Calculating acid when adjusting boil pH.

[beerkench]

Is there a way of calcultating the amount of lactic acid needed to adjust boil pH to the required level?
I use Bru N' Water for the mash but I'm not sure if it can be used on the boil. (Maybe the sparge section?)

My usual technique is to add 1ml at a time of acid until the desired level reached but a rough calculation would be more effective for me to avoid checking sample after sample of wort.

 

[ajdelange]

Is there a way of calcultating the amount of lactic acid needed to adjust boil pH to the required level?

Ya, sure. Kolbach found that it takes 0.29 g of lactic acid to reduce the pH of wort by 0.1 pH for each kg of malt that it was made from. Given that lactic acid is about 94% dissociated at pH 5 that would imply buffering of 10*290/(0.94*90.08)= 34.25 mEq/kg•pH. THE 88% stuff we use is just about 11 N at pH 5 so there it is.

My usual technique is to add 1ml at a time of acid until the desired level reached but a rough calculation would be more effective for me to avoid checking sample after sample of wort.

Kolbach's number can only be taken as representative. You will still have to proceed carefully but at least you'll have a rough idea.

 

[beerkench]

Ya, sure. Kolbach found that it takes 0.29 g of lactic acid to reduce the pH of wort by 0.1 pH for each kg of malt that it was made from. Given that lactic acid is about 94% dissociated at pH 5 that would imply buffering of 10*290/(0.94*90.08)= 34.25 mEq/kg•pH. THE 88% stuff we use is just about 11 N at pH 5 so there it is.




Kolbach's number can only be taken as representative. You will still have to proceed carefully but at least you'll have a rough idea.

Hi Aj,

Thanks for the help. Could you please however break that down for an idiot like me? If it's 0.29g per 0.1pH per KG malt. What's that in ml?


Cheers.

 

[ajdelange]

Thanks for the help. Could you please however break that down for an idiot like me? If it's 0.29g per 0.1pH per KG malt. What's that in ml?

Well the 0.29 grams is 290 mg. The molecular weight of lactic acid is 90.08 mmol/mg and, at pH 5 lactic acid is 94% dissociated. Thus 290 mg of lactic acid, the amount required to move 1 kg 0.1 pH yields 290/(0.94*90.08) mEq and the amount required to move 1 kg 1 pH is 10 times that

10*290/(0.94*90.08) = 34.25 mEq/kg•pH which is the apparent buffering capacity of wort. If you have a wort from x kg of malt and you want to move it ∆pH you will need x*34.25*∆pH mEq of acid. 88% lactic acid is about 11 N so you would need x*34.25*∆pH/11 mL.

 

[beerkench]

Well the 0.29 grams is 290 mg. The molecular weight of lactic acid is 90.08 mmol/mg and, at pH 5 lactic acid is 94% dissociated. Thus 290 mg of lactic acid, the amount required to move 1 kg 0.1 pH yields 290/(0.94*90.08) mEq and the amount required to move 1 kg 1 pH is 10 times that

10*290/(0.94*90.08) = 34.25 mEq/kg•pH which is the apparent buffering capacity of wort. If you have a wort from x kg of malt and you want to move it ∆pH you will need x*34.25*∆pH mEq of acid. 88% lactic acid is about 11 N so you would need x*34.25*∆pH/11 mL.

Still this is a bit too much for me. Could you give a an example to help me get it?
I used around 5kg of malt for the last brew session and the pH at the start of the boil was around 5.55, and the aim was to get it down to 5.2.
How many ml of 88% lactic would be needed according to Kolbach's formula?

 

[ajdelange]

I used around 5kg of malt for the last brew session and the pH at the start of the boil was around 5.55, and the aim was to get it down to 5.2.
How many ml of 88% lactic would be needed according to Kolbach's formula?

x = 5 kg

∆pH = 5.55 - 5.2 = 0.35

Protons required to move a wort made with 1 kg of malt 1.0 pH = 34.25 mEq/kg•pH

Normality (strength) of 88% lactic acid at pH 5.2: 11.04 mEq/ml

x*34.25*∆pH/11 = 5*34.25*0.35/11 = 5.4 mL

 

[beerkench]

x = 5 kg

∆pH = 5.55 - 5.2 = 0.35

Protons required to move a wort made with 1 kg of malt 1.0 pH = 34.25 mEq/kg•pH

Normality (strength) of 88% lactic acid at pH 5.2: 11.04 mEq/ml

x*34.25*∆pH/11 = 5*34.25*0.35/11 = 5.4 mL

Thank you sir, now I get it.

