Version 2.0 of Palmer's Residual Alkalinity Spreadsheets

[Colin Kaminski]

Plus the extensive discussion at viewtopic.php?f=6&t=17060.

I'd guess the "correct" range of RA for stouts is something like 0 - 400 based on the diversity of waters with which stouts are brewed all over the world (see Lewis's book). My stouts come in between 70 and 80 SRM and are brewed with untreated water with an RA of about 50.

All of my beers are made with a RA of -60 to +60 ppm. Pale Ales around -33. I like to mash round beers in at 5.2 and hoppy beers at 5.4. I like Sulfate to Chloride ratios of 1/1 for round stouts and up to 9/1 for very crisp dry pales. The house pale is about 4/1. Mg to Ca ratios also influence flavor so be careful adding too much Gypsum without adding Epsom.

Here are some balanced equations for reducing alkalinity with acids that I will add so people can find them:

Acetic Acid = C2H4O2+Ca(HCO3)2>Ca(CO2H3)2+2CO2+2H2O
Hydrochloric Acid = HCl+Ca(HCO3)2>CaCl2+2CO2+2H2O
Phosphoric Acid = 2H3PO4+3Ca(HCO3)2>Ca3(PO4)2+6H20+6CO2
Sulfuric Acid = H2SO4+Ca(HCO3)2>CaSO4+2CO2+2H2O

I need to work out Lactic acid. Since I don't use it I keep forgetting to balance the equation. Remember adding acid will create CO2 and in large tanks you will need to either pump over or sparge with air to avoid crashing pH and damaging equipment. Be careful with Hydrochloric or Sulfuric as they can boil during dilution and cause accidents. Always add acid to water and never add water to acid. When using Sulfuric and Hydrochloric add acid slowly in an ice bath. Gloves and a splash shield are a great idea.

Calculating your water only gives a starting point. It takes me about 10 iterations of a recipie to get my water to taste right.

[ajdelange]

You need 1 acetic acid molecule for each bicarbonate ion:
2HAc + Ca(HCO3)2 --> CaAc2 + 2CO2 + 2H2O; Ac- = CH3COO-
also (as noted in the previous post)
2HCl + Ca(HCO3)2 --> CaCl2 + 2CO2 + 2H2O
which is enough to establish the pattern for monoprotic acids
2HA + Ca(HCO3)2 --> CaA2 + 2CO2 + 2H2O; A- = any singly charged anion for example lactic:
2HLac + Ca(HCO3)2 --> Ca(Lac)2 + 2CO2 + 2H2O; Lac- = CH3COCOO-

For a diprotic acid such as sulfuric you have
H2SO4 + Ca(HCO3)2 --> CaSO4 + 2CO2 + 2H2O
which is enough to broaden the general rule a bit further to
1. One equivalent of acid must be supplied for each equivalent of bicabonate
2. Each equivalent of bicarbonate eliminated (converted to CO2 and water) is replaced by 1 equivalent of the anion of the acid used.

For a polyprotic acid such as phophoric or citric things are a little different and depend on target pH. If you acidify to pH 5.2, for example, 98.89% of the phosphoric acid yields up 1 hydrogen ion converting to the monobasic anion H2PO4- but the remaining 1.032% yields 2 protons becoming HPO4--. As each proton "destroys" 1 bicarbonate the balanced equation would look like this:
2H3PO4 +1.095Ca(HCO3)2 --> 0.9889Ca(H2PO4)2 + 0.0206CaHPO4 + 2.190CO2 + 2.190H2O
which is approximately
2H3PO4 + Ca(HCO3)2 --> Ca(H2PO4)2 + 2CO2 + 2H2O
IOW, phosphoric, while it has 3 protons, behaves like a monoprotic acid if you use it to get to pH 5.2 (or lower). If you use it to go to say, pH 7, 60% yields 1 proton while 40% yields 2. The equation in the previous post shows all 3 protons released. This can only happen when the pH is close to 14.

This is true of phosphoric and citric acid because they have pK's close to the target pH values. It is not true of sulfuric acid because it's pK's are both appreciably less than the target pH range. Same for hydrochoric and lactic.

Details on how much acid is requird to neutralize a given amount of alkalinity (actually how much acid is required to set the pH of a water of a given alkalinity to a desired pH) can be found at http://www.wetnewf.org under Acidification of Water (note: I'm on the genny tonight as there has been substantial snow in the area and the connection to my ISP is going in and out so probably better to wait until tomorrow if you want to check this out).

[Colin Kaminski]

Thank you AJ. I built a spread sheet that graphs buffers and acids if you know the pKs. It gives me a seat of the pants feel for how many hydrogens are released etc. It generates the data for a given pH as well but I never know the exact pH for the whole process and what is going to happen. I'll check out your site later.

A question I have long had is if you are using phosphoric and it only gives up one hydrogen, what happens to the PO4 later in the brewing process? Will the Ca force off the other H so as to pake CaPO4?

PS without all the papers you have sent me over the years I would be clueless about water! Thank you! Thank you!

[ajdelange]

What happens when you add phosphoric is

H3PO4 --> f1*H3PO4 + f2*H2PO4- + f3*HPO4-- + f4*PO4--- + (f2 + 2*f3 + 3*f4)H+

For a ph of 5.2: f1 = 0.0007406; f2 = 0.9885396; f3 = 0.0107198; f4 = 6.9011E-10 so that a tiny portion remains phosphoric acid, 98.85% gives up a single proton and becomes the monobasic ion, 1% gives up 2 and becomes the dibasic ion and a teeny portion gives up 3 and becomes the tribasic phosphate ion (PO4---). Thus the answer to the question "what happens to the phosphate?" is "it becomes monobasic phosphate". Now if calcium is present in sufficient quantity it can coalesce with the PO4--- ions, few though they be, and precipitate hydroxyl appataite. If this happens, the equilibrium is upset and some HPO4-- gives up its proton to become PO4--- to restore it. This results in a deficit of HPO4-- and so some of the H2PO4- gives up a proton to become HPO4-- etc. The freed hydrogen ions lower the pH and this decreases f3 and f4 which is a second mechanism for restoring equilibrium.

