...and the main thing to remember about what a "mole" is, is that it's just the same amount of "particles" (atoms or molecules), no matter WHAT it is that you have a mole OF. A mole of H2 (2 grams--molecular weight, in grams) has the same number of molecules as a mole of chlorine (35.5 grams [right?]). So if you reacted 2 grams of molecular hydrogen with 76 grams of molecular chlorine (Cl2), you would end up with 2 moles of HCL (weighing 78 grams)--and no leftovers.

(simple eq. for finding mols from given mass of compound)

# moles compound = mass of compound x (1 mol/molar mass of compound)
where 1 mol = 6.02 x 10^23 atoms

the molar mass can be calculated from the compound's stoichiometry, but it's easier just to google the compound, molar masses are typically listed.

Thanks everyone. That what sucks about using just a book; you have no one to ask if you don't get something

so if this works, shouldn't this also work with KOH.
DMT-Cl + KOH ---> DMT + KCl + H2O ?

Yes, this is correct. But note that these reactions (with either NaOH or KOH) is only a small part in the process of freebasing. The main part is that enough NaOH (or other base) is addedd to keep the pH high enough and thus the spice in a freebased form.

The above equations are quite simplistic and do not provide a mechanism of how dmt is freebased. They just show how a salt of dmt reacts with a base.

Hi everyone. I noticed on Noman's tek, it says Why? We all know acids neutralize bases, but in SWIM's chemistry classes it has been highly stressed that neutralization is an exothermic process, so pouring vinegar on one's skin to get rid of a lye burn would be highly painful (remember the Fight Club scene). SWIM thinks that rinsing with cold water is the best solution, is that correct?

Why do aromatic solvents pull more stuff than aliphatic ones.

because a lot of alkaloids have aromatic moeities. like dissolves like.
solvent accessibility is greatest near the aromatic structure

Yes. I use phosphoric acid (ph down) as an acid and KOH (caustic potash or ph up) as a base, and it works great.

the amount of heat released is very small for a that kind of volume, so it shouldn't get hot.
pouring vinegar on your skin to get rid of a lye burn would be a lot less painful than the lye burn itself.
if you rinse with cold water, you'll feel the soap that has formed from your natural oils, its best to neturalise.

I have a basic question.

1.) How can I determine what the ph of a solution is if I know what's in it?

For example, if I have an acid mixed with water, and I know the ration of acid/water, I should be able to determine how many moles of water I have and how many moles of acid I have, and from there determine what the ph is, correct?

Is it perhaps more complicated than that?

I'm open to hearing the long explanation if anyone's willing to give it to me. Also, if anyone can suggest some begginer level reading that covers these aspects, I'd be very appreciative.

not quite..

the acid is a protonated aqueous phase (it's all water).
instead, know what the concentrations of the solvent/solute are.
also know the Ka (or pKa) values of the solvent and solute.

dilutions are made by adding acid to water, so volumes are considered.
10-fold dilution is prepared by adding 10 mL acid to 90 mL water.

Yes, and No.

Knowing the moles is the first step to calculating the pH. Knowing the total volume of the water (i.e. water plus the acid or base) is also essential since pH has to do with the concentration (moles per volume). Finally, temperature is also good to know since pH is also temperature dependent but we can generally ignore it for the average kitchen chemistry and the average theoretical calculations.

As long as we know the concentration of the acid or base we can calculate the pH. The calculations are easy for strong acids (Hydrochloric acid and sulfuric acid) and strong bases (sodium and potassium hydroxide) but become fairy difficult for weak acids (acetic acid, fumaric acid, phosphoric acid, citric acid etc) and weak bases (ammonia). The calculations get also complicated with conjugate bases (sodium carbonate, sodium bicarbonate, sodium acetate and so on) as well as with conjugate acids (ammonium chloride, ammonium sulfate, dmt sulfate).

Would you like to go into the calculations? Would you like to choose a real-life example of something to calculate? that would help alot!

hello and thanks,
I would very much like to go into the calculations. My strongest math skills are linear rate equations. Although I used to understand more complicated things such as integration/differentiation/minima/maxima & titration, that was several years ago and i remember very little of it.

This is related to my "Questions on cold water technique" thread (post #2).

I have a very conrete example:
If one wished to create a solution of phosphoric acid and water at a ph of 2.5, how would they go about it without using PH paper?

Assuming that one has the following materials on hand.

