5ml
How much water was the 5ml to, What style of beer, and what were your other mineral additions?
11.34 G with 5grams gypsum and 5grams CaCl2. Brewing a blonde on a no-sparge system, but planning on taking 2 gallons of the 11.34 and cold-sparging with it. Is it ok to use those 2 gallons in a sparge?
No, nothing at all. Of course the pH will drop dramatically because there are no grains in the water to absorb the acid. The amount of pH drop depends entirely on the alkalinity of the water in this case. You may, of course, be adding some of the lactic acid to remove this alkalinity. The fact that your pH went below 4.4 (the value to which alkalinity is measured) indicates that you are adding more than enough to remove the alkalinity and that, presumably, the rest is there to neutralize the alkalinity of the grains. When they are added in the pH will rise again to, we hope, the value you used to calculate your lactic addition.
Yes, per EZ water and Bru'N, it should bring be to an estimated 5.3 pH
This all looks generally in the ball park and fine to me. AJ gave a good explanation of it.
29 ml of lactic acid (generally high % acid) or 29 ml of phosphoric (generally low % acid).
Exactly! When you treat/mineralize your water, the impact on pH can be very high, since water by itself has very little buffering capacity, so huge swings can be expected from small additions. Once you add your grains, their buffering capacity is huge, and brings the pH to just the right level as was calculated for you.
What about the water I pulled aside to cold-spare with? It's at the same pH. Will it effect anything or have the same reactions with the grain?
Yeah..... 29 ml of an 88% acid to 11 gallons of water seems like an extraordinary amount. I am generally adding 2-5ml to 10-11 gallons depending on beer... and that is only for light colored beer.
Easy enough to WAG it. He's using a bit over 9 gal mash water - call it 35 L. If the alkalinity is 2 mEq/L (not uncommon) he'd need about 0.9*70 = 63 mEq for that. 88% lactic acid is about 11.8 N to mash pH so he's adding 29*11.8 = 342 mEq. With 63 going for the water that leaves 279 for the grains. Assuming a quart and a half per pound he's have about 36/1.5 = 24 lbs of grain which is 11.9, call it 12 kg. Assuming the malt has a DI pH of 5.7 and he wants 5.3 then he'd need approximately 40*12*.4 = 192 mEq for the grains assuming that they have buffering of -40 mEq/pH•kg. That's only 255 total so 342 does seem a lot unless the water's alkalinity were, say 4 mEq/L in which case the total requirement would be 192 + 126 = 328. But if the water were that alkaline he wouldn't want to be using acid to neutralize it.
Oops. I've made the same mistake I have before and that is responding as if we were in the Brewing Science forum where I spend most of my time to the point that when I respond elsewhere I often forget that I'm not there. I would expect many, if not most, readers of that forum to understand what I posted and clearly some here do too as well.
That sort of says you aren't really interested in understanding further so I won't go into further explanation beyond saying that mEq/L is the natural way of expressing alkalinity. If you have alkalinity of 100 ppm as CaCO3 the lab analyst measured it by adding 2 mEq of acid to each liter of your water and then multiplied that by 50 to get the ppm as CaCO3.
Not true at all. I struggle reading through books to get answers as words run together very fast, so I target items like this that I can absorb in small chunks. Put me in real world, hands on stuff and I can do it very well and learn it very fast. I see it once, I can do it. If I have to read an entire book, it is struggle to focus after a few pages.