 

[ajdelange]

An example is often the best way...

 

[phishie]

x = 5 kg

∆pH = 5.55 - 5.2 = 0.35

Protons required to move a wort made with 1 kg of malt 1.0 pH = 34.25 mEq/kg•pH

Normality (strength) of 88% lactic acid at pH 5.2: 11.04 mEq/ml

x*34.25*∆pH/11 = 5*34.25*0.35/11 = 5.4 mL

I know this is reviving an old(ish) thread but anyways...

What would be the equivalent equation for 85% phosphoric acid?

 

[ajdelange]

The normality of 85% phosphoric acid at pH 5.2 is 14.75 mEq/ml.

 

[phishie]

The normality of 85% phosphoric acid at pH 5.2 is 14.75 mEq/ml.

Thank you!

So mimicking past example with my data...

x amount of grain = 4.762 kg

∆pH = 5.4 - 5.2 = 0.2

Protons required to move a wort made with 1 kg of malt 1.0 pH = 34.25 mEq/kg•pH

Normality (strength) of 85% phosphoric acid at pH 5.2: 14.75 mEq/ml

x*34.25*∆pH/14.75 = 4.762*34.25*0.35/14.75 = 3.87 mL

For future reference, as a complete chemistry novice I found these links very helpful for understanding mole and normality:
http://www.lagoonsonline.com/laboratory-articles/molarity.htm
http://www.lagoonsonline.com/laboratory-articles/normality.htm
http://www.lagoonsonline.com/laboratory-articles/normal-solutions.htm

Although, those links don't cover concepts of wort pH, being the missing piece we're here for.

Thus 290 mg of lactic acid, the amount required to move 1 kg 0.1 pH yields 290/(0.94*90.08) mEq

Is that "290/(0.94*90.08)" equation for this based on this equation "Volume of concentrated acid needed = (grams of acid needed)/(percent concentration x specific gravity)" that I pulled from the third link above? I'm unclear based on using the lactic acid molecular weight 90.08 rather than specific gravity.

 

[ajdelange]

Well the 0.29 grams is 290 mg. The molecular weight of lactic acid is 90.08 mmol/mg and, at pH 5 lactic acid is 94% dissociated. Thus 290 mg of lactic acid, the amount required to move 1 kg 0.1 pH yields 290/(0.94*90.08) mEq and the amount required to move 1 kg 1 pH is 10 times that

10*290/(0.94*90.08) = 34.25 mEq/kg•pH which is the apparent buffering capacity of wort. If you have a wort from x kg of malt and you want to move it ∆pH you will need x*34.25*∆pH mEq of acid. 88% lactic acid is about 11 N so you would need x*34.25*∆pH/11 mL.

It's really unfortunate that the operators of this site have chosen to prohibit editing of posts by their authors once a certain time period has expired. I see absolutely no advantage to doing that and one big disadvantage in that sometimes, as in this case, an error is discovered which can't be fixed. Some poor schnook reads as far as No. 4, thinks he has the answer and doesn't read the rest of the thread and is mislead despite the original writer's desire to correct his mistake (and, at the same time, of course, remove the permanent record of his foolishness). But to the point: 290 mg of lactic acid in 290/90.08 = 3.21936 mmol. At pH 5 each mmol releases 0.94 mEq of protons so the 290 mg of lactic acid releases 0.94*290/90.08 = 3.0262 mEq of acid.

The second calculation should, similarly, be 10*0.94*290/90.08 = 30.262 mEq.

The volume of acid required for a particular job is V = mEq_required/Normality and

Normality (%, pH) = dissociation_factor(pH) * (%/100) * density(%) / Mol_wt

The dissociation factor depends on the pH and the density of the percent concentration. Thus normality depends of pH and concentration.

