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Message |
    
Little Dipper
Member
Username: Littledipper
Post Number: 245
Registered: 02-2004
Posted From: 69.221.233.226
| | Posted on Sunday, June 17, 2007 - 11:25 pm: | 
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In the new edition of BYO magazine (July/August) there's a question about water chemistry on page 17. The guy gives his water profile as follows and basically asks what happens to the bicarbonate when he boils it:
Ca: 103
Mg: 45
Na: 119
SO4: 8
Cl: 207
HCO3: 444.8
pH: 7.8
Mr. Wizard goes on to give the equation of:
Ca + 2HCO3 --> CO2 + H2O + CaCO3
Then he tells him that if he were to boil, he could remove up to 314ppm of biocarbonate if all the calcium were consumed in the reaction and 229ppm if all the Magnesium were consumed. He then explains that in reality, the reactions are not 100%, etc.
How did he come up with these results? When I look at that equation, I would think that for every ppm of calcium, you could remove 2 ppm of HCO3, so that would be 103 calciums to 206 biocarbonates.
Can anyone explain how he got these numbers? Thanks. |
Chris Colby
Intermediate Member
Username: Chriscolby
Post Number: 489
Registered: 03-2003
Posted From: 66.68.185.220
| | Posted on Monday, June 18, 2007 - 12:35 am: |
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Hey Little Dipper,
This is a good question. The answer is that you need to take the formula weight (FW) of the molecules into consideration.
Calcium ions (Ca++) have a molecular weight of 40 and bicarbonate (HCO3-) has a formula weight of 61 (carbon = 12, hydrogen = 1 plus three oxygens at 16 = 48).
So, for every 40 ppm (or mg/L) of Ca++, you remove [2 X 61 =] 122 ppm (mg/L) of HCO3-. (Dividing by 40 you get, for every 1 ppm (mg/L) of Ca++, you remove 3.05 ppm (mg/L) HCO3- (not 2 ppm) because HCO3- is heavier than Ca++.)
Taking the letter writier's water into consideration, the 103 ppm of Ca++ he has dissolved in his water should remove [103 X 2 X 61/40 =] 314 ppm of HCO3-, assuming the reaction goes to completion.
Chris Colby (Editor BYO)
Bastrop, TX |
Little Dipper
Member
Username: Littledipper
Post Number: 246
Registered: 02-2004
Posted From: 69.221.233.226
| | Posted on Monday, June 18, 2007 - 12:44 am: |
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That makes sense. Thanks, Chris, for your quick reply.
So does boiling longer push the reaction closer to full completion? When I boil my water, I usually just let it go for 10 minutes or so.
Thanks again. |
Chris Colby
Intermediate Member
Username: Chriscolby
Post Number: 490
Registered: 03-2003
Posted From: 66.68.185.220
| | Posted on Monday, June 18, 2007 - 12:57 am: |
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I would guess that the longer you boil, the more HCO3- you would precipitate out, as long as Ca++ is still present. But, I don't have any numbers handy that give that rate or what percentage of the total carbonate you'd drop out.
From what I've heard, aeration plays a huge role. If you boil hard and aerate the water before (and maybe even during) the boil, you'll get a lot more of the carbonate to fall out.
Chris Colby
Basatrop, TX |
Bill Pierce
Moderator
Username: Billpierce
Post Number: 7279
Registered: 01-2002
Posted From: 24.57.224.220
| | Posted on Monday, June 18, 2007 - 03:34 am: |
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It shouldn't be necessary to aerate the water during the boil. For one thing, at boiling temperature water can contain virtually no dissolved O2. For another, the boil itself should provide physical agitation and accelerate the reaction.
I do agree that aeration (vigorous stirring or pouring the water back and forth between vessels) prior to boiling might be a good idea. |
Tom Meier
Advanced Member
Username: Brewdawg96
Post Number: 595
Registered: 03-2003
Posted From: 67.88.124.194
| | Posted on Monday, June 18, 2007 - 07:08 am: |
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preboiling requires seeding of excess calcium to drive the reaction. I think this is well covered in the old guard textbooks.
If you read the old AJ Delange posts, he also gives how much bicarb that preboiling can remove.. it cant go below a certain ppm without additional calcium added.. I can't remember the specifics but they are easy enough to find. |
Little Dipper
Member
Username: Littledipper
Post Number: 247
Registered: 02-2004
Posted From: 206.114.61.199
| | Posted on Monday, June 18, 2007 - 02:35 pm: |
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So now I'm a little confused. Let's say, just for argument's sake, that when I pre-boil my water, everything goes to completion. My water profile is as follows:
Ca: 104.5
HCO3: 187
Mg: 30.9
So just looking at the Calcium, I could remove up to (104.5x2x61/40)=318.725 ppm of HCO3 (all of my bicarbonates), correct? So a boil would, in theory, leave me with zero bicoarbonates and about 61 ppm of calcium [(187*40/61/2)=61.3]?
Thanks. I'm trying to wrap my mind around this. |
Bill Pierce
Moderator
Username: Billpierce
Post Number: 7282
Registered: 01-2002
Posted From: 24.57.224.220
| | Posted on Monday, June 18, 2007 - 02:58 pm: |
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I could do a little more research, but my recollection is that about 10 mg/L (the same as ppm) of dissolved calcium is essentially untouchable. This is why recommendations for treatment with slaked lime often suggest the addition of calcium (via gypsum or calcium chloride) before beginning the process. |
Little Dipper
Member
Username: Littledipper
Post Number: 248
Registered: 02-2004
Posted From: 206.114.61.199
| | Posted on Monday, June 18, 2007 - 03:23 pm: |
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Does this help:
http://homeroastnbrew.info/hbd/2004/4501.txt |
Bill Pierce
Moderator
Username: Billpierce
Post Number: 7285
Registered: 01-2002
Posted From: 24.57.224.220
| | Posted on Monday, June 18, 2007 - 03:28 pm: |
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I wouldn't argue with A.J. deLange. In my mind he's practically the gold standard for brewing water treatment information and advice. |
KeepBrewing
Intermediate Member
Username: Kb7
Post Number: 285
Registered: 05-2002
Posted From: 24.184.80.79
| | Posted on Friday, June 22, 2007 - 12:43 am: |
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Hate to dredge up an old post but I remember I had a link to this beer geek site and thought I would throw it in.
http://www.antiochsudsuckers.com/tom/brewingwater.htm |
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