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Volume of lactic acid in weight of acidulated malt

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IceChisel

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Acidulated malt recommended usage is 1%-5% of the total grain bill or up to 10% for a sour.

Is there a formula which relates the weight of the acidulated malt in a recipe to a volume of lactic acid?
 

Big Monk

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Acidulated malt recommended usage is 1%-5% of the total grain bill or up to 10% for a sour.

Is there a formula which relates the weight of the acidulated malt in a recipe to a volume of lactic acid?
1 kg of Sauermalz is equal to ~31.3 g of Lactic acid at a pH of 5.45. So you can expect a basic relationship that holds relatively true for typical target pH if you assume 3.13% Lactic Acid.

In your case, just replace the Sauermalz with base malt in your software and use whatever acid you like to hit pH, if you need it.
 
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Silver_Is_Money

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At a mash target of pH 5.4 the substitution of 1 mL of 88% lactic acid will be right close to the acid equivalent of ~35.4 grams of Acid Malt, or ~1.25 ounces of Acid Malt.
 

Silver_Is_Money

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When you factor in the caustic (with respect to pH 5.4) nature of the base malt carrier, Big Monk and I are saying the same thing here.

He says acid malt has 337.7 mEq/Kg of acidity, and 3.13% effective lactic acid:

At pH 5.4 88% lactic acid has 11.451 mEq/mL acid strength.
337.7 mEq / 11.451 mEq/mL = 29.4909 mL of 88% lactic acid
29.4909 mL x 1.206 g/mL density = 35.566 grams
35.566 grams x 0.88 = 31.30 grams of pure lactic acid (or precisely what Big Monk told you)
But:
Most of the 1 Kg. of acid malt is a base malt of about 14 mEq/Kg of caustic with respect to pH 5.4
Therefore:
337.7 mEq acid - 14 mEq caustic = 323.7 mEq/Kg. of "NET" lactic acid

And I cut right to the chase and say that acid malt has "effectively" 323.7 mEq/Kg of acidity, or 3% effective lactic acid:

323.7 mEq / 11.451 mEq/mL = 28.2683 mL of 88% lactic acid
28.2683 mL x 1.206 g/mL density = 34.0916 grams
34.0916 grams x 0.88 = 30.0 grams of pure lactic acid (or 3.00% of 1 Kg.)

I merely factor out the caustic impact of the base malt carrier up front, whereas Big Monk does not.

Note also that: (3.13% / 3.00%) x 323.7 mEq = 337.7 mEq

Therefore I can roughly conclude:
Same same but different. (to quote a popular Chinese saying)

Moving along:

323.7 mEq/Kg x 35.3755 grams/1000g/Kg = 11.451 mEq
1 mL of 88% lactic acid = 11.451 mEq
11.451 = 11.451
Thus 35.3755 g. of acid malt = 1 mL of 88% lactic acid

And:

35.3755 grams / 28.34952 grams/ounce = 1.24873 ounces of acid malt = 1 mL of 88% lactic acid
 
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Silver_Is_Money

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Given that 88% lactic acid ranges from 88% to as high as apparently up to a maximum in the ballpark of 92% concentration, and that acid malt likewise varies similarly in acid strength, all of the above information (from myself and Big Monk) is merely ballparking it.

NOTE: Due to laws which are intended to assure that you get what you pay for, it is considered legal to receive somewhat more than you pay for, but illegal to receive less. That is why it is quite likely that any given lot of nominally 88% lactic acid (and likewise for all other acids) is likely to be somewhat stronger in acidity than advertised.
 
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IceChisel

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I understand what you are saying.

Thanks for explaining in finer detail.

I would probably use the higher figure as the acid is not consumed (it simply exists) until the malt is mashed and by "consuming" it before hand mash pH calculations would likely be thrown off albeit a negligible amount.
 

Silver_Is_Money

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I would probably use the higher figure as the acid is not consumed (it simply exists) until the malt is mashed and by "consuming" it before hand mash pH calculations would likely be thrown off albeit a negligible amount.
As soon as the acid malt is crushed the caustic (with respect to pH 5.4) inner grist and the surface acid coating are brought into contact. The grists caustic is thereby to some measure consumed by the acid well before the mash, neutralizing a portion of the acid in the process. So I must disagree that the acid is not consumed until the mash. Either way, this consumption process is rapidly completed during the mash. The bottom line is that if you presume the higher acid figure you must never fail to factor in the caucticity of the base malt carrier. More things to juggle. Keep it simple.
 