 

[phishie]

It's really unfortunate that the operators of this site have chosen prohibit editing of posts by their authors once a certain time period has expired. I see absolutely no advantage to doing that and one big disadvantage in that sometimes, as in this case, an error is discovered which can't be fixed. Some poor schnook reads as far as No. 4, thinks he has the answer and doesn't read the rest of the thread and is mislead despite the original writer's desire to correct his mistake (and, at the same time, of course, remove the permanent record of his foolishness). But to the point: 290 mg of lactic acid in 290/90.08 = 3.21936 mmol. At pH 5 each mmol releases 0.94 mEq of protons so the 290 mg of lactic acid releases 0.94*290/90.08 = 3.0262 mEq of acid.

The second calculation should, similarly, be 10*0.94*290/90.08 = 30.262 mEq.

The volume of acid required for a particular job is V = mEq_required/Normality and

Normality (%, pH) = dissociation_factor(pH) * (%/100) * density(%) / Mol_wt

The dissociation factor depends on the pH and the density of the percent concentration. Thus normality depends of pH and concentration.

Thank you for the follow-up, updating my previous example:

x amount of grain = 4.762 kg

∆pH = 5.4 - 5.2 = 0.2

Protons required to move a wort made with 1 kg of malt 1.0 pH = 30.262 mEq/kg•pH

Normality (strength) of 85% phosphoric acid at pH 5.2: 14.75 mEq/ml

x*30.262*∆pH/14.75 = (4.762*30.262*0.2)/14.75 = 1.954 mL

*Edit*

I previously read the formula as kg*protons_req*(delta_pH/normality) but finally realized it was meant to be read (kg*protons_req*delta_pH)/normality.

 

[phishie]

How do you determine normality of 88% lactic acid as 11 and normality of 85% phosphoric acid as 14.75?

Trying to determine this myself I get 11.811 and 14.615 respectively. Is it because I am not factoring in a dissociation factor? If I assume as much then backing into the dissociation factor of phosphoric acid would be 1.0092.

So the volume of 85% phosphoric acid required to move the pH of wort made from 1 kg grain by 1 = 30.262 mEq / 14.75 = 2.0517 mL.

Or for my brewday this weekend, the volume of 85% phosphoric acid required to move the pH of wort made from 4.762 kg grain by 0.2 =

(4.762 * 30.262 * 0.2) / 14.75 = 1.954 ml

I'm having trouble trying to expand that further to the more detailed equation you mentioned before. Can you show me how Normality (%, pH) = dissociation_factor(pH) * (%/100) * density(%) / Mol_wt = 14.75 for 85% phosphoric acid?

 

[ajdelange]

The density of 88% lactic acid is 1.20525 grams per mL. As it is 88% acid by weight a mL contains 0.88*1.20525 = 1.06062 grams of lactic acid. Lactic acid's molecular weight is 90.08 so the millimoles per mL is 1000*1.06062/90.08 = 11.7742. Finally, the number of protons released per mmol of lactic acid to pH 5.2 is 0.956289152mEq/mmol so the normality is 0.956289152*1000*1.06062/90.08 = 11.2595. The process for phosphoric is the same except that the density and dissociation of phosphoric acid are used.

I don't actually go through all this calculation. I have an Excel spreadsheet into which I type =LacNorm(pH, %) and it returns the normality of lactic acid of strength % at the specified pH.

 

[phishie]

Thank you, I really appreciate the explanation. I only got as far as the 11.77 number (I had found 1.209 density of LA on Google) and then got stuck.

Without any serious background in chemistry it's a lot of Googling, guessing, fumbling, and deduction but it's fun to learn rather than just take the answer and move on.

"..the number of protons released per mmol of lactic acid to pH 5.2 is 0.956289152mEq/mmol"

What do I refer to this 0.9562 number as? I keep coming across the "acid dissociation constant i.e. Ka" but that doesn't seem to be the same thing.

 

[ajdelange]

"..the number of protons released per mmol of lactic acid to pH 5.2 is 0.956289152mEq/mmol"

What do I refer to this 0.9562 number as?

I'm not sure it has a name, IOW it seems "the number of protons released..." is about as good as we can do.

I keep coming across the "acid dissociation constant i.e. Ka" but that doesn't seem to be the same thing.

