Appreciable depends on which glasses you're looking through..
Acetone's application in washing/purifying cocaine HCl rely not only on cocaine's solubilty in it, but rather than it's most commonly used cuts's are more soluble in it or the fact that they theoretically should be a contaminant(aka not the main substance).
The wash with anhydrous *as cold as practically possible* acetone calls for 10-20ml per GRAM cocaine HCl..
I can't find scources for cocaine HCl' solubilty in acetone. But I can tell you, from experience, that cocaine HCl's solubility in water@room temp is greater than +0,1g/ml - I suppose we can agree it's solubilty in acetone must be lower..
So for theory's sake we put in
- 1g cocaine HCl (80% purity)
- we suppose there is 20% of this sample that isn't cocaine HCl and rely on the, in this case, only contaminant being more so or equally as soluable
- lidocain HCl is the contaminant. It is soluble at ~31g/l [0,031mg pr. ml * 10 = 0,31mg]
- there is no mechanical loss, for the sake of theory
- we wash the 1g powder with 10ml 'the colder, the better' anhydrous acetone
-for the sake of this theoretical math bitch, mg & ml is used interchangeably
- Yield = 0,8g = 800mg
1000 - 800 = 200mg potential contaminant
10ml * 0,31mg/ml = 0,31mg lidocaine HCl
In this case the, remaining 169mg dissolved in acetone, would then be
Our finished product is now ~99,99% pure, according to this theory
Input-weight - 1000mg
- 200mg (subtracted, total potential impurity)
= 800mg yield
- 31mg Lidocaine HCl
-169mg Cocaine HCl loss to solvent
= 169mg dissolved cocaine HCl in 10ml acetone = ~0,169mg/ml = 0,0169mg * 10. [meaning cocaine HCl ending up being more than 3 times more soluble, which in reality probably is at least closer to equal to- or opposite from in solubilty between the 2 salts in acetone compared to with water.
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