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Old 01-09-2014, 03:35 PM   #61
cervid
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Or, looked at another way, if your water has chloramine at 1 mg/L and you are treating 20 L you have 20 mg of chloramine to deal with requiring 20*3.127 mg metabite which is 20*3.127/20 mg/L.
Just looking at this now.. Doesn't that equation just come back to 3.127 mg/ml? That was an arbitrary equation in that, if I had 2 mg/ml of chloramine, then the ppm would double..

So.. couldn't I just take the mg of ions in either of those charts and multiply by a factor of whatever the mg/ml concentration of chlorine or chloramine is, respectively?

So, in your above example equation, I would be adding 2.70 mg/ml of SO4, when using the tablet in the prescribed amount we have been discussing? If my concentration of chloramine was 2 mg/ml, I would be adding 5.4 mg/ml of SO4, for instance?

Edit: I just heard back from my water guy regradring the municipal well. They chlorine where I am is between .2 and .5 ppm. Considering I will almost always be diluting my water, I'm not that worried about it.


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Old 01-09-2014, 05:35 PM   #62
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Let's do an example which I hope will make it clear. Your water report says your water contains 2 mg chloramine per liter. You are preparing 40 L (about 10 gal) for brewing. The table says that each mg of chloramine requires 3.127 mg of K2S2O5 and produces 2.70 mg of sulfate and cancels 1.43/50 mEq of alkalinity. Thus for each liter of water treated you will need 2*3.127 = 6.254 mg of K2S2O5 per liter, will obtain 2*2.70 mg of sulfate per liter and remove 2*1.43/50 mEq alkalinity per liter equal to 2*1.43 ppm as CaCO3. Since you are working 'per liter' here the alkalinity can be written a 2*1.43 ppm as CaCO3. The total dose of metabite is, as you are treating 40 L 40*6.254 = 250 mg or about 1/2 a campden tablet.

Or you can multiply the 40 L by 2 mg/L to calculate that you must treat 80 mg chloramine. As each requires 3.127 you will need a total of 80*3.127 = 250 mg. Sulfate produced will be 80*2.70 mg. The mg/L, the number likely to be of more interest is 80*2.70/40 = 5.40 mg/L. 80 mg chloramine neutralizes 80*1.43 mg as CaCO3 alkalinity. Per liter that is 80*1.43/40 = 2*1.43 mg/L as CaCO3 (ppm as CaCO3). Thus it doesn't matter whether you prefer to work with total mg or mg/L. You get the same answer and well you should.


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Old 08-05-2014, 08:58 PM   #63
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Excellent info AJ.

I have been doing all grain for many years and have always bought bulk water from the grocery for a few dollars that claims to be chlorine/chloramine free. At my level much of this information is over my head at this point in the game anyway.

I know my local water supply has chorine/chloramine, would you suggest I find a printout of my municipal water supply and adjust with camden tablets as needed for the time be? Any other pointers? I appreciate your time and thanks in advance.
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Old 08-05-2014, 09:51 PM   #64
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You can usually be guided by the results of the standing over night test. If you pass that then you don't need to do anything. If you don't then use half a tablet in 20 gallons of water treated. That will cover most situations. If you really want to be precise obtain a free and total chlorine kit from Hach or Lamotte or a pool supplier, measure the chlorine and chloramine and then use the tables to determine how much of a Campden tablet. Using too much campden won't hurt anything.
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Old 08-05-2014, 10:31 PM   #65
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Awesome, thanks!
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Old 11-19-2014, 04:13 AM   #66
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Okay, heres my question for one of you who's a chemistry wizz.

I have a 30L HLT but usually require about 35L for my all grain batches. I also add Potassium Metabisulphite at a rate of approximately 0.0666g/L.

If I were to fill my mash tun to the 25L mark, treat with K2S205, mash in, and then add 10L more of water [to my mash tun to use for sparging] **edit**...

- would the additional water have to be pretreated before being added to the HLT.

OR

- could the additional water be added and then the entire mixed water supply be treated with another hit of K2S205 proportionate to the amount of water added.

I think the latter is correct but have no solid grounds for my reasoning.
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Old 11-19-2014, 12:50 PM   #67
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I think either way would be okay, but if you've already mashed in, I think you would be more likely to ensure the campden tablet is dissolved in if you added it to the second, separate water addition.
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Old 11-19-2014, 11:19 PM   #68
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Thanks Matt, I realize I wasn't clear. The additional water would be added to the water already in my hot liquor tank.
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Old 11-24-2014, 04:37 PM   #69
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Possibly a stupid question, but I am not a chemist, so here goes: What effect will treating water with campden have on the ph of the water (and really, the ph of the mash, eventually)?
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Old 11-24-2014, 05:50 PM   #70
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If you go back to #1 here you will see that for each mg/L chlorine neutralized by metabite 2.11/50 = 0.042 mEq/L of hydrogen ions are released while for each mg/L chloramine it is 0.028 mEq/L. These are not appreciable as they correspond to neutralization of water alkalinity of respectively 2.1 and 1.4 ppm as CaCO3 and a typical mash has buffering of 30 - 40 mEq/kg-pH.


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