No, it's not the dissociation constant. It is computed from the dissociation constant(s) for the acid in question. The sticky at the top of this forum on calculating bicarbonate and carbonate will show you how. A convenient way to explain this stuff is by keeping track of the charges on the anions of the molecules of things in a mixture whose charges change with pH. Pure lactic acid is a neutral molecule, HLac. At pH 3.86 half of lactic acid molecules will have come apart into H+ ions and Lac- ions: HLac <---> H+ + Lac-. If we started with a mmol of lactic acid molecules we would find half a mmol of negatively charged Lac- ions. We would say Qlac = -0.5 mEq (a mEq is a mmol of charge). At pH 5.2 Qlac = -0.956 and so on. If the charge on lactic acid went from 0 to -0.5 mEq it must have lost 0.5 mEq of positively charged things i.e. protons.

 

[beervoid]

I've read over this thread 4x times but I'm not identifying the formula for phosphoric acid 85%
I would like to make a spreadsheet to calculate this.

Is this the right formula?

x amount of grain = 4.762 kg

∆pH = 5.4 - 5.2 = 0.2

Protons required to move a wort made with 1 kg of malt 1.0 pH = 30.262 mEq/kg•pH

Normality (strength) of 85% phosphoric acid at pH 5.2: 14.75 mEq/ml

x*30.262*∆pH/14.75 = (4.762*30.262*0.2)/14.75 = 1.954 mL


Would it also be possible to get the formula for phosphoric 10% ?

I have made an excel sheet but the forum doesn't allow me to upload it.
Hope it's ok I share a link to it?
https://www.sendspace.com/file/1vzbdf

I don't think I have put the calculation of normality (strength) in it yet so if one wants to aim for let's say 5.0 that nr would change me thinks?

Thank you!

 

[SanPancho]

Kolbach's number can only be taken as representative. You will still have to proceed carefully but at least you'll have a rough idea.

@ajdelange im curious to know if this estimation takes hops into account? In normal settings youd have a bit of bittering hops, which i wonder if kolbach used since the units are only in malt kg. Maybe not such a big deal as bittering seems less important back in the day, in terms oh his calculations.

itd seem modern hoppy beers pose their own issues. If i recall, each dry hop #/bbl ups the ph by like .15pH units so in theory you could account for that as well in a modern hoppy recipe on the hot side too?

If my thinking is correct then hopefully this would make adjustment of kettle pH a much less frustrating experience.

 

[ajdelange]

Jt takes them into account in the sense that Kolbach doubtless obtained that number by measurement of beer which presumably was a typical one which would have been hopped at typical levels. But keep in mind that this number is only representative. Kunze gives instructions in his book on how to titrate a sample of the wort in order to calculate the amount of acid required to effect a desired pH shift. It is, of course, not necessary to do any calculation beyond adding drops of lactic acid to, say, 100 mL of wort until the desired pH is reached and then scaling the drops to the total volume being acidified.

 

[beervoid]

Jt takes them into account in the sense that Kolbach doubtless obtained that number by measurement of beer which presumably was a typical one which would have been hopped at typical levels. But keep in mind that this number is only representative. Kunze gives instructions in his book on how to titrate a sample of the wort in order to calculate the amount of acid required to effect a desired pH shift. It is, of course, not necessary to do any calculation beyond adding drops of lactic acid to, say, 100 mL of wort until the desired pH is reached and then scaling the drops to the total volume being acidified.

This is a good method but the problem with this is that after dry hopping you dont want to open the fermenter anymore. The best moment to add any acids would be together with the dry hops which makes it difficult to calculate how much we need.
I'm currently keeping track of where my PH goes in the end product and try to find a good amount to acidify the wort when dry hopping.
I meassured my beer pre dry hop at 4.1ph and added 9oz of dry hops. My ph ended up at 4.36.
I've noticed at least with my particular yeast strain that if I make sure my ph is about 5 going into the fermenter my ph will be good at around 4.5 post dry hop and fermentation

 

[SanPancho]

The number i noted above is the cited number in studies/presentations i believe. .14/#/bbl. cryo hops are reported anecdotally to have less impact though, likely due to less vegetative matter. And if you add before fg is reached then obviously yeast will still do a bit on their own.

