Thanks, Jim. I definitely welcome critique.
Yes, there are at least two effects that I didn't take into account in that article, and I plan to correct them in a future version where I also add in potassium and possibly other ions. One effect is, as you mention, that the demand for calcium is not as high as is added because magnesium (and to a smaller extent, strontium) takes the place of some calcium ions in the calcium carbonate in skeletons and in precipitates. The article used a ratio as is present in limewater/kalkwasser, and so tends to have a bit too much calcium.
The second effect is that I did not adjust the calcium level for required downward salinity adjustments. That tends to increase the demand for calcium and so the recipe would need more for this reason.
These two things tend to offset, but not completely, and there's a bit of excess calcium left over. If a user used the recipe in exactly equal amounts, maintained alkalinity at a fixed level, and assuming calcification was the only user of calcium and alkalinity, then calcium will rise over time.
So how much would it rise? Less than using limewater/kalkwasser would make it rise because of the salinity correction, but we can calculate how much it is.
Recipe one is 594 grams baking soda (7.07 moles) and 500 grams of calcium chloride dihydrate (3.40 moles). Dow says the calcium chloride dihydrate is typically 77-80% calcium chloride, which is not exactly the dihydrate, but most folks seem to be using other materials these days, so let's just use the dihydrate exactly.
So there is a 3.8% (0.27 mole) excess of alkalinity without taking any magnesium into account.
How much magnesium gets used is a guessing game since it depends a lot on the corals and coralline algae present. But in the article I estimated 9 grams of magnesium to go along with the above calcium and alkalinity amounts, so let's stick with that. 9 grams of magnesium is 0.37 moles of magnesium.
If you incorporate that 0.37 moles of magnesium, you must leave out 0.37 moles of calcium. So the total (calcium + magnesium) incorporated is still 3.4 moles, and now 0.37 moles of calcium is left over.
That 0.37 moles of left over calcium can be partly used by the excess alkalinity in the recipe (the excess alk came by switching from Dowflake to a true dihydrate, which may or may not actually happen, but it is an approximation). So 0.37 moles excess - 0.27/2 moles taken up by alkalinity = 0.24 moles = 9.4 grams of calcium left over that cannot be accounted for in the calcification.
Now, however, we need to adjust to salinity changes to correct back to the starting salinity. I calculated in the article that 4 gallons leads to a 29% change in salinity in a 50 gallon aquarium, or 29/4 = 7.25% per gallon added.
If you are at 420 mg/L calcium, a 7.25% correction is about 30.5 mg/L of calcium reduction, which in 50 gallons (=50 gallons x 3.785 = 189 L) gives 30.5 mg/L x 189 L --> 5,760 mg.
So the true excess calcium left over after correcting for salinity changes is 9 grams minus 5.8 grams = 3.2 grams per 50 gallons.
That amounts to 3,200 mg/189 liters = 17 mg/L calcium. Not an insigificant amount.
There will be some other minor tweaks to the final value which I've not calculated yet, but to roughly correct for both these effects, the original 594 grams of calcium chloride dihydrate ought to be dropped to about 582 grams.
Hopefully I did that right.