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IceChisel

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Taking that acidity and alkalinity away before the mash pH prediction algorithms deals with them could throw the whole thing off.

Malt is basically dried to 3-4% moisture and as such when ground it starts the mash process. It's even worse for malt that's been "conditioned" with water before grinding.
 

Silver_Is_Money

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Taking that acidity and alkalinity away before the mash pH prediction algorithms deals with them could throw the whole thing off.
Not if the algorithm itself fully anticipates and accounts for this. So the only valid answer is "It depends".
 

Big Monk

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Not if the algorithm itself fully anticipates and accounts for this. So the only valid answer is "It depends".
Regardless of whether all parties agree on the calcs, we can use charge conservation here to eliminate any need for tit for tat accounting of acid/base characteristics.

Since the pH we desire (in the sense of estimation) is an equilibrium pH, i.e. proton deficits and surfeits sum to zero, we only need to account for all the constituents and drive QTotal to zero to have an accurate estimation.

Hammering down calcs, however, is always the crux!

I would say, in the grand scheme of things, that the mEq contribution of the base malt carrier for the Sauermalz can be so small, that we/may not want disregard it. Although it takes no extra effort to code it in and add it to the sum of Q, so let's check it out.

For instance, if I were to mash 5.67 kg of Euro Pils with pH DI of 5.84 in 32 l of RO/Distilled water, i'd expect a grist pH of pretty close to pH DI (in this case 5.834). If i wanted to take that mash to pH 5.40 with Sauermalz, I'd expect to have to use ~0.27 kg of it. Under these conditions (i.e. pH target of 5.40) i'd expect the mEq contribution of the malts used to be:

European Pilsner = ~ 91 mEq
Sauermalz = ~ 90 mEq

The remainder comes from the water itself, and is a very miniscule amount. In my experience, I tend to subtract an equivalent amount of base malt off a recipe for the addition of Sauermalz. In any case, if we were to add an equivalent weight of the same base malt into the recipe, ~ 0.27 kg, we'd get ~ 4.3 mEq. Frankly, if your meter doesn't read to the hundredths, you can't even track that, but it is something on the order of +/- 0.018 pH so why not account for it?
 

Silver_Is_Money

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4.3 mEq / 0.27 Kg = 15.926 mEq/Kg of caustic in 1 Kg. of this base malt. But only ~96.87% of this base malt is actually base malt and the rest is lactic acid, so 15.926 x 0.9687 = 15.43 mEq of caustic per Kg. Now if (as I showed above) the software effectively accounts for 14 mEq/Kg of caustic from the base malt, then the difference for this particular base malt is 15.43 - 14 = 1.43 mEq/Kg. as opposed to 4.3 mEq. And then we must factor in that we are only adding 0.27 Kg, and not 1 Kg.

So I'll stick with "It Depends".
 
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Big Monk

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4.3 mEq / 0.27 Kg = 15.926 mEq/Kg of caustic in 1 Kg. of this base malt. But only ~96.87% of this base malt is actually base malt and the rest is lactic acid, so 15.926 x 0.9687 = 15.43 mEq of caustic per Kg. Now if (as I showed above) the software effectively accounts for 14 mEq/Kg of caustic from the base malt, then the difference for this particular base malt is 15.43 - 14 = 1.43 mEq/Kg. as opposed to 4.3 mEq. And then we must factor in that we are only adding 0.27 Kg, and not 1 Kg.

So I'll stick with "It Depends".
Firstly, define caustic for me. I’ve never seen it used with respect to what we are talking about.

Secondly, we have to assume it's both a 100% base malt and has some lactic acid content. I've always thought of it as an added weight of Lactic, i.e. if we took the raw source malt used to produce it and the finished Sauermalz, the Sauermalz would be "heavier". I could be wrong. I have been before!
 

Silver_Is_Money

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Caustic is the opposite of acidic. It is what is generally referred to as basic, but I find that particular terminology to be potentially confusing to the general audience when we are discussing a base malt substrate. And all of this is with respect to a targeted mash pH of 5.4 as our effective "neutral" point.