Dosing in the keg isn’t a big concern as long as it’s before you carb. Unscrew the prv while adding a little co2 pressure. Acid goes in the hole with syringe. Easy peasy.

But yes, knowing how much to add means acid goes with the hop charge and would be easier and safer. Your numbers dont seem to square with .14/#/bbl unless your yeasst weren’t finished. Or your batch size is large. At 5 gal thats the equivalent of 4.5#/bbl i believe. Lots of hop matter for such a minimal ph change. Not sure where the discrepancy lays.

 

[beervoid]

The number i noted above is the cited number in studies/presentations i believe. .14/#/bbl. cryo hops are reported anecdotally to have less impact though, likely due to less vegetative matter. And if you add before fg is reached then obviously yeast will still do a bit on their own.

Dosing in the keg isn’t a big concern as long as it’s before you carb. Unscrew the prv while adding a little co2 pressure. Acid goes in the hole with syringe. Easy peasy.

But yes, knowing how much to add means acid goes with the hop charge and would be easier and safer. Your numbers dont seem to square with .14/#/bbl unless your yeasst weren’t finished. Or your batch size is large. At 5 gal thats the equivalent of 4.5#/bbl i believe. Lots of hop matter for such a minimal ph change. Not sure where the discrepancy lays.

Thats a good trick! Noted and will try that next time.
From the 9oz 4oz is pellets and 2.5 (5oz) is cryo so perhaps that is the reason it didnt go up that much.

 

[SanPancho]

Ah, ok. Then thats just under 2#/bbl if we ignore the cryo. That’s actually in the ballpark then.

But my point if i wasnt clear is that i wonder if that .14/#/bbl figure is pretty solid or if its +/- 25% or something with wide variation. I would think itd be fairly constant since its thought to be the plant matter that causes the effect but it’s possible that it could vary a bit by hop variety, processing, crop year, etc.

And yeah, that syringe in the prv hole is pretty sweet. Picked that up here on the forum from somebody a while back. Works well for all kinds of stuff like acids, finings, even priming sugar+yeast if you mix it in water. Bought a large syringe at like 100ml for that kind of stuff.

 

[SanPancho]

so just tested the kolbach numbers on a 5.35 wort trying to preacidify to 4.5 for kettle sour. came in at 4.45. great prediction on unhopped wort. i assume for knockout ph purposes i'll just have to assume the ph creep from the hops and "overshoot" my pH a bit.

 

[Rob2010SS]

So, haven't heard of or dabbled in messing with boil pH yet.

What are the reasons to do this?
Is the pH target the same as mash pH targets?

 

[mabrungard]

There are reasons to keep pH a bit higher in the boil. DMS reduction occurs at a higher rate when pH is higher, say 5.4. Lowering pH after the boil, then enhances the yeast’s ability to reduce the beer pH and enhance the crispness of beer.

 

[Silver_Is_Money]

Generally in order to wind up with a beer that is at a healthy pH of 4.0 to 4.3 after fermentation is complete (and before carbonation) you need to be at a pH of roughly 5.0 to 5.3 post the boil. The yeast will then typically move the wort to roughly 4.0 to 4.3 pH by the end of the fermentation cycle.

Wort pH will naturally fall on its own during the boil by about ballpark 0.2 points (give or take a bit). So if you are at 5.45 pH pre boil (for example), you will likely measure ballpark 5.25 pH post the boil. But sometimes the yeast does not drop the pH to 4.0-4.3 from this yeast launching point, so many people like to help the yeast along by dropping the cooled post boil wort to 5.0 - 5.1 pH via acidifying the cooled wort right before pitching the yeast.

There is no longer any grist present pre or post boil, so I'm struggling to understand how any past history of any once extant grist buffer factor or of any initial grist weight consideration validly enters into the calculation of acid required to move the wort to pH 5.0-5.1 at this late stage in the game.

