitate into a soluble form so it does not hinder the NPS.
When I've done extractions too concentrated and with a minimum of lye I end up sitting there for hours like Wut? feeling like I'm waiting for oil to separate from custard. I avoid that by not doing extractions too concentrated.

In mescaline chemistry I've seen people argue that saturating with lye makes the mescaline float to the top as an oil, I'm content to use one twentieth the quantity of lye and do two more pulls Wink
 
Th Entity
#16 Posted : 1/8/2019 7:14:41 PM

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Thank you downwardsfromzero for explaining this to me.
 
rOm
#17 Posted : 1/9/2019 7:38:53 PM

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why so much lies ???

aah ok I was trolling softly.
I'll self ban for 24H.
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Jees
#18 Posted : 1/9/2019 8:51:36 PM

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The rising viscosity of a far too concentrated NaOh solutions can prevent the easy floating up of the naphtha too. I've ran into that kind of problems by mantra more lye makes better separation. If you're really overdoing it you might run into that kind of problems, it's something else than emulsions but you get same kind of problem. Had to dilute the experiment to get the viscosity down again for a better separation, and it did.
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Th Entity
#19 Posted : 1/10/2019 6:54:41 AM

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I've explained it like that to myself: (why emulsions form less if we add salt or lye or both) Yes polar(water) and non polar(oil) liquids dont mix but i remember reading that CTM said once that they kinda do just in so small amounts that we dont really see it. So i understanded it like that. If we
dissolve more things in the polar liquid (water) we get it more saturated in this case with NaCL,NaOH and whatever dissolves in the water from the bark, and the water now holds a lot of stuff in it now, and it becomes more saturated so theres not a lot of "space" in the water to hold more stuff so it "pushes" the non polar more easily away from it, this i think results in faster/easier separation of the two layers and less chance of forming emulsion, i think thats why people add salt, so the aq solution gives more easily the alkaloids after basifying and to prevent emulsion at same time.

Is my thinking right? im not a chemist this just makes sense to me i know its more complicated.
And i deffinetly dont explain it very good because of "language barrier". Please tell me if im right or wrong so i can edit/delete this reply so there is no spreading of false information. Smile
 
AcaciaConfusedYah
#20 Posted : 1/10/2019 5:34:27 PM

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Th Entity wrote:
I've explained it like that to myself: (why emulsions form less if we add salt or lye or both) Yes polar(water) and non polar(oil) liquids dont mix but i remember reading that CTM said once that they kinda do just in so small amounts that we dont really see it. So i understanded it like that. If we
dissolve more things in the polar liquid (water) we get it more saturated in this case with NaCL,NaOH and whatever dissolves in the water from the bark, and the water now holds a lot of stuff in it now, and it becomes more saturated so theres not a lot of "space" in the water to hold more stuff so it "pushes" the non polar more easily away from it, this i think results in faster/easier separation of the two layers and less chance of forming emulsion, i think thats why people add salt, so the aq solution gives more easily the alkaloids after basifying and to prevent emulsion at same time.

Is my thinking right? im not a chemist this just makes sense to me i know its more complicated.
And i deffinetly dont explain it very good because of "language barrier". Please tell me if im right or wrong so i can edit/delete this reply so there is no spreading of false information. Smile



Loveall mentioned above, about the viscosity of an aqueous solution at varying concentrations of NaOH. So, imagine an oil - like cooking oil. It would be considered more viscous than water. I guess a way to explain this may be: if you take 20 mL of DI water, and pour it from a cup, beaker, anything - then it's easily poured, hits the surface, and splatters/spreads at a relatively fast rate. When it was poured, nearly all of the water could readily flow from the vessel to the surface. OK, now, take 20 mL of canola oil, or vegetable oil. Pour it from the same vessel. You'll notice that the movement of the oil is "slower", the splatter/spread pattern is more condensed and does not spread as readily, or at the same rate that the water had spread. It will likely be more uniformed spreading/splattering than the water. It may be difficult to pour all of the oil from the vessel, as it moves "slower."

Though this is not an elaborate "chemistry derived" observation or explanation of viscosity, it may help you understand the difference between a liquid that is comparably less viscous (water) to another more viscous liquid (oil).

If I am interpreting the data correctly, presented by Loveall, then higher amounts of NaOH in an aqueous solution will increase the viscosity of water. This makes sense - especially if you have ever worked with solutions of NaOH that are >20 M in concentration. The solution does become more viscous.

Why is this relevant? It is easier to separate liquids that are less viscous than more viscous. So, if your aqueous layer becomes too viscous, it will potentially create some issues in a hasty separation (this is going to be solvent-dependent). If a person is using a particularly viscous solvent, like d-limonene, then the separation will occur at a decreased rate. It is likely not much of an issue, though it does require additional patience.

And, yes - there are usually "small" amounts of mixing between the organic and the aqueous layers. In a general procedure that you would find in an academic journal, most chemists will use a drying agent, like anhydrous magnesium sulfate or anhydrous sodium sulfate, to remove any water that may have transitioned to the organic layer. Due to lack of the equipment needed to properly dry and filter the organic layer from the drying agent, it does not seem to be a step that is added to the "TEKs" that you'll see on the Nexus. However, the amount of water that is "dissolved" into the organic layer is usually not in sufficient concentrations to negatively impact an extraction.

Smile
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ACY
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