Water Report Help

[mcar1919]

Hello all. Longtime lurker, first-time poster here at hbt. First of all, a brief but gracious "thank you" to the many of you here that have thus far unknowingly helped expand my brewing knowledge immensely (Yooper, Revvy, Kaiser, deathbrewer, AJ Delange, and Martin Brungard just to name a few).

I have been brewing all-grain for a couple of years now with generally decent results, but have recently decided that I want to improve my processes and brewing by improving my water chemistry modifications. I apologize for posting yet another request for help interpreting a water report here in the forums, but I genuinely want to learn more on the subject and would greatly appreciate any guidance as it pertains to my water profile.

From what I understand about water chemistry so far, the average pH of my water (9.19) seems rather high to begin with. What are your thoughts on using half (or even entirely) RO water instead, or will adding some acid malt and/or some lactic/phosphoric acid be enough to salvage my tap water undiluted? Obviously I'm aware this will largely depend on the composition of my grain bill, but I generally brew light/pale styles--commonly with wheat--so getting my mash pH low enough has likely been an issue for many of my beers prior. I do have Yooper/AJ's stickied RO water primer bookmarked so I know I could use that as a fall back plan, but if my tap water can be utilized effectively with some slight modifications, obviously that would be a more desirable solution. I am open to using either Bru'n Water or the EZ Water Calculator, but have included screenshots of my data inputted into Bru'n Water for reference. I would greatly appreciate a second (or third) pair of eyes to make sure that I have inputted values correctly, used the Ion Concentration Conversion Calculator correctly where applicable, and that everything looks "normal." The ion balance is .02 meq/L so if I entered everything correctly, it at least balances out. One quick question I have is my water report doesn't include Nitrite anywhere that I can see, so I put the value at 0 in my Water Input Report. Does that make sense or is there another approximate value that you would recommend inputting instead?

I've read AJ Delange's posts elsewhere where he asserts that using acidulated malt at 1% of grist lowers the mash pH by ~ .1, and I think he/others have stated not to use it at a rate of much more than 7% of grist. Can someone confirm this? For the sake of round numbers, let's say I had a grain bill of 10 pounds. I would use 1.6 ounces of acidulated malt over and above my original 10 pounds to lower the mash pH ~.1, correct?

A couple specific questions I have regarding acidifying the sparge water on that page are 1. if I'm using Phosphoric Acid 10%, for example, does using 21 mL of it in ~4 gallons of sparge water sound correct? 2. I understand Phosphoric Acid is pretty flavor neutral, but am I going to get any undesirable side effects from using that much?

Any help/advice that you can offer would be greatly appreciated and welcomed. If there's any other information that you need/would like to help tailor your response, please let me know and I would be happy to provide it. (Sorry for the wall of text, and if you've made it this far, thank you!) Thanks again for all your help and I look forward to your responses.

 

[ajdelange]

From what I understand about water chemistry so far, the average pH of my water (9.19) seems rather high to begin with.

Water pH in itself is not a problem. What is a problem is the alkalinity of the water which at 97 is enough that you will need to deal with it. The only problem with high pH is that 9.7 is in the region where the simplified treatment of alkalinity used by the popular spreadsheets and calculators starts to break down. The errors are not that big, however.

What are your thoughts on using half (or even entirely) RO water instead,...

Using a 1:1 dilution with RO cuts the alkalinity in half, 2:1 reduces it to 1/3 and all RO renders it practically 0. All of these are good options for you.

...or will adding some acid malt and/or some lactic/phosphoric acid be enough to salvage my tap water undiluted?

The spreadsheets and calculators blithely tell you to just neutralize any alkalinity without emphasizing the important fact that every mEq of alkalinity removed this way is replaced by 1 mEq of the anion of the acid used. Brewers sometimes wind up with more acid_ate than they want from using the acid both to neutralize their water's alkalinity and their malts.

One quick question I have is my water report doesn't include Nitrite anywhere that I can see, so I put the value at 0 in my Water Input Report.

Good! You do not want nitrite in your water as it is toxic to yeast. It is OK to have a fair amount of Nitrate though and some waters contain quite a bit.

Does that make sense or is there another approximate value that you would recommend inputting instead?

Funny that they would report NO2- and not NO3-.

I've read AJ Delange's posts elsewhere where he asserts that using acidulated malt at 1% of grist lowers the mash pH by ~ .1, and I think he/others have stated not to use it at a rate of much more than 7% of grist. Can someone confirm this?

I can confirm that I stated this but I am parroting what is on Weyermanns website - not really asserting. It does make sense that 3% should not be objectionable as all German beers are going to contain about that amount or equivalent in Sauergut. I do theorize that as we can hear a 3 db sound level change (doubling in SPL) or see a 1 stop light level change (also 3db and, thus, a doubling) that our taste buds may work the same way and that we may find 6% Sauermalz noticeably more lactic. This, plus the fact that Weyermann has, also on their website, a recipe for Berliner Weiße that uses 8%, suggests that we probably ought to stay at 6% or below unless we want to taste lactate.