 

[Silver_Is_Money]

Example Scenario:

1) 6 gallons of cooled wort collected post boil
2) Cooled Pre boil wort pH = 5.5
3) Cooled post boil wort pH = 5.3
4) Target cooled post boil wort pH = 5.0 prior to pitching yeast

6 gallons x 3.7854 ~= 22.7 Liters
Moles of hydronium ions present (per Liter) for the initial post boil pH 5.3 wort = 10^-5.3 = 0.000005012
Moles of hydronium ions desired (per Liter) when at the target of pH 5.0 = 10^-5.0 = 0.00001

Moles of current Hydronium ion deficiency post boil (per Liter) = 0.00001 - 0.000005012 = 0.000004988
22.7 Liters x 0.000004988 moles/Liter deficiency = 0.000113228 moles of overall H+ deficiency

It seems to me that the goal is to supply sufficient acid to cover this overall amount of moles of H+ deficiency in order to move the wort from 5.3 pH to 5.0 pH before pitching the yeast. This should be a function of the molar concentration (or normality) of the acid.

 

[Big Monk]

Example Scenario:

1) 6 gallons of cooled wort collected post boil
2) Cooled Pre boil wort pH = 5.5
3) Cooled post boil wort pH = 5.3
4) Target cooled post boil wort pH = 5.0 prior to pitching yeast

6 gallons x 3.7854 ~= 22.7 Liters
Moles of hydronium ions present (per Liter) for the initial post boil pH 5.3 wort = 10^-5.3 = 0.000005012
Moles of hydronium ions desired (per Liter) when at the target of pH 5.0 = 10^-5.0 = 0.00001

Moles of current Hydronium ion deficiency post boil (per Liter) = 0.00001 - 0.000005012 = 0.000004988
22.7 Liters x 0.000004988 moles/Liter deficiency = 0.000113228 moles of overall H+ deficiency

It seems to me that the goal is to supply sufficient acid to cover this overall amount of moles of H+ deficiency in order to move the wort from 5.3 pH to 5.0 pH before pitching the yeast. This should be a function of the molar concentration (or normality) of the acid.

I have a more empirical idea.

About 3 years ago, I developed a spreadsheet based Sauergut calculator for my longtime collaborator The Beerery. He had been using Sauergut for a while and we wanted to incorporate that capability into our shared spreadsheet.

We went to Kunze, whose sections on biological acid are great and very current, yet based firmly in observations and values from Kolbach’s time.

The basic dosing information is as follows:

A biological acid solution with Lactic acid % of 0.8% will drop the pH of the mash by 0.1 when dosed at 60 ml/kg of malt used and 0.1 pH of the boil when dosed at 30 ml/kg of malt.

Just a quick note before equations and math and theorizing: you can’t treat Sauergut like Lactic acid in the normal sense, i.e. you can’t plug low acid percentages into a Lactic acid calculator and expect sensible values for Sauergut amounts. However, you can use the biological acid’s Lactic acid percentage and the above equations to reliably predict Sauergut amounts. Remember this as it may come in handy.

I can’t access the original equation right now but I will post it here when I can.

Move forward 3 years and The Beerery and myself were talking about how to help him with his automation and auto Sauergut dosing and we wanted to be able to enter 2 pH values, find the Δ and then give us the ml amount of Sauergut required. The original equation gave a Δ pH output.

So I did some algebra and reorganized the equation:

Sauergut (ml) =(((Δ pH/0.1)*(60*Total kg))/Sauergut Acid %)*0.008

The 0.008 factor corrects for values higher than the 0.8% in the above dosing rates.

Now full disclosure: I have never proposed or tried what I am about to theorize.

Let’s say that we would like to take a Grist pH of 5.75 down to 5.40 and we are using Sauergut with a Lactic Acid % of 1.8%. Our Grist consists of 5.67 kg of malt:

Sauergut (ml) = (((0.35/0.1)*(60*5.67))/0.018)*0.008 = 529.2 ml

Now remember when we said before that modeling Sauergut with Lactic Acid equations using normality and such don’t work well? Well we shouldn’t have that problem here because these dosing rates and this equation doesn’t factor that stuff in.

Here we enter the purely hypothetical as I’ve not tested or discussed this before.

Let’s say that for the same mash as above, we measure 5.40 pH in the mash and near flameout in the boil we measure 5.30 pH. We would like to use 88% Lactic Acid to drop the pH to 5.00 at flameout. We are going to change the equation slightly to reflect the 30 ml/kg above.

