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itsame

Rising Star
Hey all,

10~15 years ago I used to be much like "one of you" - AKA a walking encyclopedia of chemicals, basic neurology, and a hardcore explorer.

I spent about 7 years drug-free and loved it. Recently the old systems have been kicking back online (AKA from "the old me" ) and I find myself coming here to seek wisdom of why a friend's tek didn't work and how to fix it.

Also hope to be able to contribute more as I learn
 
Welcome :) Sounds like you've got a good depth of life-experiences to build on and share.

Hope you're able to find what you seek here. With some searching and reading I'm sure you could help your friend improve his process greatly
 
Thanks for the warm welcome. Is it ok if I turn this thread into a diagnostic question?
 
itsame said:
Thanks for the warm welcome. Is it ok if I turn this thread into a diagnostic question?
Click to expand...
Best ask already.

The more precise the description of what was used, what was done, and what observations were made, the easier it is to point you in the direction of useful information.

And welcome to the Nexus!
 
OK, so SWIM tried to take the strongest performing parts of multiple teks to try and get the highest first yield with the least amount of work. They went with straight to base method.
1500ml water : 250g MHRB, very finely shredded : ~280g Lime
Two of these setups in 1 gallon jars.
Mix lime, water, MHRB
Let sit for ~36 hours, mix a lot lot lot over this time. Water turns iced tea color, MHRB turns gray/black
After this time, filter out MHRB with a cloth a little finer than cheese cloth.
Add 250ml naphtha to filtered base water. *SWIM intended to hot bath the naphtha, but spaced out on performing that step*
SWIM read two teks saying to gently roll to avoid emulsion, and one where it says to mix a lot and (paraphrase) “it will emulsify, this is fine, it will separate in a few hours, or maybe a day.”
SWIM shakes vigorously, creating something like 2½ distinct layers in the jar.
Bottom layer: base water, Mid layer is sediment that settled in the bottom of the naphtha. It is whispy and off-white colored. If you look closely at it, it looks like oddly-shaped bubbles, Top layer: clear naphtha with no particles floating in it.
SWIM took as much naphtha from the top of both jars and put in the freezer for 24 hours.
Filter naphtha, no sediment in filter, let the dish evaporate, completely clean. Nothing at all. SWIM tests the pH of the base water, just in case. It maxes out the pH paper at what looks like at least 13.5
SWIM read somewhere that it can get trapped if there’s not enough water, so SWIM took the top and mid layers, (naphtha and whatever is sitting at the bottom of the naphtha), and consolidated them to a quart jar. Then added ~1000ml water to see if it would change anything. Same layering problem. Has not had an opportunity to test naphtha again from this.

SWIM wonders if that sediment is the spice, or if it’s just fats from the plant. Oh, Saved and reused naphtha. *NOTE: Klean Strip VM&P Naphtha evapped clean for SWIM*

SWIM is just trying to wrap their head around how they came up with literally nothing and how to save it.
 
This procedure wont work with lime, aka Calcium Hydroxide, (CaOH) as the base, because it is not very water soluble, nor basic enough to break down shredded bark. Can be used in a drytek however.

What you need is lye, aka Sodium Hydroxide, NaOH, Caustic Soda.

You should be ok to add this, slowly and carefully to your jars of mhrb, lime and water. The solution should turn black. Then pull again with fresh naphtha and consider doing freeze precipitation to avoid evaporating large amounts of petro-chemicals.
 
OK, so SWIM tried
Click to expand...
Ah, that's the problem. This guy, SWIM, is always screwing up extractions. You would have thought he might have learned something by now :lol:

Anyhow, when using lime one should only add enough water to make a doughy paste. This is then mashed around with the NPS - using a long-pronged fork helps to maximise contact between the phases.

In your ("SWIM's") case with all its surplus water, you can create NaOH in situ by adding the requisite quantity of sodium carbonate. This reacts with Ca(OH)2 to produce NaOH and calcium carbonate (chalk).

The other thing to try would be mixing the bark/lime sludge with the NPS.

Next time don't leap in and waste 250g of bark if you don't know what you're doing 😉 Start with 50g.
 
downwardsfromzero said:
Anyhow, when using lime one should only add enough water to make a doughy paste.
Click to expand...

I had the biggest facepalm moment when I read this... I had forgotten about reading that before.

