Hi!
I have just started a similar, but much simpler thing: instead of adding vinegar to kalkwasser or calcium reactor, I decided not to use any reactor, but simply start dosing Calcium acetate on a daily basis. I have been thinking about this for quite some time, and then decided to do some calculations a few days ago.
My calculation was simple. I have a 250l (about 65g) mixed reef tank which is heavily stocked by fish, so I have to dose vodka: about 5ml. daily. Calcium supplementation was using the Balling method, about 50ml. of each solution daily.
So, the Balling liquid #1 contains 55.5 of Calcium chloride, which means, 50ml of the solution will contain about 2.78g of calcium chloride, or about 0.76g of calcium.
Vodka is about 40% alcohol, in 5ml. there is 2ml of spirit, i.e. about 1.58g., that is, 0.82g of carbon.
Now, consider Calcium acetate monohydrate (CH3COO)2Ca*H2O, molar weight 176.18g. It consists of Calcium ion and acetate. However, when dissolved in tank water in very dilute form, acetate, or any organic matter, will be eaten up by bacteria. The corals too will consume it, to make Calcium carbonate. So, for simplicity of calculations, we can write Calcium acetate up in a simple form as CaH6C4O4. If we "subtract" from this CaCO3 that was consumed by corals, what remains is some non-existent C3H6O. What is of interest to us here, that in the remaining compound there are three atoms of carbon.
No, Calcium acetate monohydrate has 22.75% of Calcium. The remaining 3 atoms of carbon comprise 20.45% of molecule's weight. That is, if we take 3.4 g of the stuff, we'll yield 0.77g of calcium, and 0.7g of "free" carbon available to bacteria (to replace vodka).
So, for my setup, Calcium acetate can be a perfect substitute to replace both balling and vodka dosing: I only need to add 3.4g. of it daily. Calcium acetate powder is available through chemical shops, however in my city they did not have it. Instead, I purchased Calcium oxide and 70% acetic acid. I mixed up 100g of the acid with equal amount of water, then added about 40g. of Calcium oxide, stirring gently. Calcium oxide dissolved completely (the stuff becomes very hot), then I added a small excess of it, to neutralize the vinegar completely, waited until the excess Calcium hydroxide sediment settled, gently poured the resultant pink liquid into a different bottle, adding water to 300ml. Thus I obtained 300ml of liquid containing about 110g of Calcium acetate, i.e. 3.6g per 10ml. Thus, for my setup, I can dose just 10ml. of this liquid daily, replacing 50ml of each of the three balling liquids, and the 5ml. of vodka. I think, it will be much simpler to dose just one simple solution instead of four!
I decided to start slowly, and dosed just 3ml. on the first day, and 6ml. on the second. I will start dosing the full 10ml. dose tomorrow. I have not seen any negative effects so far.
Now, I think that I am using a significantly higher dose of carbon than an average aquarist. That is because, as I said, I have too many fish in my system. The ratio of Calcium to "excessive" Carbon in Calcium acetate is 1.11, and if someone doesn't want top add as much carbon to the system (which can result in bacterial boom, bacterial slime threads, cyano, and a general drop of Redox potential in the tank), a chemist friend at a Belorussian aquarium forum suggested that Calcium formiate Ca(HCOO)2
is a better substitute to Calcium acetate. The ratio of calcium to "excessive" carbon is 3.34, so 2.5g of Calcium formiate solution will add the same 0.77g of calcium to the system, but carbon addition will be a modest 0.23g - an equivalent of the amount contained in just 1.4ml. of vodka.
While Calcium acetate solubility is up to 400g/l, Calcium formiate will only dissolve 160g/l, but still the figure is quite good for practical usage. I will try to get some formic acid next week to make some Calcium formiate for my experiments. The stuff should also be available through major chemical stores and I heat that it must be very cheap - at least, in some countries they use tons of Calcium formiate to fight ice formation on the roads...
P.S. I am not a chemist, and I last studied the subject in my primary school. Chemistry guys, please point me if there are any flaws in my logic or calculations.