Applying water content to potency value on anhydrous basis

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Hello,

Would you be accounting for water content twice if you applied the water content correction factor to a value that has been described on the anhydrous basis already?

For example, I am looking at the USP reference standard Clindamycin Phosphate, and the label indicates "For quantitative applications, determine water content titrimetrically at the time of use, and use a value of 833 ug of clindamycin per mg of material on the anhydrous basis".

If clindamycin already has a value of 833 ug/mg on the anhydrous basis, wouldn't this already account for the water content? Would determining the water content and correcting the 833 ug/mg value using the water content be applying the water correction factor twice?
You mean like 99.9% USP glycerol can contain up to 5% water?

I hope you're not asking for someone to explain USP !!!!
... this is not my field, so be careful with this answer. But no, I think you're misunderstanding what the supplier is saying.
They have given you Clindamycin phosphate, which has a molecular weight, including the phosphate, of 505 g/mole. Clindamycin alone, without the phosphate, is 425 g/mole. If you take 505mg of Clindamycin phosphate and dry it as far as you possibly can, it will assay to the same as 425mg of Clindamycin, which is about 84% of the mass. So, to a reasonable approximation, if you want to express your results in terms of Clindamycin-content, ignoring phosphate, your standard is about 84% pure when totally dry (the "impurity" being the remaining phosphate). Actually they've given you instructions to use 83.3% purity as your baseline, which might be appropriate if there's a very small amount of additional non-volatile contaminant in the sample.
So what they mean is, the dry stuff we gave you is 83.3% Clindamycin. If you've taken the top off and it's got wet, then the percentage is lower still, and it's your job to compensate for that; we are just telling you what to allow for on top of the water.
3 posts Page 1 of 1

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