Let's not forget the original question:
How much alkalinity is added by the addition of 1 mg CaCO3 per liter?
I say it is 1.2 ppm HCO3- and 1 ppm as CaCO3, but all brewing water spreadsheets I have checked so far disagree with me.
I guess I have to shoot John Palmer an e-mail regarding this.
Kai
Dear Kai,
I thought about this some more, and had to look up the actual definition of alkalinity (it's really a silly term, as a chemist we usually use more logical terms such as molarity, pH, etc). Here is a simple definition:
"Alkalinity can be measured by titrating a sample with a strong acid until all the buffering capacity of the aforementioned ions above the pH of bicarbonate or carbonate is consumed. This point is functionally set to pH 4.5. At this point, all the bases of interest have been protonated to the zero level species, hence they no longer cause alkalinity."
I think this is important because it changes the way we should think about things. Now I've made the assumption that we're dealing with a buffered system in my following calculations, because at the end of the day, your mash will be a buffered system. So this may not be the answer you're looking for, or what Palmer and co. use when calculating water chemistry. Basically, your total alkalinity will be the total numbers of moles of 2*CO3(2-) + HCO3- + OH- multiplied by the molecular weight of CaCO3 (because alkalinity is expressed in units of ppm CaCO3) * 10^3). So I right off the bat ignored the equilibrium between carbonate and bicarbonate because the pKa of that reaction is about 10.5, meaning that at pH 6.5, over 99.99% is in the bicarbonate form. The interesting thing is that pKa of carbonic acid is 6.37, meaning that in a buffered system, the residual alkalinity will change significantly with pH. Then, I considered a buffered system since it's much easier to do calculations because we can deal with a constant pH. Then with simple algebra, you can calculate the concentration of both CO3(2-) (assumed to be zero because it is much less than HCO3-, and negligible in these calculations), HCO3-, and OH-. This sum is then converted from moles to the units of alkalinity. Unfortunately, I don't think I can attach a spreadsheet, though I'd be willing to email it if anyone is interested. I have copied and pasted the cells below, and they are ugly. I also have calculated alkalinity at several pH's for a system when adding 1 g CaCO3 per liter of water below, including contribution from OH- and from CO3-.
pH 7:
Total: 810 ppm CaCO3
CO3-: 810 ppm CaCO3
OH-: 0.01 ppm CaCO3
pH 8:
Total: 977.1 ppm CaCO3
CO3-: 977.2 ppm CaCO3
OH-: 0.1 ppm CaCO3
pH 5.2:
Total: 63 ppm CaCO3
CO3-: 63 ppm CaCO3
OH- : 0.0001 ppm CaCO3
I think that these numbers demonstrate two very interesting points. Firstly, for those using pH 5.2 stabilizer, the residual alkalinity in your water is highly neutralized. Another point of note is that even at pH 8, the total alkalinity is almost completely arising from bicarbonate, and not OH-. Anyhow, it seems to me that calculations of this sort are actually quite complex, and calculating the "alkalinity" contribution of CaCO3 is silly. Instead, we should focus on the concentrations of the individual ions. I apologize for the length of this post. Hope at least one of you enjoys.
-Matt
pKa H2CO3 6.37
Ka = [hco3-][h+]/[h2co3]
pka 6.37
Ka 4.2658E-07
pH 7
10^-pH 0.0000001
1 g / L
moles CaCO3 0.009991308
molarity 0.009991308
moles H2CO3 9.9913e-3 - X
moles H+ 10^pH
moles HCO3- X
product top X*10^pH
cross mult Ka*(9.991e-3)-Ka*X
cross mult 2 10^pH*X
solve x 10^pH*X + Ka*X = Ka*(9.991e-3)
simplify X(10^pH+Ka) = Ka * (9.991e-3)
X X = [Ka * 9.991e-3]/[10^-pH+Ka)
X 0.00809391
alkalinity CO3 moles 0.00809391
alkalinity HCO3- 810.0951586
moles OH 0.0000001
alkalinity OH 0.0100087
total alkalinity 810.1051673