TheLastHairbender
Established Member
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So I know there has been a good deal of talk about using AHK-Cu lately, and a number of people seem to be happily beta testing various combinations of the pure stuff from last autumn. I'm one of those - sitting on 28g right now but have not yet taken the initiative to work out the recipe for optimal application.
The best treatment of the topic I've seen to date are: mix a few grams into a bottle of Tricomin or ethanol/PG/H20. I'm a little hesitant to start experimenting based on such imprecise chemistry. I wanted to take a look at the original study and determine exactly how many grams need to be mixed into an amount of the vehicle appropriate for daily application. Most advice on the topic seems to center around mixing a 2.5% - 5% w/v solution of AHK-Cu, but without any mention of the volume you should be applying. I think the right approach is to determine how many moles of AHK-Cu you want to deliver and from there determine the amount (in weight) of AHK-Cu required to provide those moles, from which you can solve for the strength of solution you'd need for a given volume of solution you plan to apply.
Looking at the Korean study, they treated cells with solutions of varying concentrations, from 10^-12 M to 10^-7 M. They found positive results for concentrations from 10^-12 M to 10^-9 M, and negative results for concentrations of 10^-8 M and 10^-7 M (not sure how tenuous the differences between those amounts are...factors of ten for each increment to the exponent, right? So those are pretty discrete intervals if I understand correctly). The best results were obtained for a concentration of 10^-9 M.
I plan to apply 2 mL of solution daily and want to determine the number of grams of a compound I need to add to achieve the desired quantity in moles. The molecular weight of the compound appears to be 416.94u.
I thought I just needed to multiply the number of moles desired, 1.0 x 10^-9, by the molar mass, 416.94 g/mol, to determine the mass, in grams, required to deliver said moles. This works out to a tiny amount... 4.41694 x 10^-7 (.00000041694) grams per day. This seems way too small. Maybe I actually want to deliver more moles than that...? The study was on units of 30 follicular dermal papilla cells...so maybe I need to multiply the number of moles they delivered to their sample by the number of samples I'd expect to have on my scalp…about ~30,000 samples? I'm out of my league here and want to make sure I do this properly if I do it at all...unfortunately the lack of precedent in this area and the one study available (in which the treatment process doesn't seem to be well described) doesn't give me much confidence that I'm doing this right. There's actually a good chance I'll end up saying this isn't a good idea, but I wanted to give it a go to see if what I'm after can be scientifically determined with relative ease.
People who have attacked this less scientifically report adding approximately 2g to 60 mL (57g) of solution, which yields a 3.5% w/v concentration, and if 2 mL are applied daily yields a mass of active ingredient of .067 grams applied daily. Way higher than the 4.1694 x 10^-7 grams per day I solved for based on the number of moles I wanted to deliver. Which is the right way to think about this? I think the problem is that I don't exactly know how much I need to deliver...I would think the study should reveal that information, in g/kg or some equivalent that might be useful. Here is a link to to the study in question, maybe that quantity jumps out at someone more than it does to me:*http://www.dermoday.com/dosyalar/1277308389.pdf
If anyone knows their stoichiometry please advise. If you're just going to say make a 2.5% w/v solution and not tell me how much volume that assumes you're applying then I'm going to know you don't know what you're talking about, as 1 mL of a 2.5% solution is a lot different than 4 mL of 2.5% solution. I'd really prefer to hear from someone who understands the molarity calculations, if such a person exists on HairLossTalk.com. I've asked a few other people who might be in the know from a chemistry perspective; whether they're interested in staking my health on their back-of-the envelope calculations I'm not sure yet, but I'll post back whatever I hear. I appreciate anyone's input in the meantime.
The best treatment of the topic I've seen to date are: mix a few grams into a bottle of Tricomin or ethanol/PG/H20. I'm a little hesitant to start experimenting based on such imprecise chemistry. I wanted to take a look at the original study and determine exactly how many grams need to be mixed into an amount of the vehicle appropriate for daily application. Most advice on the topic seems to center around mixing a 2.5% - 5% w/v solution of AHK-Cu, but without any mention of the volume you should be applying. I think the right approach is to determine how many moles of AHK-Cu you want to deliver and from there determine the amount (in weight) of AHK-Cu required to provide those moles, from which you can solve for the strength of solution you'd need for a given volume of solution you plan to apply.
Looking at the Korean study, they treated cells with solutions of varying concentrations, from 10^-12 M to 10^-7 M. They found positive results for concentrations from 10^-12 M to 10^-9 M, and negative results for concentrations of 10^-8 M and 10^-7 M (not sure how tenuous the differences between those amounts are...factors of ten for each increment to the exponent, right? So those are pretty discrete intervals if I understand correctly). The best results were obtained for a concentration of 10^-9 M.
I plan to apply 2 mL of solution daily and want to determine the number of grams of a compound I need to add to achieve the desired quantity in moles. The molecular weight of the compound appears to be 416.94u.
I thought I just needed to multiply the number of moles desired, 1.0 x 10^-9, by the molar mass, 416.94 g/mol, to determine the mass, in grams, required to deliver said moles. This works out to a tiny amount... 4.41694 x 10^-7 (.00000041694) grams per day. This seems way too small. Maybe I actually want to deliver more moles than that...? The study was on units of 30 follicular dermal papilla cells...so maybe I need to multiply the number of moles they delivered to their sample by the number of samples I'd expect to have on my scalp…about ~30,000 samples? I'm out of my league here and want to make sure I do this properly if I do it at all...unfortunately the lack of precedent in this area and the one study available (in which the treatment process doesn't seem to be well described) doesn't give me much confidence that I'm doing this right. There's actually a good chance I'll end up saying this isn't a good idea, but I wanted to give it a go to see if what I'm after can be scientifically determined with relative ease.
People who have attacked this less scientifically report adding approximately 2g to 60 mL (57g) of solution, which yields a 3.5% w/v concentration, and if 2 mL are applied daily yields a mass of active ingredient of .067 grams applied daily. Way higher than the 4.1694 x 10^-7 grams per day I solved for based on the number of moles I wanted to deliver. Which is the right way to think about this? I think the problem is that I don't exactly know how much I need to deliver...I would think the study should reveal that information, in g/kg or some equivalent that might be useful. Here is a link to to the study in question, maybe that quantity jumps out at someone more than it does to me:*http://www.dermoday.com/dosyalar/1277308389.pdf
If anyone knows their stoichiometry please advise. If you're just going to say make a 2.5% w/v solution and not tell me how much volume that assumes you're applying then I'm going to know you don't know what you're talking about, as 1 mL of a 2.5% solution is a lot different than 4 mL of 2.5% solution. I'd really prefer to hear from someone who understands the molarity calculations, if such a person exists on HairLossTalk.com. I've asked a few other people who might be in the know from a chemistry perspective; whether they're interested in staking my health on their back-of-the envelope calculations I'm not sure yet, but I'll post back whatever I hear. I appreciate anyone's input in the meantime.