For the sake of round numbers, let's say I had a grain bill of 10 pounds. I would use 1.6 ounces of acidulated malt over and above my original 10 pounds to lower the mash pH ~.1, correct?

Strictly speaking it means that you should use 9 lbs 14.4 Oz of base and 1.6 Oz sauermalz but the way you have calculated it is fine. Also note that the 0.1pH/% is a rule of thumb. You will not get exactly that drop.

A couple specific questions I have regarding acidifying the sparge water on that page are 1. if I'm using Phosphoric Acid 10%, for example, does using 21 mL of it in ~4 gallons of sparge water sound correct? 2. I understand Phosphoric Acid is pretty flavor neutral, but am I going to get any undesirable side effects from using that much?

Your alkalinity is about 100 i.e. 2 mEq/L and you'll need about 90% of that or 1.8 mEq/L to get the water to mash pH. As 10% acid is about 1 N that means 1.8 mL of it to treat 1 L of the water. Four gal is about 15 L so that says 24 mL of the acid for 4 gal - about what you calculated. This goes back to the question raised earlier. You will be adding 24 mEq of phosphate to your beer plus the phosphate you will have to add to handle the alkalinity of your grains. This may be too much. As you say phosphate is pretty flavor neutral but IMO it would be better to get rid of the sparge (and mash) water alkalinity by other means.

 

[mcar1919]

Thanks for the thorough and prompt reply! The first follow up question I have is if pH and alkalinity can be calculated similarly to gravity points? i.e with pH, if I dilute my water 50/50 with RO I would have, say 5 gallons of my tap water at 9.2 (9.2 x 5 gal= 46) and 5 gallons of RO water at 7.0 (7.0 x 5= 35) and then (46 + 35= 81)/ 10 = 8.1 pH, or doesn't pH work that way?

Furthermore, your last paragraph, "Your alkalinity is about 100 i.e. 2 mEq/L and you'll need about 90% of that or 1.8 mEq/L to get the water to mash pH. As 10% acid is about 1 N that means 1.8 mL of it to treat 1 L of the water. Four gal is about 15 L so that says 24 mL of the acid for 4 gal - about what you calculated. This goes back to the question raised earlier. You will be adding 24 mEq of phosphate to your beer plus the phosphate you will have to add to handle the alkalinity of your grains. This may be too much. As you say phosphate is pretty flavor neutral but IMO it would be better to get rid of the sparge (and mash) water alkalinity by other means" goes way over my head. This is totally my problem not yours, but is there anyway you can explain how you landed on these values or refer me to a resource where I can read about it? The mEq and "1 N" are units of measure that I am unfamiliar with so that is probably further contributing to my trouble following along. Thanks again, I sure appreciate your response.

 

[ajdelange]

The first follow up question I have is if pH and alkalinity can be calculated similarly to gravity points? i.e with pH, if I dilute my water 50/50 with RO I would have, say 5 gallons of my tap water at 9.2 (9.2 x 5 gal= 46) and 5 gallons of RO water at 7.0 (7.0 x 5= 35) and then (46 + 35= 81)/ 10 = 8.1 pH, or doesn't pH work that way?

Not for pH. For alkalinity it does work that way. If you have 10 gal of water with alkalinity if 2 mEq/L and 5 of 1 mEq/L the mix will have alkalinity pretty close to (20 + 5)/15 = 1.667 mEq/L. Or if you take water with alkalinity 2 and dilute it 1:1 with RO (0 alkalinity) it goes down to 2/2. For a 2:1 dilution to 2/3, for 3:1; 2/4 etc.

Furthermore, your last paragraph ... goes way over my head. This is totally my problem not yours, but is there anyway you can explain how you landed on these values or refer me to a resource where I can read about it? The mEq and "1 N" are units of measure that I am unfamiliar with so that is probably further contributing to my trouble following along.

First let me show you how to save yourself some typing. At the lower part of each post's window is a button labeled "QUOTE". If you push it the text of the post you are looking at appears in the Reply window thus:

[QU0TE=ajdelange;7533947]
Your alkalinity is about 100 i.e. 2 mEq/L and you'll need about 90% of that or 1.8 mEq/L to get the water to mash pH......[/QUOTE]
Edit out the parts you don't want to quote. Copy and paste the bracketed bits to enclose other parts of a post.


Back to the regularly scheduled program:

Your alkalinity is about 100 i.e. 2 mEq/L

A mEq (milliequivalent) is a quantity of protons (hydrogen ions). The tough part of brewing water/mash chemistry is actually a relatively simple matter of keeping track of those protons. If you are going to make a mash with water that comes out of the tap at some pH much higher than you want the pH of the mash to be you will have to add protons to it. The alkalinity is a measure of the number of protons needed to get 1 L of your water to pH 4.5. It will take 2 mEq for each liter. When the lab measures your alkalinity they come up with 2 mEq/L but instead of putting that in the report they multiply by 50 and report the alkalinity as 100. This is handy for some purposes one of which is not brewing where this has confused more people than the Donald's campaign promises.

