fun with dimensional analysis
hello everyone
i am a bit unsure how to turn the base into hcl, i have read in the old 25c nbome big and dandy that
"Calculate the molarity of your compound in the water and use the equivalent amount of HCl"
puh so i have 30-35% HCl, which doesnt make it to easy for me.
my calculation is:
HCl = 36,4 g/mol
25c-nbome base: 335,8 g/mol (?)
so for 15mg of 25c i should take 150 mg of HCl (100% )
so my HCl is 30% so i take 450 mg of HCl (30% ) for 15 mg of 25c?
then i can add 1.5g (1.5ml) of destiled water and i have a solution of ~ 2ml with 15mg of 25c
so 0.1 ml would be 133 microg of 25c?
then in each side of the nose 0.2 ml and i have 532mic and off we go?
is it that simple and am i right?
what pH will said solution have? 30% Hcl has pH of -1 i think iv read?
with 450mg of 30% HCl in 1.5 g H2O we will have 7.5% of HCl still sounds pretty acidic to me.
anyone can help?
i am sorry for this stupid question, or at least i feel pretty stupid with it

its not 2 years ago i had to calculate this stuff for university... and now, all gone...
thanks for all,
love and light,
lord yupa
heh. Don't get down on yourself dude, dimensional analysis is supposedly a basic tool for chemistry and physics, but I have not seen that it gets the attention is deserves in coursework. It kicked my ass in basic gen chem way back when and I vowed never again to be defeated by that mathematecal approach. Funny thing is, I see it now as more of a conceptual approach to looking at relationships in the world (kinda like algebra), instead of being a "math" approach. Anyhoo...
Given that you want to add a stoichiometric amount of protons to your 25C-NBOMe base, good ole' dimensional analysis will save the day, every day:
This equation provides how much volume of your 30% HCl sol'n you need to form the HCl salt from 15 mg 25C-NBOMe base.
NOTE: It's prolly not suggested to work directly with your 30% HCl sol'n here, and it's definitely not necessary (It might even destroy your 25C material if it's added directly to it, as that is seriously strong acid.
Another complication is that you can't initially dissolve the 25C base in water and then add the acid, because the base won't dissolve in water.
So, it's suggested to first dilute your HCl by 100 Fold (now at 0.3% HCl), so that you would wind up adding 100 times as much of the diluted HCl solution to the 15 mg 25C powder than you would have if you had used the 30% HCl sol'n that is described in the following equation.
Still required of course is that you subsequently dilute the 25C/HCl sol'n further in distilled water to achieve the desired final concentration.
Volume of 30% HCl req'd for stoichiometric protonization of 15 mg 25C-NBOMe (check the units, they all cancel except the desired one- volume). Google did my math so someone might want to double check it. The equation ensures ample protonation by erring on the high side with a little excess of HCL so there is sure to be no base left over.
(0.015 g 25C-NBOMe)*(mol 25C-NBOMe/335.8 g 25C-NBOMe)*(1.1 mol HCl/1 mol 25C-NBOMe)*(36.4 g HCl/mol HCl)*(1 L 30% HCl Sol'n/300 g HCl) = 5.96 microliters 30% HCl sol'n (about 6 microliters 30% HCl sol'n)
tl;dr version => 0.015/335.8*1.1*36.4/300 = 5.96 × 10^-6 L
Clearly, as I alluded to above,
this volume is miniscule and nearly impractical to work with, and also the strength of this reagent might decompose your material. So, dilute your HCl sol'n to 0.3% and add 596 microliters (meh, just use 600 ul or 0.6 ml) of this to the 15 mg solid instead. [NOTE: It might not fully dissolve and will form a slurry, that's ok, you are going to dilute it further, see below]
So, to prepare a final dilution so that it's ready for liquid insufflation,
it's suggested to dilute to 7.5 mg 25C-NBOMe/ml. Thus, in your case, suggested to add an additional 1.4 ml distilled water to the ~0.6 ml of 25C + 0.3% HCl sol'n that had just been prepared to arrive at a final volume of 2 ml and a concentration of 7.5 mg 25C-NBOMe/ml. Some vortexing agitation, mild warming and patience might be required to get it into solution, but this method is tried and true.
Then, for example, a 0.05 ml (50 microliter) droplet provides: (0.05 ml)*(7500 mcg 25C-NBOMe/ml) = 375 mcg
Two 375 mcg droplets (2 x 0.05 ml) will probably be what your looking for, but suggested to start with 1 X 0.05 ml (50 microliters) to provide 375 mcg if never tried before.
NOTE: many people here are enamored with 0.1 ml graduated oral syringes for use with the liquid insufflation route. To follow that, method, you will essentially just double the volume that contains 375 mcg 25C-NBOMe from 0.05 ml to 0.1 ml. Dilute the 2 ml final volume described above to 4 ml and then and then use the techs described by others.