Another question for those who know chemistry:

How much citric acid is needed to chelate calcium and magnesium in hard water?
My analysis below says 1 g/L. But is this analysis correct?

Water is consided "very hard" if it contains over 180 ppm of metal ions. Let's assume we have 200 ppm. That's .2 g/L. Let's assume it's all calcium. The atomic weight of calcium is 40 g/mole. So have .2/40 moles/L of calcium in the water. Citric acid has a molecular weight of 192 g/mole. And assume each molecule of citric acid can chelate one atom of calcium. That means we need (.2/40)*192 = 0.96 g/L of citric acid to chelate that calcium.

Magnesium is lighter, weighing 24 g/mole, so we'd need (.2/24)*192 = 1.6 g/L of citric acid to chelate solely magnesium. But I'll assume that hard water contains mostly calcium (is this true?), so closer to 1 g/L of citric acid is sufficient.

Or did I goof?

BTW, I'm ignoring the problem of the Fenton reaction with iron which slowly destroys the ascorbate. That reaction occurs over days or weeks. My immediate goal is to prevent the developer from becoming cloudy after 45 minutes due to calcium and magnesium precipitating out of hard water. Also, to maintain pH, the citric acid will need to be countered by considerably more sodium metaborate, which will improve buffering.

Mark Overton