..and you'll need about 90% of that or 1.8 mEq/L to get the water to mash pH.

It will take 2 mEq/L to get your water to pH 4.5 but you don't want it at 4.5, you want it at mash pH (5.4 or so). The exact amount required to do this depends on the actual target pH and to some extent on the original pH of the water but it is going to be close to 90% of the alkalinity.

As 10% acid is about 1 N that means 1.8 mL of it to treat 1 L of the water.

The protons needed to lower the water pH have to come from somewhere and one of the candidate somewheres is a bottle of phosphoric acid. To figure out how many mL to add we need to know how many protons are in a mL of the stuff in the bottle. The "normality" of an acid is that quantity. Ten percent phosphoric acid is 1.09N at mash pH and thus each mL supplies 1.09 mEq protons.

Four gal is about 15 L so that says 24 mL of the acid for 4 gal - about what you calculated. This goes back to the question raised earlier. You will be adding 24 mEq of phosphate to your beer plus the phosphate you will have to add to handle the alkalinity of your grains.

The rest of the story is the acid requirement for the grains. They have to be acidified to mash pH too. For a typical light beer mash with 4 gal water the acid requirement might be 45 - 50 mEq. This means your beer would have a total phosphate content of 24 + 45 plus the amount from the sparge. This might or might not be too much but if you use 1:1 RO, for example, then the waters contribution of phosphate is halved (because, if you remember back to where we came in, the alkalinity is cut in half and the proton requirement to neutralize it is cut in half).

 

[mcar1919]

First let me show you how to save yourself some typing. At the lower part of each post's window is a button labeled "QUOTE". If you push it the text of the post you are looking at appears in the Reply window thus:

I simply copy/pasted your original quote rather than retype the entire thing, but thanks for the tip nonetheless!

A mEq (milliequivalent) is a quantity of protons (hydrogen ions). The tough part of brewing water/mash chemistry is actually a relatively simple matter of keeping track of those protons. If you are going to make a mash with water that comes out of the tap at some pH much higher than you want the pH of the mash to be you will have to add protons to it. The alkalinity is a measure of the number of protons needed to get 1 L of your water to pH 4.5.

Awesome explanation here; thanks! So protons are essentially acidic, or at least are used to neutralize the alkalinity for our purposes? Out of curiosity, any idea why somebody decided a 4.5 pH would be the magic number that alkalinity would be based off of? It would seem to me that returning the water to a neutral 7.0 would make more sense, but I'm sure there is a reason for it.

It will take 2 mEq for each liter. When the lab measures your alkalinity they come up with 2 mEq/L but instead of putting that in the report they multiply by 50 and report the alkalinity as 100.

You are talking about my specific scenario here, correct? I believe that you just rounded the average value of my water's alkalinity to 100 (from 97) for the sake of round numbers? The more precise value would be 1.94 mEq (97/50), correct? Just wanting to make sure I'm understanding.

It will take 2 mEq/L to get your water to pH 4.5 but you don't want it at 4.5, you want it at mash pH (5.4 or so).

This part makes perfect sense to me now (assuming my above assertions were correct).

The exact amount required to do this depends on the actual target pH and to some extent on the original pH of the water but it is going to be close to 90% of the alkalinity.

This part confuses me a little bit again. It seems to me like you have been saying that alkalinity and pH have a direct correlation, but in saying that, "to some extent on the original pH of water" also affects the mEQ/L of acid/protons required to neutralize the alkalinity, it seems to contradict that. Is anything below a neutral pH of 7.0 considered to have negative alkalinity then since the water is technically acidic at that point?

Also, I know you made mention of 90% of alkalinity (and resulting 1.8 mEq/L since that is 90% of the 2.0 mEq/L, of course) in your previous post as well, but I don't quite follow where that 90% came from in the first place or how that translates to a pH value. I need to neutralize/dilute 90% of the alkalinity away to get the water down to mash pH levels, must be what you are getting at? I assume the acidity of the grains is going to help bring the mash pH down over and above any acid treatment though too, correct? I know that the target mash pH we are shooting for is in that 5.3-5.4 range and sparge water ideally at 5.8-6.0 or so, but what does that equate to in terms of mEq/alkalinity? If 2 mEq/L of acid/protons would bring the water down to a pH of 4.5 from where it stands at 9.2, then 1.8 mEq/L of protons would have to bring it down to 5.0 or slightly under still I would think? This is where I'm getting lost, I think, when we are flipping back and forth between different units of measure from alkalinity/mEq to pH; there must be some formula that I am missing or something here to convert one to the other or vice-versa

The protons needed to lower the water pH have to come from somewhere and one of the candidate somewheres is a bottle of phosphoric acid. To figure out how many mL to add we need to know how many protons are in a mL of the stuff in the bottle. The "normality" of an acid is that quantity. Ten percent phosphoric acid is 1.09N at mash pH and thus each mL supplies 1.09 mEq protons.

