DoubleMash _ MashPH

Hey, I have a question I can't seem to find the answer to.

Tomorrow I am planning on brewing a barleywine, and need to do a double mash. I have a ph meter, and have gotten pretty good at hitting my mash ph and my beers have never been better (thanks mostly to this forum).

My question is how do I calculate lactic additions to hit my mash ph.

My general plan is to treat most of my water as normal (for a standard batch), Then mash and sparge.

Reload the mash tun with malt - mash, sparge. I will probably use my standard chemical additions but no lactic in the water for this stage.

I know I can hit the mash ph on the first charge with this method....but will it drop too low on the second load of malt? Does it matter? Or should I try to hit the mash-ph on the second mashing instead of the first?

I realize the answer is to perform a mini-mash and test with my ph meter. But I was kind of hoping someone had already done the legwork and could answer.

Thanks

Rb

I'm not sure I understand exactly what you intend to do. I assume that you are going to use the wort from the first charge as the liquor for the second charge. Assuming this to be the case then the basic difference between the two is that the acid addition for the first mash must be sufficient to overcome the proton deficits of both the malts and the water whereas in the second the water will have been taken care of and you only need enough to overcome the proton deficits of the malts.

An ostensible approach is to figure out how much of the lactic acid you used went to neutralize the alkalinity. For example, if your water's alkalinity is 2 mVal/L and you are working with 20L then ~40 mEq (it's actually more like 35) of the lactic acid went for the liquor and you could use 40 mVal less for the second mash. The catch would come in if the total acid you used was less that 40 mVal - say only half this i.e. the first mash hits proper pH with 20 mVal total acid. This means that darker malts are supplying some of the acid necessary to overcome base malt proton deficit. You would then calculate 20 - 40 = -20 which tells you that for the second mash you would not only with hold the acid but in addition add 20 mVal base.

Thus the detailed answer depends on the details of the mash and liquor. If the alkalinity of the liquor is very low (RO water) then the acid additions for both first and second mashes would be the same.

Thanks AJ. You understood me perfectly.

What I was really wondering was if it might be necessary to add acid to this first mash and a base to the second. And apparently it might be. So I will run through the calculations with my water profile tonight and see what I come up with.

Thanks again