1.) Water.
2.) A measuring cup to measure liters.
3.) An eye dropper to measure milliliters.
4.) Phosphoric acid solution at 25%.
5.) A milligram scale.

You need to know the acid's pKa which is 2.15.
then use this equation.


where square brackets means concentration, in moles per litre.

Eeew, you went for a triprotic acid!!!

That is an "extreme" example, because phosphoric acid has 3 pKa numbers. Phosphoric acid dissociates like this to provide protons (or H+, the stuff that makes a solution acidic) in the water:

H3PO4 <-> H2PO4- and H+ :: pKa=2.15 (as shoe said)
H2PO4- <-> HPO4-- and H+ :: pKa=7.2
HPO4-- <-> PO4--- and H+ :: pKa-12.38

Phosphoric acid is a weak acid and it does not fully dissociate in solution. Hence the "<->". If it were a strong acid, like hydrochloric acid I'd write: HCl -> H+ and Cl. The pKa means the pH at which half of the species will be dissociated and half will be non-dissociated, e.g at a pH of 2.15 50% of the acid will be H3PO4 and 50% will be H2PO4-

As one dissolves the acid in water, the acid will be mostly dissociating and the more the pH goes down the less phosphoric acid will dissociate, until it reaches the pH of 2.15, where 50% will be dissociated.

Now, if you want to calculate how much acid to add in water to get a pH of 2.5 it will be easier to consider only the first dissociation reaction: H3PO4 <-> H2PO4- and H+
because the second (H2PO4- <-> HPO4-- and H+) and the third (HPO4-- <-> PO4--- and H+) do not practically happen in this low pH. The next step is to use the formula that shoe provided, by substituting 2.5 for the pH and 2.15 for the pKa. We then solve the equation for the logarithmic ratio:

2.5 = 2.15 + log([H2PO4-]/[H3PO4]) ->
0.35 = log([H2PO4-]/[H3PO4]) ->
(we solve for the logarithm) [H2PO4-]/[H3PO4] = 10^0.35 [[["^" means "to the power"]]]
[H2PO4-]/[H3PO4] = 2.24.

The ration therefore of dissociated to non-dissociated acid is 2.24. This cannot be solved further since we know neither how much [H2PO4-] nor [H3PO4] is. We can however find [H2PO4-].

The concentration of the dissociated [H2PO4-] is exactly the same as the [H+], as shown in the dissociation equation H3PO4 <-> H2PO4- and H+. According the the stoichiometry of the reaction, there are same amount of species [H+] as there are species of [H2PO4-]. At a pH of 2.5 there are -log[H+] of pH, or:

2.5 = -log[H+] ->
-2.5 = log[H+] ->
[H+] = 10^(-2.5) ->
[H+] = 0.0031 moles/litre

or, in other words, [H+] = [H2PO4-] = 0.0031.

We now go back to the [H2PO4-]/[H3PO4] ratio and by substituting we get:

[H2PO4-]/[H3PO4] = 2.24 ->

0.0031 / [H3PO4] = 2.24 ->
[H3PO4] = 0.0013.

So in the dissociation equation we have:

H3PO4 <-> H2PO4- and H+
0.0013 and 0.0031 and 0.0031 or if we convert these into grams /litre taking into account that the molecular weight of H3PO4 is 98, the molecular weight of H2PO4- is 97 and the molecular weight of H+ is 1:

H3PO4 <-> H2PO4- and H+
(0.1274 and 0.3007 and 0.0031) grams per litre,

or a total of 0.4312 grams of phosphoric acid in 1 litre of water. And finally, if one has a 25% solution of phosphoric acid, that means 25 grams phosphoric acid/100ml water therefore if one gets in order for one to get (100*0.4312)/25 = 1.725 ml of a 25% phosphoric acid in a total of 1 litre solution to get a pH of 2.5

I understand that may be far more complicated than the half-assed Henderson-Hasselbach equation given by shoe, so please feel free to ask back if something is obscure or does not make much sense!

This is an invaluable resource, excellent guide to general chemistry:
http://en.wikibooks.org/wiki/General_Chemistry

And for O. Chem (more advanced):
http://en.wikibooks.org/wiki/Organic_Chemistry

Man, I wish I didn't take chemistry last year in school. These basics are the only thing you need to know...

I wasted so much time on worthless stuff like redox reactions, Stochiometry etc...

I like redox reactions, and equilibrium and everything