How much alkalinity does 1 ppm of CaCO3 (Chalk) really add?

A few weeks back I decided to write another brewing water calculation spread sheet. The formulas were mostly taken from the literature and existing spread sheets. Then I decided to add a cation (positively charged ions) to anion (negatively charged ions) balance check just to see if the water profile that I created made sense. This is when I noticed an imbalance when creating brewing water from scratch by using distilled water and salts. The resulting water should not show an imbalance and every cation should have matching anion. But it was showing an imbalance when chalk was used. So I gave the fomulas used for chalk a closer look.

 And found that 1 mol (a unit that is proportional to the amount of molecules/ions of a particular substance) of CaCO3 was assumed to add one mol of bicarbonate to the water. And that in most spreadsheets and calculators the bicarbonate contribution was later used to calculate the alkalinity as CaCO3. But that didn't seem right. If CaCO3 adds only one bicarbonate, it also needs to add one hydroxyl ion (OH-):

(1)  CaCO3 + H20 -> Ca2+ + HCO3- + OH-

Since this would liberate hydroxyl the pH of the water would need to rise. If that is not happening then chalk can also be dissolved in the presence of CO2

(2)   CaCO3 + H2O + CO2 -> Ca2+ + HCO3- + HCO3-

In this case each mol of chalk would add 2 moles of bicarbonate. Yet another reaction is possible in the presence of acid and free protons

(3)  CaCO3 + H+ -> Ca2+ + HCO-

(4)  HCO- + H+ -> H2O + CO2

If neither of these reactions hapen the chalk won't dissolve. And that is clearly happening in brewing: If you add chalk to the brewing water it just turns the water cloudy and it will eventually settle. 

But does it really matter if the chalk dissolves or not? No. Because the bigger picture is that we added the chalk to give the water+chalk mixture more "alkalinity" I.e. acid buffering capacity. That acid buffering capacity is needed to reach a targeted mash pH once the malt, and with it acid buffers, has been added. At that point reactions (3) and (4) can take place. Whichever reaction is happening (1)..(4), chalk can neutralize 2 equivalents of acid and for all intents and purposes 1 ppm of chalk should therefore raise the alkalinity by 1 ppm as CaCO3. 

But that is not what most water treatment spreadsheets assume. They assume that 1 mmol/l CaCO3 adds 1 mmol/l HCO3- (bicarbonate) which drops one negative charge on the floor and caused the imbalance that I noticed. And then they go ahead and convert the ppm HCO3- to alkalinity as ppm CaCO3 by multiplying with the factor 50/60. In the end the addition of 1 ppm CaCO3 raises the alkalinity by only 0.5 ppm as CaCO3. This certainly seems wrong and I thought I had it all figured out until I decided to confirm this theory with an experiment.

The experiment is seemingly simple. Make small mashes with 3 different waters that are supposed to have the same residual alkalinity and test their pH. The first water (A) would be reverse osmosis water and serve as the control. The second water (B) would be reverse osmosis water with chalk and calcium chloride added such that the added residual alkalinity is 0 if the chalk contributes 2 alkalinity equivalents. The 3rd water (C) has chalk and calcium chloride added such that the added residual alkalinity is 0 if chalk contributes only one alkalinity equivalent. Whichever water that causes a mash pH to match the RO water mash pH the closest would have used the correct formula for alkalinity contributions by chalk. Here is a summary of the waters used:

  • water A: reverse osmosis tap water
  • water B: RO water + 80 ppm CaCO3 + 290 ppm CaCl2*2H2O; this increases the Ca2+ content by ~110 ppm
    • if 1ppm CaCO3 adds 1 ppm alkalinity as CaCO3 then the water's residual alkalinity (RA) increases by 0.0 over the RO water's RA
    • if 1 ppm CaCO3 adds 0.5 ppm alkalinity as CaCO3 then the water's RA decreases by 2.2 dH (German Hardness) or 40 ppm as CaCO3
  • water C: RO water + 150 ppm CaCO3 + 150 ppm CaCl2*2H2O; this increases the Ca2+ content by ~110 ppm
    • if 1ppm CaCO3 adds 1 ppm alkalinity as CaCO3 then the water's residual alkalinity (RA) decreases by ~4.4 dH or 80 ppm as CaCO3
    • if 1 ppm CaCO3 adds 0.5 ppm alkalinity as CaCO3 then the water's RA remains unchanged compared to the RO water

200ml of each water were taken and heated to ~64C in the microwave. Then 50g of crushed pilsner malt were added to each water sample and stirred in. The mashes were occasionally stirred and a 15ml sample was taken from each mash after 5 min and cooled to 22C when it was measured with a pH meter. The results were surprising:

  • mash A : pH = 5.76
  • mash B : pH = 5.69
  • mash C : pH = 5.77

According to these results the chalk added only 0.5 ppm alkalinity as CaCO3. And the pH shift for mash B is even in the range that would have been expected from the 2.2 dH RA drop. According to Kolbach the shift is 0.03 pH units for each dH which would be 0.066 and the results show ~0.07.

I couldn't believe it and started to ponder why that would be the case. Why is the added CaCO3 only neutralizing 1 equivalent of acid and not 2? Maybe it has something to do with the chalk not being dissolved.

So I conducted another similar experiment. This time between a control, water with suspended chalk and water with dissolved chalk. The chalk would be dissolved with CO2 which is brought into solution through shaking. Here is what I did. I added 0.24 g chalk and 0.88g calcium chloride to 1.5 l of reverse osmosis water. This is twice the salts added to water B in the previous experiment because I wanted to pronounce the effect of the residual alkalinity difference. I then shook this water and the added salts in a 2l soda bottle until the calcium chloride was dissolved. Immediately after shaking, without giving the chalk a chance to settle, I poured off 200ml for sample B. I then removed another 300ml in order to increase the head space. This headspace was then filled with CO2 and the bottle closed. When I started shaking the bottle, it immediately contracted which was a sign of the CO2 going into solution. After some shaking I let the bottle sit until the water became crystal clear again. This was not the result of the chalk settling but it being dissolved in the water. I then took 200ml of that water for samle C:

  • water A: reverse osmosis
  • water B: RO + 160 ppm CaCO3 + 580 ppm CaCl2*2H2O
    • RA = -4.4 dH or 80 ppm alkalinity as CaCO3 if chalk adds 1 alkalinity equivalent
    • RA = 0 dH or 0 ppm alkalinity as CaCO3 if chalk adds 2 alkalinity equivalents
  • water C: water B + CO2
    • RA = -4.4 dH or 80 ppm alkalinity as CaCO3 if chalk adds 1 alkalinity equivalent
    • RA = 0 dH or 0 ppm alkalinity as CaCO3 if chalk adds 2 alkalinity equivalents

I then heated both samples to 68C, added 50g crushed pilsner malt to each and rested (with occasional stirring) them for 10 min. After that I took 15 ml samples and cooled them to 20-21C:

  • mash A : pH = 5.67
  • mash B : pH = 5.47
  • mash C : pH = 5.66

So it appears that dissolving the chalk in the mash water changes its alkalinity potential. undissolved chalk has less alkalinity potential than dissolved chalk since mash B showed a much lower mash pH which could only have been the result of a lower RA than the 2 other mashes.

But why is this? Does not all the chalk dissolve in the mash as commonly assumed? And if yes why is that? And would it always be 50%? Shouldn't there be enough acid for this to happen via reactions (3) and (4)?

For now I don't have an answer to this.