So here you just rounded down to 1.0N from 1.09N (again for the sake of round numbers since the .09 must be relatively insignificant)? Therefore we needed 1.8 mL of phosphoric acid per L of water treated since each mL of phosphoric acid contributes slightly over one proton/liter (N), correct? If we used, say, hydrochloric acid (not saying I would use this, just for the sake of demonstration) at 2.0N, then would we only need .9 mL of it per L of water treated or am I misunderstanding the relationship between mEq/L and normality/N?

The rest of the story is the acid requirement for the grains. They have to be acidified to mash pH too. For a typical light beer mash with 4 gal water the acid requirement might be 45 - 50 mEq. This means your beer would have a total phosphate content of 24 + 45 plus the amount from the sparge. This might or might not be too much but if you use 1:1 RO, for example, then the waters contribution of phosphate is halved (because, if you remember back to where we came in, the alkalinity is cut in half and the proton requirement to neutralize it is cut in half).

I don't follow where the, "45-50 mEq" of acid required for a typical light beer mash comes from? I know that lightly kilned malts don't contribute as much acidity as darker ones, but they still should contribute some acidity, I thought, just not as much? Also, the phosphate content of 24 that you allude to, shouldn't it be 27 (1.8mL x ~15L =27)? Otherwise I'm really lost there too. I get that my alkalinity would be halved if I diluted 1:1 with RO water to approximately 48.5 (or .97 mEq), but what does that leave me with in the way of pH? I know you mentioned in a previous post as well that alkalinity can be calculated in the same method as gravity points, but pH cannot be calculated this way. This obviously ties back into my earlier question about a formula of some sort that I must be missing to convert the two. What alkalinity and pH values do brewers normally start out with since my water profile is obviously an outlier from the "norm?" Thanks again for all of your help/answers. It is definitely starting to "click," but I'm not quite there yet!

 

[ajdelange]

Awesome explanation here; thanks! So protons are essentially acidic,

They are the stuff of acidity. An acid is defined, for our purposes, as something that gives up protons.

or at least are used to neutralize the alkalinity for our purposes?

And a base is defined as something that absorbs protons. Hence acids and bases neutralize each other.

Out of curiosity, any idea why somebody decided a 4.5 pH would be the magic number that alkalinity would be based off of?

When talking potable water the main source of alkalinity is dissolved bicarbonate ion, HCO3-. This is a base as it can absorb protons
HCO3- + H+ ---> H2CO3. The balance between HCO3- and H2CO3 in a solution depends on the pH. At pH 4.5 98.7% of the CO3 is in H2CO3 and the rest in HCO3-. Note that H2CO3 decomposes into CO2 gas and water
H2CO3 ---> CO2 + H2O. The CO2 leaves the solution. Thus pH 4.5 is a pH at which 98.7% of the CO2 has been removed. It is the value that the ISO has decided is the proper value for use in determining the alkalinity of potable water. Thus ISO alkalinity is the amount of acid required to remove 98.7% of the CO3 species from your water. Ward Labs, used by most of the people here, actually uses 4.4 (which goes further after that last bit).

It would seem to me that returning the water to a neutral 7.0 would make more sense, but I'm sure there is a reason for it.

At pH 7 you would have converted only 20% of the bicarbonate and thus not measured 80% of it.

You are talking about my specific scenario here, correct? I believe that you just rounded the average value of my water's alkalinity to 100 (from 97) for the sake of round numbers? The more precise value would be 1.94 mEq (97/50), correct? Just wanting to make sure I'm understanding.

Yes, that's correct.

This part confuses me a little bit again. It seems to me like you have been saying that alkalinity and pH have a direct correlation,

I'm actually saying just the opposite - that pH and alkalinity are almost independent of one another. I can make you water with alkalinity of 100 ppm (2 mEq/L) alkalinity at pH 6 and I can do it at pH 8.

...but in saying that, "to some extent on the original pH of water" also affects the mEQ/L of acid/protons required to neutralize the alkalinity, it seems to contradict that.

It will take 2 mEq of acid to change the pH of a water at pH 6 to 4.5 if it has alkalinity 2 and it will take 2 mEq to change the water with pH 8 to pH 4.5 if it has alkalinity 2 but it will take slightly different amounts of acid to change its pH to mash pH. That is what depends slightly on the original pH.

Is anything below a neutral pH of 7.0 considered to have negative alkalinity then since the water is technically acidic at that point?

Alkalinity is defined as the acid required to take a liter of water from whatever pH it comes into the lab to pH 4.5 (ISO). Therefore, only samples that present at pH < 4.5 would have negative alkalinity.

