Lactic acid sub for acid malt

[Surly]

I will be brewing 15.5 gallons of a low gravity beer.

The 31 lb grain bill includes 22 lbs Vienna and 10 Pilsner.

Of the 26 gallons of water to be used I will use 10 gallons of RO during mash in and 10 of RO at spare, adding 6 gallons of my well water to that sparge.

I have traditionally added 2% acid malt to my grain bill. Not having acid malt I would like to sub with lactic acid.

Is there a good standard Lactic Acid (88%) addition I could be using?

 

[ajdelange]

1 kg of sauermalz supplies, to pH 5.4, about 310 mEq of acidity. It takes about 27 mL of 88% lactic acid solution to do the same (also to pH 5.4).

 

[ajdelange]

Because it was asked in the Primer I'll give some more details about how to calculate the equivalence. To begin, each sauermalz is going to be a little different in how much acid it delivers. What follows is from a measured sample of Weyermanns which I think we can probably consider typical. To a given pH 1 kg of it delivers

292.09*(pH - 3.62) - 68.443*(pH - 3.62)^2 +5.3985*(pH - 3.62)^3 mEq/kg.

For lactic acid, compute

r = 10^(pH - 3.86)
Then f0 = 1/(1 + r)
and then f1 = r*f0

The density of 88% lactic acid is 1.2053 so a mL contains 0.88*1205.3 milligrams grams equivalent to 0.88*1205.3/90.08 mmol/ml. The protons delivered by 1 mL of lactic acid is then
f1*0.88*1205.3/90.08 mEq/mL


By scaling the mEq/kg for the malt and the mEq/mL for the acid to whatever units you want you can determine the acid equivalent in drams/shekel or whatever suits you.

 

[thekraken]

Because it was asked in the Primer I'll give some more details about how to calculate the equivalence. To begin, each sauermalz is going to be a little different in how much acid it delivers. What follows is from a measured sample of Weyermanns which I think we can probably consider typical. To a given pH 1 kg of it delivers

292.09*(pH - 3.62) - 68.443*(pH - 3.62)^2 +5.3985*(pH - 3.62)^3 mEq/kg.

For lactic acid, compute

r = 10^(pH - 3.86)
Then f0 = 1/(1 + r)
and then f1 = r*f0

The density of 88% lactic acid is 1.2053 so a mL contains 0.88*1205.3 milligrams grams equivalent to 0.88*1205.3/90.08 mmol/ml. The protons delivered by 1 mL of lactic acid is then
f1*0.88*1205.3/90.08 mEq/mL


By scaling the mEq/kg for the malt and the mEq/mL for the acid to whatever units you want you can determine the acid equivalent in drams/shekel or whatever suits you.

[Coming over from the Primer thread]
Okay, from that information we can derive a single equation for subbing 88% lactic acid solution for acid malt

= (0.458483440561761*(pH - 3.62)^3 - 5.8127224455624*(pH - 3.62)^2 + 24.8065996394711*pH - 89.7998906948854)*(10^(pH - 3.86) + 1)*10^(-pH + 3.86)

ml / kg.

What should the pH variable be if I were just trying to follow the primer's kiss spirit? The only reason I ask is because I already have a bottle of lactic acid I wanted to use up before I just start using sauermalz.

[Edit to figure out if I'm doing something wrong] I tried to calculate your figures in post 2 (calculated to pH 5.4) using the equations from post 3 but I needed a pH value around 5.18 to get post 2's values.

 

[ajdelange]

What should the pH variable be if I were just trying to follow the primer's kiss spirit? The only reason I ask is because I already have a bottle of lactic acid I wanted to use up before I just start using sauermalz.

I guess 5.4 or 5.5 would do.

[Edit to figure out if I'm doing something wrong] I tried to calculate your figures in post 2 (calculated to pH 5.4) using the equations from post 3 but I needed a pH value around 5.18 to get post 2's values.

