Hi all, quick question here... Swim is working on updating his limonene tek, and was wondering if anyone knew how much NaOH it takes to neutralize vinegar? Let's say you have a litre of 5% vine gar solution, how much NaOH would it take to neutralize all the acetic acid? In other words how much NaOH is being lost to the neutralization and converted into sodium acetate, before it starts remaining in the solution as NaOH?
Thanks!
SyZ

It would roughly take 35 grams of NaOH to neutralise one litre of 5% acetic acid. This is assuming that 5% acetic acid means 5% by volume, not weight. 100% (or glacial) acetic acid is liquid at temperatures above 16 Celsius.

But please note that acetic acid boils at 118 Celsius. This is to say that:

1) One can actually boil off the acetic acid instead of neutralising with NaOH

2) when making an acid exrtaction by boiling bark with say 5% acetic acid, one does not have a clue how much acetic acid is left in the brew (esp. after reducing the volume by boiling etc etc.) to neutralise

Hope that helps!

why do that? if boiled off until dry, then what's the point of trying to reconstitute it? The dry stuff would have no vinegar to be neutralised.

Cool thanks yeah that helps. I've always wondered when it says on the vinegar bottle that it's been "diluted to 5% acidity" whether that means by volume or by weight...
Basically the reason for this post is that Swim's been trying to solve a long-standing debate between him and some friends about how to go about prepping the bark in the first stage of an extraction.
Swim used to go STB, but he had a friend that swore he got better yeilds by first cooking the bark in a 50/50 vinegar/dH2O soln in a crock pot for several hours.
Swim thinks there may be some merit to this... doing the pre-cook with gar may soften up the cell walls, making them more easily broken down by the NaOH when the base is added. Swim noticed back when he was doing STB and just adding NaOH soln straight to dry bark powder that clumps of bark powder would form and were hard to break up, so there's at least some merit to cooking the powder just to get it all nice and liquidy before basifying. The question in swim's mind now is whether all one should do is maybe cook it in pure dH2O, or whether the gar helps "soften up" the cellulose (and pull out some alkaloids in the process), or whether it's just getting in the way by neutralizing NaOH.
Of course the only real way to tell would be to conduct some controlled experiments. Swim's just having me gather some preliminary data to help him formulate his experimental design. So thanks for your help!
That's a little weird tho that it would take 35g NaOH to neutralize a litre of vinegar, as if memory serves the Mars tek calls for having like a gallon of 50/50 water/gar which is then neutralized with only 50g NaOH and still enough is left over to freebase dimitri... so perhaps "diluted to 5% acidity" refers to weight and not volume? Do you happen to know how much NaOH would be needed to neutralize then if this were the case?
As for what you said Dagger about evaping the gar and reconstituting it, yes, this is what swim does later on when he salts the medicine out of the limo using gar, he evaps it and reconstitutes it in a little dH2O. He uses that as is for oral administration, or if he wants smokable freebase he just adds a little sodium carbonate solution to it and pulls with iso, then evaps the iso and redissolves the alks in dry iso, filters out sodium carbonate residue, and re-evaps the iso onto some powdered herbage... but that's a different story
Thanks everyone for all your help!

working with an A/B has as the biggest advantage in SWIMs opinion the fact that one deals with nice liquid in the basifying step, instead of a messy slur. But in terms of yield, SWIM already did a side-by-side A/B and STB and yields were basically the same. SWIM doesnt remember the amounts but it was pretty funny as it was nearly exactly the same yield. Maybe other people feel differently but thats how it went for SWIM, so he generally goes for the STB way

I am pretty confident that 5% vinegar is by volume, not weight. Pure acetic acid is liquid at temperatures above 16 C and this is how most people tend to use it. Just take an amount of the liquid acetic acid and dissolve it in water. But this should not be really a big point. Even if "5%" actually meant "by weight" in the case of vinegar, it wouldn't make a big difference.

As of Mars tek, the 50g NaOH sounds fine. One should also take into account that vinegar is boiled off during the acid extraction steps, so there is less vinegar to be neutralised when it comes to basifying. But if one gets a gallon 50/50 water/gar and wants to neutralise it (that is no bark, no boiling no nothing, just the water/vinegar) it should take around 70g NaOH to neutralise it to pH 7 .

These are my calculations for people who wish to double check me:

5% vinegar means 50ml of acetic acid in 1 litre of water.

50ml acetic acid weights 50 * 1.049 = 52.45 grams. (1.049 is the density of vinegar)

52.45grams of acetic acid is 52.45/60.05 = 0.873 moles.(60.05 is the molecular weight of acetic acid)

And 0.873 moles of acetiic acid require 0.873 moles of NaOH to be fully neutralised.

Finally, 0.873 moles of NaOH corresponds to 0.873 * 40 = 34.93g of NaOH.