Lactic Acid (ml) =(((0.3/0.1)*(30*5.67))/0.88)*0.008 = 4.64 ml

Viola?!?!

Seems logical to me.

EDIT: Keep in mind this only works for Lactic acid and I did not notice AJ’s more robust treatment above. Take my post as an interesting tidbit. Ive been curious for some time if the Sauergut calcs could be used in the opposite direction so not an altogether worthless exercise after all if it works.

 

[ajdelange]

6 gallons x 3.7854 ~= 22.7 Liters
Moles of hydronium ions present (per Liter) for the initial post boil pH 5.3 wort = 10^-5.3 = 0.000005012
Moles of hydronium ions desired (per Liter) when at the target of pH 5.0 = 10^-5.0 = 0.00001

Moles of current Hydronium ion deficiency post boil (per Liter) = 0.00001 - 0.000005012 = 0.000004988
22.7 Liters x 0.000004988 moles/Liter deficiency = 0.000113228 moles of overall H+ deficiency

It seems to me that the goal is to supply sufficient acid to cover this overall amount of moles of H+ deficiency in order to move the wort from 5.3 pH to 5.0 pH

As has been explained to you dozens of times before acid requirement can be calculated from the difference in hydrogen ion concentrations as determined from the pH's ONLY when the solution has 0 buffering capacity. This is never true in aqueous solutions and is, therefore, not the case with wort. To determine how much acid is needed one must effectively titrate a sample of the wort between the initial pH and the desired pH as was explained earlier in this thread.

 

[Silver_Is_Money]

AJ, when Kolbach determined that nominally 0.29 grams of 88% lactic acid will move 1 Kg by 0.1 pH, what mash efficiency was he presuming, and does ones efficiency during the mash thereby matter?

I.E., if for a pH target of 5.1 we can derive that 10*290*0.948/90.8 = 30.278 and therefore the protons required to move a wort made with 1 kg of malt by 1 full pH unit = 30.278 mEq/kg•pH for the case of targeting 5.1 pH, does this mEq/kg•pH value have any relationship which correlates to ones nominal mash efficiency, such that if Kolbach's mash efficiency was nominally 85% when he initially derived this association and my mash efficiency is only nominally 73%, when using 30.278 mEq/kg•pH should this for my case be multiplied by 73%/85% and thereby be corrected to ~26 mEq/kg•pH?

Short version: Rather than using ones starting Kg's of grist, would it perhaps be better to somehow incorporate post boil and cooling specific gravity whereby to determine the worts presumed buffering capacity more precisely, and then from there determine the mL's of 88% lactic acid required to move the wort from its currently measured post boil and cooling pH to a post boil and cooling target of 5.1 pH?

And if there is any such mash efficiency correlation to buffering, would it not determine the buffering capacity characteristics of ones grist during the mash as well as for post mash?

 

[ajdelange]

AJ, when Kolbach determined that nominally 0.29 grams of 88% lactic acid will move 1 Kg by 0.1 pH, what mash efficiency was he presuming, and does ones efficiency during the mash thereby matter?

I may be having a senior moment here but I don't remember any conclusion like that from Kolbach. He noted in his "Enfluß.." paper that reduction of the alkalinity of the water (I assume that means mash and sparge) by a certain amount would result in a drop in knockout pH of a certain amount. Not reference was made in that paper to the nature of the mash itself. Given where he worked I assume this observation was made for typical German lagers.

And if there is any such mash efficiency correlation to buffering, would it not determine the buffering capacity characteristics of ones grist during the mash as well as for post mash?

I would assume that there is a correlation between mash efficiency and wort buffering but have no idea as to how strong it would be. It seems to me the best way to acidify wort is to do what Kunze recommends, i.e. measure the buffering of the wort (very easy to do) in ml/pH•Lactic acid and compute the lactic acid required for the adjustment accordingly.

 

[Silver_Is_Money]

I may be having a senior moment here but I don't remember any conclusion like that form Kolbach.

It came from post #2 in this thread.

I agree that the best way is to measure it, but few to none will be doing that.

