Is it ok to do that step wihout ph papers and just let It sit longer
to be sure?
Well, you can work out the pH of your solution without using pH papers. Its not that hard, but it takes some number-crunching, and might be confusing if you haven't done it before. But its definitely worth crunching through if you can manage it, because knowing the pH of your solution is very important. If its too low (ie, the solution isn't alkaline enough), then you won't be able to convert the DMT in the bark to freebase. Its really important that the pH is as high as possible, because a highly alkaline solution wants to take the DMT's vulnerable protons for itself to make water and release heat, leaving the DMT in its de-protonated or "freebase" form.
Essentially, first you work out the concentration of OH- in your solution. NaOH has a molar mass of roughly 40g/mol, so you can work out the molarity of your solution using that value. Molarity is number of moles of solute per liter of solution. You just have to know how many grams of NaOH you added to how many liters of solution, and using molar mass you can calculate the solution's molarity.
[X]="concentration of X", expressed here in mols/Lsoln
Since there's 1 OH- for every NaOH, its a 1:1 ratio, so [OH-] = [NaOH]
Let's assume for hypothetical purposes that your NaOH solution [OH-] turns out to be 2.9 x 10^-4
M. I just pulled this number out of my ass arbitrarily for example purposes, mind.
pOH = -log[OH-] = -log(2.9x10^-4) = 3.54
Since this is pOH, and we want pH, we need to consider that pH+pOH=14.00
pH + pOH = 14.00
pH = 14.00 - pOH
pH = 14.00 - 3.54 =
10.46
The final answer shows that the solution is basic (pH > 7), which is consistent with an NaOH solution. Always ask yourself, "does this answer sound reasonable?" In this case, it does.
Alternatively, you can use the ion-product constant of water, Kw=[H+][OH-]=1.0x10^14, to calculate [H+], and then you can calculate the pH from the [H+].
Some useful relationships regarding pH:
Kw=[H+][OH-]=1.0x10^-14
pH=-log[H+]
pOH=-log[OH-]
pH+pOH=14
[H+]=10^-pH