Also, I know you made mention of 90% of alkalinity (and resulting 1.8 mEq/L since that is 90% of the 2.0 mEq/L, of course) in your previous post as well, but I don't quite follow where that 90% came from in the first place or how that translates to a pH value. I need to neutralize/dilute 90% of the alkalinity away to get the water down to mash pH levels, must be what you are getting at?

That's it. The alkalinity, as we have noted, is the acid required to get to pH 4.5. The acid required to get to a higher pH is a bit less. There are formulas at
https://www.homebrewtalk.com/showthread.php?t=473408 that you can use to calculate how much less if you want to try that out.

I assume the acidity of the grains is going to help bring the mash pH down over and above any acid treatment though too, correct?

Partly. Colored malts are somewhat acidic with respect to mash pH but base malts are alkaline IOW you must supply acid to base malt to lower its pH to correct mash pH.

I know that the target mash pH we are shooting for is in that 5.3-5.4 range and sparge water ideally at 5.8-6.0

Actually the sparge water should be at the mash pH.

or so, but what does that equate to in terms of mEq/alkalinity? If 2 mEq/L of acid/protons would bring the water down to a pH of 4.5 from where it stands at 9.2, then 1.8 mEq/L of protons would have to bring it down to 5.0 or slightly under still I would think? This is where I'm getting lost, I think, when we are flipping back and forth between different units of measure from alkalinity/mEq to pH; there must be some formula that I am missing or something here to convert one to the other or vice-versa

There is no simple formula that connects pH and alkalinity. The equations that do connect them are at https://www.homebrewtalk.com/showthread.php?t=473408
The fundamental principle at work in a mash is that everything you put into it has a proton deficit with respect to the mash pH. If you have water with alkalinity of 2 mEq/L its proton deficit WRT nominal mash pH is about 1.8 mEq/L. That means you would have to make up that deficit (supply acid) to the extent of 1.8 mEq for each liter of water to bring it to mash pH 5.4. Weyermanns Pilsner malt has a deficit of about 9.6 mEq/kg. Again you must cover that deficit by supplying 9.6 mEq of protons for each kg of that malt you use to bring it to pH 5.4. Crisps 600L chocolate malt has a proton deficit of -55 mEq/kg. That means a kg of it would yield 55 mEq of protons to pH 5.4. Lactic acid has a deficit of -11.4 mEq/mL to pH 5.4. The name of the game is to use enough lactic acid and/or chocolate malt to cover the deficits of the water and the Pilsner malts. The sum of all the deficits is 0.

So here you just rounded down to 1.0N from 1.09N (again for the sake of round numbers since the .09 must be relatively insignificant)? Therefore we needed 1.8 mL of phosphoric acid per L of water treated since each mL of phosphoric acid contributes slightly over one proton/liter (N), correct?

It's 1 mEq (a very large number of protons) per milliliter.

If we used, say, hydrochloric acid (not saying I would use this, just for the sake of demonstration) at 2.0N, then would we only need .9 mL of it per L of water treated or am I misunderstanding the relationship between mEq/L and normality/N?

That is correct and many brewers in England do use HCl (or actually a blend of it and H2SO4)

I don't follow where the, "45-50 mEq" of acid required for a typical light beer mash comes from? I know that lightly kilned malts don't contribute as much acidity as darker ones, but they still should contribute some acidity, I thought, just not as much?

As noted above they don't contribute any. They require about 10 mEq/kg. That's where the 45 - 50 comes from. It's based on about 5 kg of malt in a brew.

Also, the phosphate content of 24 that you allude to, shouldn't it be 27 (1.8mL x ~15L =27)?

Yes. Getting kind of bleary eyed here but I think so.

I get that my alkalinity would be halved if I diluted 1:1 with RO water to approximately 48.5 (or .97 mEq), but what does that leave me with in the way of pH?

When you added the RO water did you add any acid or base? If not then the pH won't shift (it will in fact but only by a tiny amount from effects that we aren't considering here).

I know you mentioned in a previous post as well that alkalinity can be calculated in the same method as gravity points, but pH cannot be calculated this way. This obviously ties back into my earlier question about a formula of some sort that I must be missing to convert the two.

If you think of pH as a voltage (it is, in fact, related to the chemical potential of hydrogen ions) then 'current', the flow of protons, occurs from mash components with low pH to those with high pH until everything is at the same pH. You are looking for a simple law like Ohms law that says current is proportional to voltage difference. There is no Ohms law here. The 'circuit' is more like an electrical circuit that has capacitors in it (and this isn't a very good analogy even then).

What alkalinity and pH values do brewers normally start out with since my water profile is obviously an outlier from the "norm?"

Your pH is a little high to the extent that many of the popular calculators will be in error because they do not accurately model the carbonic acid system but other than that it is quite normal.

 

[mcar1919]

Thanks again for the outstanding reply! I apologize I am writing this on my phone at work as time allows, so I am not able to quote and format as nicely as I'd like to, but hopefully I'll get some time to look it over again this evening.