This is the mechanism by which calcium, reacting with malt phosphate, lowers mash pH. At pH 5.2, however, f4 is so small that even though appatite is very insoluble quite a bit of calcium can be tolerated before precipitation occurs. For example, if you had water with alkalinity 200 ppm as CaCO3 at pH 7 and added enough phosphoric to reduce its pH to 5.2 you could tolerate as much as 216 mg/L calcium before precipitation occured.

[Colin Kaminski]

2H3PO4 + Ca(HCO3)2 --> Ca(H2PO4)2 + 2CO2 + 2H2O
IOW, phosphoric, while it has 3 protons, behaves like a monoprotic acid if you use it to get to pH 5.2 (or lower). If you use it to go to say, pH 7, 60% yields 1 proton while 40% yields 2. The equation in the previous post shows all 3 protons released. This can only happen when the pH is close to 14.

So when I add acid to hot liquor I get a pH of around 7.2ish. I precipitate lots of Calcium Phosphate (i know this because I have to clean it). Then I mash in to 5.2-5.4pH. So, since I have used up 40% of the second hydrogen how does the HPO4-- react with the malt? Then when I am done with fermentation I am around 4.2 pH. Is the residual phosphoric acid buffering my fermentation up by wanting hydrogens?

[mabrungard]

I am sceptical of the potential that adding phosporic acid to a calcium containing solution is going to precipitate out a calcium phosphate compound.

Typically, metalllic salts (such as Ca, Mg, Fe, etc) are solubilzed as the pH of the solution falls and can create precipitates when the pH increases. Could someone explain who this mash phosphate and calcium mechanism could occur when it flies in the face of every other metal precipitation reaction?

Colin, I appreciate your evidence of a precipitate, but have you had it analyzed? It would be much easier to believe that it is a CaCO3 precipitate that was created by driving off CO2 and causing the CaCO3 to drop out.

[Colin Kaminski]

The evidence I have is: that it has no flavor, it will not dissolve in water or phosphoric acid solutions, CaCO3 would be much easier to clean; and CIP with phosphoric or nitric acid would cut it quickly. Other than that I am just guessing. As for the mash, every collegiate text book on brewing claims that precipitating phosphate derived from the mash is the main buffer in setting mash pH. I take it on faith as I have not done my malt homework, yet.

[ajdelange]

Lets suppose Colin is working with water that comes from the mains at pH 8.3 with alkalinity and calcium hardness each equal to 100 ppm as CaCO3 (40 mg/L as the ion) and that he wants to lower the pH to 7.2. That would require 0.257 mEq/L H+. Even though f4 = 0.0000028 hydroxyl appatite is so insoluble that addition of even small amounts phosphoric acid in a solution containing this much calcium will be saturated and appatite is going to precipitate. The precipitation reaction is

6H3PO4 + 10Ca++ + 2H20 --> Ca10(PO4)6(OH)2 + 20H+

Thus each millimole of appatite precipitated requires 10 mmol (20 mEq) of calcium and releases 20 mEq of protons. We need 0.257 mEq/L H+ and can get all those by the precipitation of 0.257 mEq/L calcium which is 12.85 ppm as CaCO3. The amount of phosphoric acid consumed in this reaction is (6/20)*.257 mmol/L or (6/20)*.257*98 = 7.5 mg/L.

At this point we assume that our calcium hardness is now 100 - 13 = 87 ppm as CaCO3. At pH 7.2 that much calcium results in super saturation if the equivalent of .1 mg/L phosphoric acid is in solution but that much will remain in solution as the solution will be at saturation when at equilibrium. Now that little bit "extra" will supply extra protons by
H3PO4 --> f2H2PO4- +f3HPO4-- + f4PO4--- + (f2 + 2f3 + 3f4)H+
But these will not be significant relative to the protons from precipitation and we can ignore these. An exact solution is possible by iterative techniques.

With f2 = 0.5077914 and f3 = 0.4922015 that means that H2PO4- concentration will be 0.05 mg/L and HPO4-- concentration about the same. So to drop the pH of the hypothetical water to 7.2 would require 7.5 + 0.1 = 7.6 mg/L phosphoric acid most of which would "flow through" to precipitate calcium and only a small portion of which would remain in the water as the mono and dibasic phosphate ions. IOW the answer to Colin's question is that there isn't enough phosphate left in solution to worry about how it reacts. That aside, it reacts exactly as the inorganic phosphate from the malt reacts - it just adds to it. When the malt is added to the treated water the phosphate in it, which totally swamps the amount of residual phosphate from the water treatment, "flows through" to precipitate appatite as discussed thus lowering mash pH.

Martin: to convince yourself that this works either as a gedenken or actual experiment put about 1/4 gram of calcium chloride and 150 uL 6N NaOH in about 100 mL DI water. Ca(OH2) may form local to where the base goes in but stirring will dissolve that and you will have a clear solution. Now add 50 uL lab (85%) phosphoric acid. A precipitate will form. Add another 50 uL acid and it will dissappear.

It's late - hope I didn't bobble the numbers. If I did I'll come back and edit. In any case the principle stands.

[EDIT]After doing this excercise that the way Colin wrote the calcium bicarbonate/phophate reaction (i.e. with all 3 protons released) is indeed what happens when calcium is present in any appreciable quantity which is most of the time.