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Old 01-21-2009, 08:33 PM   #1
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Default alkalinity as CaCO3

Why does the addition of 1 g of CaCO3 to 1 liter of distilled water not cause an alkalinity of 1000 mg/l as CaCO3? If I plug this into a water spread sheet (like Palmer’s) I get an alkalinity of 490.

I always assumed that 1 mg/l alkaliniy as CaCO3 is the same level of alkalinity that 1 mg/l CaCO3 contributes to the water.

Kai
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Old 01-21-2009, 08:55 PM   #2
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Originally Posted by Kaiser View Post
Why does the addition of 1 g of CaCO3 to 1 liter of distilled water not cause an alkalinity of 1000 mg/l as CaCO3? If I plug this into a water spread sheet (like Palmer’s) I get an alkalinity of 490.

I always assumed that 1 mg/l alkaliniy as CaCO3 is the same level of alkalinity that 1 mg/l CaCO3 contributes to the water.

Kai
Seems to me like 1 g of CaCO3 to 1 liter of water would produce 1000 mg/L as CaCO3. Could Palmer be taking into account the fact that calcium carbonate isn't very soluble in pure water? I believe the solubility is only 40-50 mg/L.
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Old 01-21-2009, 09:13 PM   #3
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Palmer and others, don't take solubility into account as the salts will eventually dissolve in the mash.

1 g to 1 l is a little much. But the same problem exists with 10 mg to 1l which should give an alkalinity of 10 ppm CaCO3.

Kai

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Old 01-21-2009, 09:27 PM   #4
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Quote:
Originally Posted by Kaiser View Post
Why does the addition of 1 g of CaCO3 to 1 liter of distilled water not cause an alkalinity of 1000 mg/l as CaCO3? If I plug this into a water spread sheet (like Palmer’s) I get an alkalinity of 490.

I always assumed that 1 mg/l alkaliniy as CaCO3 is the same level of alkalinity that 1 mg/l CaCO3 contributes to the water.

Kai
Dear Kai,

It's been a little while since I was a TA for gen chem, but basically this is because of the various equilibria involving carbonate. For example, the 490 figure that you got from the spreadsheet will actually depend upon the pH of the water, and the pressure of CO2 in the surrounding atmosphere. This is because of the following equilibria:

CaCO3(s) <---> Ca(2+)(aq) + CO3(2-)(aq)

H2O + CO3(2-) <----> HCO3(-) + OH(-)

H2O + HCO3(-) <----> H2CO3 + OH(-)

H2CO3 <---> CO2 + H2O

So the second equilbrium, assuming a moderate pH, will lean almost completely to the right, so you'll really have bicarbonate in solution, and not as much carbonate. From here, the bicarbonate can be protonated and turned into carbonic acid, which will quickly dissociate into water and CO2. There is then an equilibrium between CO2 in solution and CO2 in the atmosphere. If the pressure of CO2 in the surroundings is high, it won't escape from the solution (though this is not a factor in practical brewing practice). Thus, as the pH is dropped, more CaCO3 can dissolve as the equilibrium is pushed towards bicarbonate and carbonic acid. At the same time, you're going to be losing some of the carbonate to CO2. What this means is that even if you fully dissolve 1g of CaCO3, you will not get the expected 1000 alkalinity, though you WILL achieve the expected 400 ppm of calcium if added 1g is added to a liter of water. If you really want to convince yourself, you can take the pKa's of the three acid base equilbrium, and calculate the residual alklanity based on neutral water.
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Old 01-21-2009, 11:52 PM   #5
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mkade, I have been thinking about this as well, but none of these calculators seem to use the pH for determining the equilibrium of carbonate and bicarbonate. They all seem to work on a simple calculation of the amount of ions that are added. So I did some more calculating and found this:

CaCO3 + H2O -> Ca++ + HCO3- + HO+

Now if you determine the bicarbonate content that has been contributed you find that 1 ppm CaCO3 will yield 0.61 ppm HCO3-. CaCO3's molecular weight is 100 and HCO3's is 61.

Now there is a formula that converts alkalinity as HCO3- to alkalinity as CaCO3. This assumes that HCO3- has one alkalinity equivalent and that CaCO3 has 2 alkalinity equivalents. As a result you mulyiply [HCO3-] with 50 and divide by 61. The result is 0.5ppm alkalinity as CaCO3.

But there is a problem. The actual alkalinity contributed by CaCO3 was not one equivalent but 2. The formula that calculates the CHO3 concentration forgot about the HO- that was also created for each dissolved CaCO3 molecule. According to Wikipedia (and I tend to believe the science stuff there), alkalinity is defined as the sum of all the possible H+ receptors balanced by the H+ ions:

[A] = 2[CO3--] + [HCO3-] + [HO-] - [H+]

So the alkalinity as HCO3 for the 1ppm CaCO3 addition should have been 2*0.6 ppm HCO3. One for the HCO3- and one for the HO-. Then the alkalinity as CaCO3 is also 1 ppm.

But nobody else seems to see it like that. Either everybody copied from the same source or I'm on the wrong track here. The latter is quite possible.

mkade, the equations that you listed change the type of the ions in the water but they don't change the alkalinity of the water. I.e. its ability to resist a pH change by absorbing newly added H+ ions as these H+ ions can be absorbed by CO3--, HCO3- or HO-.

Kai

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Old 01-22-2009, 12:01 AM   #6
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sorry to go off topic. This is why I love brewing. Otherwise dull and boring science stuff becomes of the utmost importance when faced with the greatest quality in a brew. I was never that interested in chemistry class. Although I can't really contribute, I will be following this thread closely!

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Old 01-22-2009, 12:11 AM   #7
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I will not explain this too much because I have to finish working up a patient and frankly I am tired of using my brain today...

You are trying to relate alkalinity as a linear function of concentration, when in fact it is logarithmic.

Henderson-Hasselbalch equation goes something like pH= pKa + log ([A-]/[HA]) By adding CaCO3 not only are you losing some to CO2 as mentioned above (leChataliers principal if I am not mistaken...) but the hydrogen ion concentration is only being decreased (alkalizing) according to the disassociation constants and the balance achieved in your water.

I have the feeling I just rambled a lot without making any sense whatsoever, so I am going to go back to reading about protein losing enteropathies before I confuse myself or anyone else....

Edit: I don't know if it was a typo, but it looked like you you were using a monoprotic representation for carbonic acid (HCO3) when it is realy polyprotic (H2CO3)...

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Old 01-22-2009, 12:29 AM   #8
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Is it necessary to add calcium carbonate to distilled water when one is using other minerals. For example, I use distilled and/or RO, and plan to use gypsum, epsom salts, and calcium chloride to arrive at a base mineral profile, then supplement with buffer 5.2 to make sure my pH is on target. In this instance, is CaCO3 a desirable, or even necessary addition?

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Old 01-22-2009, 12:32 AM   #9
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This is over my head but I'm reading and trying to learn. I dunno if this article might help but it talks a bit about the carbonic acid/bicarbonate/carbonate relationship.

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Old 01-22-2009, 12:58 AM   #10
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Oh wow....and so many years ago when I was taking chemistry survey classes......balancing equations wondering when am I ever going to have to figure moles per solution.....does an example come up years later.

Since it has been so many years that I've studied molecules, I can't be of any help with the original question. Will be interested to see what the concensus is and also wondering why this question has been asked. Is it more just for reference at how the calculating tools do it, or is it that they're under question?

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