From a logic standpoint, it would probably make the most sense to acidify the entire volume of water down to 5.4 pH and then treat the further acidification of the mash water to account for the alkalinity of the grains as a separate calculation, would it not? I feel like I have seen you allude to this before, but now I can understand why.

To make sure I am understanding your point about balancing protons WRT the grain bill, let's assume that I have already acidified the strike water for the mash down to 5.3 or 5.4 pH and we are making a "chocolate pilsner" which I'm sure would be terrible in practice, but we'll roll with it. I'm going to over-simplify numbers again here for sake of ease, but let's say we used 8 pounds of pilsner malt at 9 mEq and 1 pound of chocolate malt at -55 mEq. We'd have a defecit of 72 (8 x 9) from the pilsner and surplus of 55 (1 x 55) from the chocolate malt for a remaining deficit of 17 that we need to make up. Using the lactic acid at 11 mEq/mL, we'd use 1.54 mL (17/11) to balance it out?

I did take a look at the other hbt thread that you referenced--along with the external website that appeared to be your personal one--and they are both very informative resources. The hbt thread that you linked to gives formulas for calculating bicarbonate and carbonate ion concentrations, and then how to correct them for the presence of phosphate, but I am totally lost as to how that can be used to connect pH and alkalinity for my purposes. Would it be possible for you to demonstrate use of the formulas to show where the 1.8 mEq/L required to neutralize the alkalinity down to 5.4 pH comes from so I can follow along with your work to see how you got there, and also possibly how to calculate what my pH would be if I were to dilute my water 1:1 with RO water, for example? I am pretty confident I will understand and be able to do the math, I just need to see how it's done.

 

[ajdelange]

From a logic standpoint, it would probably make the most sense to acidify the entire volume of water down to 5.4 pH and then treat the further acidification of the mash water to account for the alkalinity of the grains as a separate calculation, would it not?

Seems logical to me but there are those here that find it totally illogical though I cannot tell you why.

To make sure I am understanding your point about balancing protons WRT the grain bill, let's assume that I have already acidified the strike water for the mash down to 5.3 or 5.4 pH and we are making a "chocolate pilsner" which I'm sure would be terrible in practice, but we'll roll with it. I'm going to over-simplify numbers again here for sake of ease, but let's say we used 8 pounds of pilsner malt at 9 mEq and 1 pound of chocolate malt at -55 mEq. We'd have a defecit of 72 (8 x 9) from the pilsner and surplus of 55 (1 x 55) from the chocolate malt for a remaining deficit of 17 that we need to make up. Using the lactic acid at 11 mEq/mL, we'd use 1.54 mL (17/11) to balance it out?

Yes, you've got it!

I did take a look at the other hbt thread that you referenced--along with the external website that appeared to be your personal one--and they are both very informative resources. The hbt thread that you linked to gives formulas for calculating bicarbonate and carbonate ion concentrations, and then how to correct them for the presence of phosphate, but I am totally lost as to how that can be used to connect pH and alkalinity for my purposes. Would it be possible for you to demonstrate use of the formulas to show where the 1.8 mEq/L required to neutralize the alkalinity down to 5.4 pH comes from so I can follow along with your work to see how you got there, and also possibly how to calculate what my pH would be if I were to dilute my water 1:1 with RO water, for example?

OK. Lets assume your water has alkalinity 2 mEq/L and pH 9 and that your lab is measuring alkalinity to pHe = 4.4. Referring to the first two steps in the prescription of the linked post you should get, respectively, 0.0398097 and 0.0068355 (use 14.165 instead of 14 in step 2) to be deducted from the 2 mEq/L alkalinity. This small correction is so small it can usually be ignored. I don't know if any of the spreadsheets use it. IN Steps 3 and 4 we get the ratios of bicarbonate to carbonate (r1) and carbonate to bicarbonate (r2) at pH 9. They are r1 = 415.2519 and 0.0421146.

From the r's we compute (Step5) the fraction of the total carbo (the molar sum of carbonic, bicarbonate and carbonate moieties) which is uncharged carbonic acid at pH 9. This is f0 = 0.0023055. There is r1 times that in bicarbonate form (Step 6): f1 = 0.9754 and r2 times that as carbonic (Step 7): f2 = 0.0403. Carbonic acid is uncharged, bicarbonate ion has a single negative charge and carbonate ions a double negative charge. As f0 + f1 + f2 = 1 a singe mole of carbo at pH 9 has total charge -f1 - 2*f2 = -1.0380139 mEq/mmol.

