Quote Originally Posted by albada View Post
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Actually, the kind of metaborate will make a substantial difference, because the number of moles of the anhydrous form (i.e., NaBO2) in 2.2 grams will differ. For 8-mol metaborate, you'll need 3.0 grams. If you have a pH meter, you can mix the ingredients directly into water (including the PG) and see if the pH is around 8.08.

Without a pH meter, you could mix the concentrate using 2.2 grams, and see how long the ascorbic acid takes to dissolve at 75C (assuming 20 ml of PG). If it takes longer than 5 minutes to dissolve most of it, then you probably have the 8-mol s.metaborate, and will need to add another .8 grams of it. Alternatively, you could mix the concentrate using 20 ml of PG and 3.0 grams of metaborate, but at a temperature of 85C for everything. If it crystallizes (white marks on bottom of beaker and then cloudiness), then you know you should start over with 2.2 grams.

Good luck,

Mark Overton
Thanks Mark. About metaborate:

tetrahydrate NaBO2-4H2O
octohydrate Na2B2O4-8H2O

The octohydrate has twice the molecular weight of the tetrahydrate. So 2.2g of it would have half the number of moles. But each mole of it is equivalent to two moles of the tetrahydrate. When dissolved, wouldn't it be the same?

(The situation seems quite different to something like, say, sodium carbonate, where the one mole of Na2CO3 can have zero, one, seven or ten molecules of water. Obviously then the water of crystallization has to be taken into account when weighing out.)