 

[ajdelange]

Ah. You are attributing my interpretation of what Kolbach observed to Kolbach. As I said in the last post he found that a reduction of liquor alkalinity by a certain amount reduced knockout pH by a certain amount. At an assumed pH and assumed beer size it is easy enough to compute the amount of lactic acid required to remove a mEq of alkalinity. The qestion you should have asked is what efficiency, pH and water to grist ratio I assumed in coming up with the 0.29 number. This was Feb 2016 and as I can barely remember what I was up to in Feb of 2019 I'm afraid I have no idea as to what I assumed then. Fair assumptions would be those associated with nominal German Lager brewing. I should, of course, have been clearer.

 

[Silver_Is_Money]

Ah. You are attributing my interpretation of what Kolbach observed to Kolbach. As I said in the last post he found that a reduction of liquor alkalinity by a certain amount reduced knockout pH by a certain amount. At an assumed pH and assumed beer size it is easy enough to compute the amount of lactic acid required to remove a mEq of alkalinity. The question you should have asked is what efficiency, pH and water to grist ratio I assumed in coming up with the 0.29 number. This was Feb 2016 and as I can barely remember what I was up to in Feb of 2019 I'm afraid I have no idea as to what I assumed then. Fair assumptions would be those associated with nominal German Lager brewing. I should, of course, have been clearer.

Got it! Perhaps the next great move forward in precision for mash pH assistant software will be to incorporate a nominal mash efficiency percentage, or perhaps an anticipated end of the mash OG for the wort (pre-sparge). I've been thinking along the lines that if two people start with the same weight of (for example) Pilsner malt of the same brand and lot #, the same volume and mineralization level of initially identical mash water, etc..., but one of the two gets 85% as to mash efficiency and the other gets only 65%, the (for this example, basic with respect to a nominal mash pH target of 5.2 to 5.6) mEq's/L of acidity of the wort for the high efficiency mash might measure noticeably greater than that for the case of the well lower efficiency mash, simply because the mash water will have permeated into far more of the hidden confines of the grist for the case of high mash efficiency.

 

[Silver_Is_Money]

Short version: From a specific gravity reading one can back calculate to an "effective" grist weight, rather than the actual grist weight. And perhaps the case may be that the effective grist weight of a wort has greater correlation to buffering capacity than does actual grist weight.

 

[Silver_Is_Money]

There may even be a means whereby to link this new line of thinking into D. Mark Riffe's recent comment regarding his proposal of a buffer multiplier that reduces buffering by an initially estimated factor of 0.6X. Perhaps instead of being a fixed 0.6X, this multiplicative factor can be linked to a degree of variability which is associated with mash efficiency. The ideal case would be that his proposed multiplicative factor is one and the same as ones mash efficiency.

 

[Big Monk]

There may even be a means whereby to link this new line of thinking into D. Mark Riffe's recent comment regarding his proposal of a buffer multiplier that reduces buffering by an initially estimated factor of 0.6X. Perhaps instead of being a fixed 0.6X, this multiplicative factor can be linked to a degree of variability which is associated with mash efficiency. The ideal case would be that his proposed multiplicative factor is one and the same as ones mash efficiency.

You wouldn’t want to use Mash Efficiency. You’d want to use conversion Efficiency

You would find actual extract weight in kg and divide that by maximum extract weight in kg and maybe use that?

I’m not 100% how effective that would be. You need to be sure to apply this “Efficiency Factor” to the proper malt characteristic.

That’s if it’s even a valid concern.

 

[Silver_Is_Money]

You wouldn’t want to use Mash Efficiency. You’d want to use conversion Efficiency

You would find actual extract weight in kg and divide that by maximum extract weight in kg and maybe use that?

I’m not 100% how effective that would be. You need to be sure to apply this “Efficiency Factor” to the proper malt characteristic.

That’s if it’s even a valid concern.

Perhaps I'm confusing or conflating the two.

Could this be looked at from the simple perspective of "points", whereby if a grist has a maximum potential of 35.5 points per pound per gallon, and one computes via SG that one achieved 27 points, the factor would thus become (for this example) 27/35.5 = 0.76? Or alternately as the grists aggregate fine grind potential derived SG vs. actually measured SG?

I believe the metric equivalent to points per pound per gallon is termed as "Liter Degrees Per Killogram".