Now we go back and run through the same steps but using pH = 4.4. You should come up with a charge of -0.0103230 mEq/mmol (or close to it) for pH 4.4. What this means is that if we have a liter of water with 1 mmol of carbo given to us at pH 9 and acidify it to pH 4.4, which is just what the analyst does, we will have to supply
-0.0103230 - -1.0380139 = 1.0276909 mEq acid per mmol carbo

to overcome the proton deficit of the carbo. But your lab used 2 - 0.0398097 - 0.0068355 mEq of acid for the carbo is a liter of your water so the carbo must be (Step 10)

C = (2 - 0.0398097 - 0.0068355)/1.0276909 = 1.90072 mmol/L

Now you go back and compute the charge on a mmol of carbo at pH 5.4 in exactly the same way. You should get -0.0944563. To shift a mmol of carbo to 5.4 from 9 would require

-0.0944563 - -1.0380139 = 0.9435577 mEq per mmol carbo. As you have 1.9 mmol of carbo it is going to require 1.9 times that equal to 1.793 mEq which is very close to 90% of 2.

So that's it.

..and also possibly how to calculate what my pH would be if I were to dilute my water 1:1 with RO water, for example?

The approach to solving that problem is a little different. You now know the composition of your water i.e. how many moles of carbo it contains. You can, from the charge computation, get the proton deficit between any two pH's in particular the pH (9) at which we assumed your water came to you and any nearby pH. You can also compute the proton deficits of hydrogen ions in water from the Step 1 formula and of hydroxyl ions (Sterp2). If you dilute 1:1 with RO you add more hydroxyl and hydrogen ions to the mix but the amount of carbo stays the same so what you must do is compute the proton deficits between pH 9 and some trial pH's close to 9 until you find one where the total sum of the deficits is 0. Using the formulas you will find this to be 8.9665. RO water is a very weak acid compared to water with 1.9 mmol carbo at pH 9 and so pH declines a little. I'm surprised it's this much.

 

[mabrungard]

Neutralizing the overall batch's alkalinity down to a level sufficient for sparging use is a logical approach. You can then add an additional dose of acid for those pale grists to meet mashing pH goals. However, this approach breaks down somewhat if the mash then needs additional alkalinity in order to avoid a lower than desirable pH. I would rather not add another salt to the water if I can avoid it.

 

[ajdelange]

You do it all gedanken and then if you find you need alkalinity in the mash you simply withhold the corresponding amount of acid from the portion of the water to be used for mashing or respond by taking the fact that you need this alkali as a signal that you have probably used too much high killed malt.

 

[mabrungard]

You do it all gedanken and then if you find you need alkalinity in the mash you simply withhold the corresponding amount of acid from the portion of the water to be used for mashing or respond by taking the fact that you need this alkali as a signal that you have probably used too much high killed malt.

Oh, come on. Then you would be treating two separate batches of water. Wasn't that the original point?

While I agree that you need to avoid excessive roast additions, there are beers that rely on and demand roast additions that exceed the water's ability to buffer the resulting mash pH. I do like Porters and Stouts that have roast components that I can taste!

 

[ajdelange]

Oh, come on. Then you would be treating two separate batches of water. Wasn't that the original point?

The original point is that you want to make the thought process and the actual treatment process simple both to understand and implement.

 

[mcar1919]

I just got a chance here tonight to sit down and run the numbers myself to see how it went.

OK. Lets assume your water has alkalinity 2 mEq/L and pH 9 and that your lab is measuring alkalinity to pHe = 4.4. Referring to the first two steps in the prescription of the linked post you should get, respectively, 0.0398097 and 0.0068355 (use 14.165 instead of 14 in step 2) to be deducted from the 2 mEq/L alkalinity.

I did get the .0398097 exactly for step 1 (bringing my alkalinity down to 1.96019), but for step 2 I got 0.0068389 using 14.165 in both instances where 14 was used, did you maybe use 14 in one and 14.165 in the other possibly (or 14 in both) or did I make a mistake? I have an adjusted alkalinity of 1.95335 mEq/L at this point.

IN Steps 3 and 4 we get the ratios of bicarbonate to carbonate (r1) and carbonate to bicarbonate (r2) at pH 9. They are r1 = 415.2519 and 0.0421146.

Again, I came up with a slightly different answer for r1 here, I got 416.869383 after doing 10^2.62 (9-6.38=2.62)

For r2, I got 0.0416869 after doing 10^-1.38 (9-10.38)

I'm sure these differences will prove insignificant in the big picture, but I'm somewhat concerned that I didn't get the exact same answers you did. What is the significance of the 6.38 and 10.38 in the equations out of curiosity?

From the r's we compute (Step5) the fraction of the total carbo (the molar sum of carbonic, bicarbonate and carbonate moieties) which is uncharged carbonic acid at pH 9. This is f0 = 0.0023055. There is r1 times that in bicarbonate form (Step 6): f1 = 0.9754 and r2 times that as carbonic (Step 7): f2 = 0.0403. Carbonic acid is uncharged, bicarbonate ion has a single negative charge and carbonate ions a double negative charge. As f0 + f1 + f2 = 1 a singe mole of carbo at pH 9 has total charge -f1 - 2*f2 = -1.0380139 mEq/mmol.