You aren't doing anything wrong. The 310 mEq/kg number was for a sample of Weyermanns sauermalz measured by Kai Troester using a simplified method from which only one coefficient is obtained approximately. The fuller formula for the sample I measured would give 333 mEq/kg which is 8% larger than the 310 number.

 

[thekraken]

Because it was asked in the Primer I'll give some more details about how to calculate the equivalence. To begin, each sauermalz is going to be a little different in how much acid it delivers. What follows is from a measured sample of Weyermanns which I think we can probably consider typical. To a given pH 1 kg of it delivers

292.09*(pH - 3.62) - 68.443*(pH - 3.62)^2 +5.3985*(pH - 3.62)^3 mEq/kg.

I drank, I slept, I woke... aanndd now I don't follow. Wouldn't a given weight of
acid malt deliver a set amount of charge? How does this pH value come into play in this particular situation?

I've attached some screenshots of my Super Complicated Brewing Water Spreadsheet v1.0 to illustrate what I'm trying to achieve.

BTW my original equation that I posted in the other thread, and my 'Method A' in this screen shot was derived from your post on thebrewingnetwork forums

 

[ajdelange]

I drank, I slept, I woke... aanndd now I don't follow. Wouldn't a given weight of
acid malt deliver a set amount of charge?

No

How does this pH value come into play in this particular situation?

The number of protons released by an acid in adjusting the pH of a solution depends on the final pH if it is anywhere near the pK of the acid. For lactic acid the pK is 3.86 (lactic acid is a weak acid) and mashing pH's, while far enough from that such that you wouldn't want to try a lactate buffer they are close enough that the pH effect is clearly seen.

In the graph below the red line represents the number of millimoles released by 1 kg of lactic acid in going to various pH values. In going to pH 5.4, for example, a kg of lactic acid would release about 11,000 mEq of protons. The blue curve is 100 times the mEq of protons released by 1 kg of the sauermalz I measured. At 5.4 the graph shows 34,000 and divided by 100 that gives 340 (it's actually 333 as we recall from earlier posts). That means about 0.031 times the sauermalz weight in lactic acid is required to deliver the same number of protons or that the sauermalz equivalent in lactic acid is about 3.1%. The green curve shows the percentage equivalence at other pH values and it varies from about 2.5 - 3.5% over the range of mash pH's for this particular sauermalz (the one I measured).

I

 

[thekraken]

Cool! I'm full of questions but I'll try and refrain so we don't derail the thread .

So then, the only problem with my first derivation was the estimate for percent acid by weight of sauermalz was too low at 1.5%? Doubling that to your estimated 3.1% gives me

By Weight:
Sauermalz_weight * sauermalz_percent_acid * 1000 (g/kg) / acid_solution / acid_solution_density
Sauermalz_weight * 0.0309 * 1000 / 0.88 / 1.2053
Sauermalz_weight * 29.132694

And the mEq method gives:

var('pH')

malt_mEq = 292.09*(pH - 3.62) - 68.443*(pH - 3.62)^2 +5.3985*(pH - 3.62)^3

r = 10^(pH - 3.86)
f0 = 1/(1 + r)
f1 = r*f0
acid_mEq = f1*0.88*1205.3/90.08

eq(pH) = (malt_mEq / acid_mEq).simplify()
print eq
    pH |--> (0.4584834405617613*(pH - 3.62)^3 - 5.812722445562402*(pH - 3.62)^2 + 24.80659963947112*pH - 89.79989069488548)*(10^(pH - 3.86) + 1)*10^(-pH + 3.86)

print eq(5.4)
    29.141336537042992

So a conversion factor of 29.14 (mL 88% solution/ kg sauermalz) close enough for the primer?

 

[ajdelange]

So a conversion factor of 29.14 (mL 88% solution/ kg sauermalz) close enough for the primer?

That's what the numbers give for pH 5.4 and the Weyermann's malt I measured so yes, that should do.