If it is 3.13% by weight acid it is 96.87% by weight base malt.
 

Big Monk

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If it is 3.13% by weight acid it is 96.87% by weight base malt.
At the risk of appearing dense or ill-informed here, what I am trying to get across, if true, is this: If I start with X kg of base malt during the Sauermalz process, one of three things happens:

1.) The base malt is allowed to naturally develop Y kg of Lactic acid, yielding X + Y kg total;

2.) The base malt is sprayed with Y kg of Lactic acid, yielding X + Y kg total;

3.) The base malt is sprayed with Y kg of Sauergut, yielding X + Y kg total.

If this is accurate, what you are saying, while technically true from a percentage standpoint, doesn't capture the whole picture, i.e. you are neglecting 3.13% * X kg of malt. Again, I could be way off base in my thinking because, frankly, I have never contemplated this.
 

Silver_Is_Money

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At the risk of appearing dense or ill-informed here, what I am trying to get across, if true, is this: If I start with X kg of base malt during the Sauermalz process, one of three things happens:

1.) The base malt is allowed to naturally develop Y kg of Lactic acid, yielding X + Y kg total;

2.) The base malt is sprayed with Y kg of Lactic acid, yielding X + Y kg total;

3.) The base malt is sprayed with Y kg of Sauergut, yielding X + Y kg total.

If this is accurate, what you are saying, while technically true from a percentage standpoint, doesn't capture the whole picture, i.e. you are neglecting 3.13% * X kg of malt. Again, I could be way off base in my thinking because, frankly, I have never contemplated this.
Do you add malts by weight or by volume. Only if by volume does your thinking apply.
 

Vale71

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Malt is basically dried to 3-4% moisture and as such when ground it starts the mash process. It's even worse for malt that's been "conditioned" with water before grinding.
No it doesn't. Oxygen plays no role in mashing and the moisture is already in the kernel. If what you say were true then malt would be basically in a state of "perpetual mash" and that is definitely not the case. You need a lot more water for amylases, which are all hydrolitic, to start working, not to mention the whole startch gelatinization issue. Conditioning, if done properly, will only increase moisture in the husks and not the kernel. The latter would only be the case with actual wet milling which is something else entirely.
 
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IceChisel

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Of course you're correct but that's not what was meant. Grinding the malt with lactic acid sprayed on it will react with that moisture no matter if It's in the kernel or in the husk.
 

Silver_Is_Money

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For clarity, we are not technically discussing mashing here, but rather we are discussing the interaction between an acid and a base (or caustic substance, to differentiate it from the term base also being confusingly used as a terminology for a class of malts irregardless of whether or not such a base malt is actually caustic or acidic with respect to mash pH). Enzymes have nothing to do with it.
 

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Maybe I'm particularly dense today but I still don't understand what you mean. Acidulated malt has acid sprayed on it at the malthouse. It is then dried in the kiln. When you mill it it in the brewhouse the acid has already been absorbed by the kernel, milling it will cause no further reactions to take place until you mash in (assuming dry milling of course).
 

Vale71

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For clarity, we are not technically discussing mashing here, but rather we are discussing the interaction between an acid and a base (or caustic substance, to differentiate it from the term base also being confusingly used as a terminology for a class of malts irregardless of whether or not such a base malt is actually caustic or acidic with respect to mash pH). Enzymes have nothing to do with it.
Actually as per the OP we were discussing how to calculate volume of lactic acid in acidulated malt. The obvious answer is to look up the weight/weight percentage on the analysis sheet, multiply by the weight of the acidulated malt in the grain bill and then divide by lactic acid's density (1.206 g/l).
The usual tangents then developed...
 

Big Monk

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Actually as per the OP we were discussing how to calculate volume of lactic acid in acidulated malt. The obvious answer is to look up the weight/weight percentage on the analysis sheet, multiply by the weight of the acidulated malt in the grain bill and then divide by lactic acid's density (1.206 g/l).
The usual tangents then developed...
I don't believe maltster's specify the weight/weight percentage on the analysis. At least not in the 30 or so Weyermann sheets I have seen.
 
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