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Old 01-09-2014, 02:35 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, 04:35 PM   #62
ajdelange
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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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