J.R. S. answered • 04/07/18
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I'm assuming that PG is propylene glycol, and the VG is vegetable glycerine and MCT oil is medium chain triglyceride oil. That being the case, I believe that PG is a better solvent for most lipid soluble material than is MCT. However, that's not your question. You mention you have the log P, but is that the log of the Ksp or some other P. If it's log Ksp, it's easy to take the antilog and obtain the Ksp. Then you can find the solubility in mole/liter and hence in grams/ml or any other units of concentration you desire.
J.R. S.
tutor
Haddy,
From the molecular weight alone, you cannot determine much about solubility. I believe the log P to which you refer is probably related to the partition coefficient of the substance you are trying to dissolve (you didn't mention what it is). Also you don't say what the two solvents are for the partition coefficient. Log P = log [solute]solvent1/[solute]solvent2.
Usually, solvent 1 is something like octanol and solvent 2 is water. I'm not sure what you log P describes. However, if the value is 6 (or 6.1), it is 1 million times more soluble in one solvent1 than in solvent2. That may not help, but without more info, that's about all we can conclude at this time. If you know the solubility of the compound in water, you can assume it will be about 1 million times more soluble in one of these other lipophilic solvents. If you have additional info, send it to me and maybe we can figure this out.
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04/09/18
Haddy Y.
Thank you again for the detailed response. I very much appreciate your guidance in this question. Here is the information I have available:
PROPERTY VALUE SOURCE
Water Solubility 0.0126 mg/mL
logP 6.1
logP 6.33
logS -4.4
pKa (Strongest Acidic) 9.13
pKa (Strongest Basic) -5.7
Physiological Charge 0
Hydrogen Acceptor Count 2
Hydrogen Donor Count 2
Polar Surface Area 40.46 Å2
Rotatable Bond Count 6
Refractivity 98.53 m3·mol-1
Polarizability 38.35 Å3
Number of Rings 2
Bioavailability 1
Rule of Five No
Ghose Filter No
Veber's Rule No
MDDR-like Rule No
Water Solubility 0.0126 mg/mL
logP 6.1
logP 6.33
logS -4.4
pKa (Strongest Acidic) 9.13
pKa (Strongest Basic) -5.7
Physiological Charge 0
Hydrogen Acceptor Count 2
Hydrogen Donor Count 2
Polar Surface Area 40.46 Å2
Rotatable Bond Count 6
Refractivity 98.53 m3·mol-1
Polarizability 38.35 Å3
Number of Rings 2
Bioavailability 1
Rule of Five No
Ghose Filter No
Veber's Rule No
MDDR-like Rule No
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04/09/18
J.R. S.
tutor
So, what can we tell from these data? First, your compound (I wonder what it is) is clearly not very soluble in water (0.0126 mg/ml). With a log P of 6.1 that means that it is 1x106.1 x more soluble in the organic (lipophilic solvent) that they used to determine the partition coefficient. Your molar solubility in water will be 0.0126 mg/ml x 1mmol/314.4 mg = 0.04 mM concentration. Unfortunately, I don't believe there is a specific equation that will give you the solubility concentration in MCT, PG, VG or any combination of these. Your best "guestimate" will be that it will probably be several hundred thousand times more than that of water, i.e. it may be in the range of 4 M. That seems high, but not surprising if the material is quite lipophilic. From the 2 hydrogen acceptor and 2 hydrogen donor counts, this appears to be an amphiprotic compound and would be more soluble in an acid or an alkaline solution than in water, but still probably not as soluble as it will be in one of your solvents. I don't know what additional light I can shed on this problem, other than to suggest you find the solubility concentration empirically. Good luck.
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04/09/18
Haddy Y.
For some reason, this page is not updating for me. I received a notification saying that you responded again but it won't pull up.
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04/10/18
Haddy Y.
Thank you so much J.R.S. I very much appreciate your patience and thorough response. Time to break out the old chem book and fill in the missing gaps of knowledge!
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04/10/18
Haddy Y.
04/09/18