I got f0=0.0022975
f1=0.9577729
f2=0.0399296
EDIT: Q(pHs)= -1.0376321 (Not sure where I came up with the prior answer, but obviously this one is much closer to what you came up with)

I am still unsure how statistically significant the difference between the two answers we came to is, but I don't think it is a huge bunch of difference. Obviously the answers are different since we had calculated different r1 and r2 values prior, plus I'm sure we calculated out to different decimal places and rounded differently here and there.

Now we go back and run through the same steps but using pH = 4.4. You should come up with a charge of -0.0103230 mEq/mmol (or close to it) for pH 4.4. What this means is that if we have a liter of water with 1 mmol of carbo given to us at pH 9 and acidify it to pH 4.4, which is just what the analyst does, we will have to supply
-0.0103230 - -1.0380139 = 1.0276909 mEq acid per mmol carbo to overcome the proton deficit of the carbo. But your lab used 2 - 0.0398097 - 0.0068355 mEq of acid for the carbo is a liter of your water so the carbo must be (Step 10) C = (2 - 0.0398097 - 0.0068355)/1.0276909 = 1.90072 mmol/L

For the Q(pHe), I came up with a charge of -0.0103628; very, very close to what you got.

For step 10, I got C=1.9014974 from 1.95335/(-.0103628 - -1.0376321) (So this is calculating total carbo/L where carbo is a combination of carbic acid, bicarbonate, and a touch of carbonate?)

So assuming my numbers are correct (unlikely), if I was the analyst, I would have to supply -.0103628 - -1.0376321 = 1.0272693 mEq acid if there was 1 mmol carbo per liter, but since I calculated above that there is really 1.9014974 mmol carbo per liter of water, I would really have used 1.0272693 x 1.9014974 = 1.953349 mEq/L acid (which was my originally calculated, adjusted alkalinity) to hit a pH of 4.4?

Now you go back and compute the charge on a mmol of carbo at pH 5.4 in exactly the same way. You should get -0.0944563. To shift a mmol of carbo to 5.4 from 9 would require
-0.0944563 - -1.0380139 = 0.9435577 mEq per mmol carbo. As you have 1.9 mmol of carbo it is going to require 1.9 times that equal to 1.793 mEq which is very close to 90% of 2.

I'll do the math again for the desired pH of 5.4 later to see how I do, but I just wanted to double-check that I was on the right track with what I just did before I spend more time calculating that out. Since I didn't say it earlier, thank you for taking the time to type so much of that out; it was reassuring to make sure I was getting the same/similar values to you as I was progressing through it.

 

[ajdelange]

Again, I came up with a slightly different answer for r1 here,...

I'm sure these differences will prove insignificant in the big picture, but I'm somewhat concerned that I didn't get the exact same answers you did. What is the significance of the 6.38 and 10.38 in the equations out of curiosity?

Those are the pK's of carbonic acid. They are a function of temperature and are nominally those values at 20 °C but in the spreadsheet I use they are calculated from temperature and I am sure the reason your answers are different from mine is that my spreadsheet is computing slightly different values (i.e. not exactly 6.38 and 10.38).

For the Q(pHe), I came up with a charge of -0.0103628; very, very close to what you got.

That means that you have the hang of things.

For step 10, I got C=1.9014974 from 1.95335/(-.0103628 - -1.0376321) (So this is calculating total carbo/L where carbo is a combination of carbic acid, bicarbonate, and a touch of carbonate?)

There is a carbic acid, bicyclo[2.2.1]hept-5-ene-2,3-dicarboxylic acid, but what we have here in potable water is carbonic acid!

if I was the analyst, I would have to supply -.0103628 - -1.0376321 = 1.0272693 mEq acid if there was 1 mmol carbo per liter,but since I calculated above that there is really 1.9014974 mmol carbo per liter of water, I would really have used 1.0272693 x 1.9014974 = 1.953349 mEq/L acid (which was my originally calculated, adjusted alkalinity) to hit a pH of 4.4?

Yep.

I'll do the math again for the desired pH of 5.4 later to see how I do, but I just wanted to double-check that I was on the right track

Yes, you are. I hope you are putting this into a neatly labeled spreadsheet so you can see how the r's, f's and Q's change with different alkalinity and pH inputs.

 

[mcar1919]

Yes, you are. I hope you are putting this into a neatly labeled spreadsheet so you can see how the r's, f's and Q's change with different alkalinity and pH inputs.

I am definitely going to do that in the near future when I get some time. The somewhat frustrating part is that I will be "guessing" the alkalinity mEq/L value using the average value from the water report, but it probably won't make too big of a difference in most cases (hopefully). Now I just have to invest in a decent pH meter so at least one of my variables is exact! I will try and remember to post my spreadsheet on here to get your opinion and see if you have any other suggestions for it. Thank you again for all your help.

Oh, I also wanted to just quickly ask too what alkalinity and pH values you use for RO water, 0 and 7.0 or something like that respectively?