Since you are a chemical expert... how much washing soda:lime is the requisite ratio? :d
 
how much washing soda:lime is the requisite ratio?
Click to expand...
They react in a 1:1 molar ratio, so the proportions correspond with the MW of each substance, taking into account the purity of each substance (lime can be as little as 75% on occasion) and the degree of hydration of the particular form of the substance (soda crystals are formally the decahydrate but they tend to effloresce [lose water]; anhydrous soda is also available and thus more reliable straight out of a fresh packet but then of course it begins absorbing both water and carbon dioxide from the air. Anhydrous soda is easy to make: simply heat bicarbonate of soda in a clean, stainless steel pan until it stops 'levitating' and no more condensation forms on a cold pan lid held over the powder. Then you will have food-grade, anhydrous sodium carbonate.)

So, first you have to put your equation together:

Lime + soda → chalk + lye [+ water]

Ca(OH)2 + Na2CO3[.xH2O] → CaCO3 + 2NaOH [+ xH2O]

Then one looks up the atomic masses for each element and adds it all up, paying attention the corresponding amounts for each compound. Give it a try and I'll be back to take a look at your result - and remember to show your workings!
 
Ca = 40.078
+ O2 = 31.998
+ H2 = 02.016
66.116

Na2 = 45.980
+ C3 = 36.033
+ O3 = 47.997
130.01

2 lime : 1 soda ?
 
itsame said:
Ca = 40.078
+ O2 = 31.998
+ H2 = 02.016
66.116

Na2 = 45.980
+ C3 = 36.033
+ O3 = 47.997
130.01

2 lime : 1 soda ?
Click to expand...
That's a start, but you have made a couple of errors: how many carbon atoms are there in a carbonate anion? {Hint: it's "CO3", not "(CO)3"}. And take another look at your calcium hydroxide calculation. What have you missed out?

Also, your workings, or the associated logical processes, to get from the MW to the proportions of materials required, are missing.

You can check the values you have obtained by looking up "calcium hydroxide" and "sodium carbonate".

Now have another try!
 
I don't know what I missed with the lime. 66.116

Carbonate adjusted for one carbon: 105.988

From what I understand, we want proportions so that the molar weight to be equal. I think I was figuring this part with atomic weights but figuring out the moles is different?

I'm confident this is wrong, but with those numbers, 1.6 lime : 1 soda

I'm trying to motivate myself to open up the ole bio textbook and look up how moles work again
 
Check your working!
(rounding to whole numbers) 40 + 32 + 2 = [????]

The result will mean that [????] grams (= 1 mole) of Ca(OH)2 will react with ~106 grams (= 1 mole) of Na2CO3.

You have your current idea of the ratios backwards because you are allowing yourself to be confused by the concept of 'moles'.
Another practically identical term would be 'equivalent weights' - perhaps that helps some more?
1 equivalent weight of lime reacts with 1 equivalent weight of soda (to form 2 equivalent weights of lye, fwiw).

I think we're getting close, though.

Once you've got the right MW for Ca(OH)2, you can work out how much Na2CO3 to add to the amount of lime you've used. The weight ratio stays the same:
Thus, you'll need to add {280 × (106 ÷ [????])} grams of sodium carbonate, because we divide the actual amount of lime by the MW of the lime, then multiply by the MW of the soda, to get the amount of soda needed to react with it.

(It's getting late here, making it harder to explain things!)
 
Yeah this is pretty intensive, I love it! haha

I'm also in the middle of moving to another city and blah blah real life stuff. I need to get my head into a place where I can do all this kind of thinking.
 
Be assured, this stuff (at this level, at least) is just simple arithmetic. It should be quite plain to see once you're more settled again.

Good luck with the move!
 
Wait so, am i right in thinking that the ratio is 74u (1 mole):106u (1 mole) ? (at least if these numbers are correct) ?

280 x 106/74 = 280 x 1.432 = 401g of soda?
 
That's about right: 280g of (pure) Ca(OH)2 will react with 401g of (pure, anhydrous) Na2CO3. But - for your bonus question - how much NaOH will be produced?

Note that these figures are idealised, not only because of "impurities" and hydration in the real world reaction, but that both the lime and the soda will continually be absorbing small amounts of CO2 from the air. So you will need to take those factors into account and - maybe just eyeball it? (but remember to wear safety goggles!) :lol:

Actually, I am serious about the goggles.
 
Safety first and thanks for the well wish for my move!

SWIM followed your advice and has had some great results. Thanks are relayed.
 
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