Water: a comprehensive ... by Palmer and Kaminski doubts

[nexy_sm]

Hi all guys,

I am opening this thread because of my lower intermediate knowledge in chemistry. I am reading a book about brewing water by Palmer and Kaminski, and there are always some doubts i would like to resolve. I hope the thread will help other readers one day.

At the moment I read a chapter about residual alkalinity (RA) and not everything was clear to me, so I found an article about that here:
http://braukaiser.com/wiki/index.php?title=Residual_Alkalinity_illustrated

There is written:
3Ca2+ + 2HPO42- <-> 2H+ + Ca3(PO4)2

If I understand things correctly, this means that 6 equivalents of Ca will give in a reaction 2 equivalents of H+?

Next they say than Kolbach discivered that 2 out of 7 Ca ions will react with phosphates and thus lower the ph. Ok, fine.
If I have a mash with 12.5 equivalents of Ca ions, that means that 3 equivalents will react with phosphates, and will lead to 2 equivalents of H+. This 2 equivalents of H+ will react with 2 eqs og HCO3- and thus lower the alkalinity for 2 eqs.

If I am right but obviousely not than the equation:
RA=Alk - (Ca/3.5 + Mg/7) needs something.

I hope somebody can explain me this. I would appreciate that.

Cheers,
nexy_sm

 

[ajdelange]

The equation for precipitation of calcium with phosphate from mash pH (where most of the phosphate is monobasic) is

10Ca++ + 6(HPO4-) + 2H2O --> Ca10(PO4)6(OH)2 + 14H+

Other salts can precipitate too but hydroxyl apatite is the least soluble and so determines whether the solution is saturated. The equation says that 20 mEq of Ca++ combine with 6 mEq of monobasic phosphate and 2 mmol of water to precipitate 1 mmol of apatite and release 14 mEq of protons. But this equation does not say anything about how many mmol of apatite will precipitate in a given situation and thus nothing about how many hydrogen ions will be released. To determine that requires solution of a rather complicated set of simultaneous non linear equations. The Kohlbach equation is based purely on empirical observations for typical lager mashes. The result is that when there is lots of phosphate present (as there is in a mash) each 35 mEq of calcium releases 10 mEq of protons, not the 14*35/20 that would be expected if each calcium precipitated. Thus it is clear that not all the calcium reacts.

 

[nexy_sm]

If I understand your answer, I my doubt was correct and has logic but things don't functionate like that, and Kohlbach only gives experimental results?

Best

 

[nexy_sm]

Hi all guys,

today I have new doubt from chapter Discussion of Malt Acidity and Alkalinity, p. 82.

In this chapter, the athors say:

In the case of a mash, the pH endpoint is the mash pH target, such as 5.4. Threfore, if a base malt has DI pH of 5.7, it is considered to be alkaline compared to target (ex. 5.4). The alkalinity of the malt is measured by titrating, i.e., adding measured amounts of acid or base to reach a set endpoint. As the acid or base is added, the pH of the solution will change as a function of the milliequivalents added. If you plot the change in pH as a function of acid or base additions (mEq), the slope of the curve is the substance's buffering capacity.
Therfore, the alkalinity or acidity of a substance is equal to the total change in pH multiplied by the buffering capacity.

Acidity/Alkalinity = (pHend - pH malt di)(Buffer Capacity)

a few sentences later stands:

This is very important: the acidity or alkalinity of a substance is defined by the change in pH multiplied by the substance's buffering capacity, or resistaance to pH change.

First of all I think that I don't understand completely the term baffering capacity. Namely, as I see the process, if you have a lot of alkalinity, ie., many bicarbonats are there and they are recombined with the added acid. In my opinon, though I am not the native speaker, buffering is something that happens actively, which means that there is no oposing to pH change, but there is quite a lot of bicarbonates to be spend. As an example, I would say it is analogue to say that when you spend 1000 $ from 10000000$, your account is oposing to it's change.

The main doubt (from the book), is:
If you have buffering capacity which is first derivative of pH change regarding added acidity, by their definition it will have the unit of [pH]/[acidity], but they say [acidity]/pH]. This means if you have a solution with pH X, and baffering capacity of Bx, then by multiplying Bx by acidity you added, you can get only pH change, but not the actual alkalinity value. I think that for calculatinf result alkalinity you must substract alkalinity chenge from start alkalinity value.

Best regards and thanks for the answer in advance

 

[ajdelange]

I think your confusion may stem from the assumption that bicarbonate is the only source of alkalinity in a mash. This is not so. Phosphates have alkalinity and malts have alkalinity. Alkalinity is defined as the amount of acid you must add to a unit amount of a substance to lower its pH from an intrinsic value (the value when the substance was presented to you) to a reference value. For example with water the intrinsic pH is the pH of the sample and the reference pH is the pH tow which the analyst chose to titrate when he measured the alkalinity. As potable water should contain no acids or bases other than carbonate and bicarbonate the alkalinity is approximately equal to the bicarbonate/61 (mEq/L). Note that when mashing one does not lower pH to the lab's alkalinity titration endpoint but to something in the range of desirable mash pH's. The water now has a different alkalinity because the reference pH is different. Water alkalinity WRT mash pH is typically about 80% of the alkalinity listed on the water report. Study of Fig. 22 on p 96 should help with this.

With malt the situation is exactly the same. The alkalinity is the amount of acid required to lower the pH from an intrinsic pH (the distilled water mash pH) and a reference pH (which is usually the desired mash pH). The curves on 21 on p92 present this kind of information. The titration curve of the base malt on figure 21 on p92 shows that to lower the pH of 1 kg of that malt from the DI pH (0 point on the curve) to pH 5.4 would be about 10 mEq/kg. Thus the buffering (the amount of acid required for a given change in pH for a unit of the malt is d(mEq/kg)/dpH ~ 10/-0.2 ~ -50 mEq/kg-pH. That is the average between those two values for infinitesimal change in pH the slope of the curve gives dMeq/dpH and those values are plotted in Fig. 20.

 

[nexy_sm]

Hi again,

ok I understood the part with Di Ph.
What I don't get now is Determining the Alkalinity of Water in the Mash, p.93.

First, the tittle itselft suggest that you have mash, which consists of water and malt and now somebody is interested in water Ph. But there is no water at this point any more, it is only mash. But ok, I realized in this chapter that alkalinity is more general term, and for different purposes 0 alkalinity has different absolute level of present carbonates and bicarbonates. If i understand correct, alkalinity of the water in this case is how much water should be titrated in order to reach optimal mash Ph?

But what I am not able to figure out for some time is this:

In general, the alkalinity if the water can be defined as the total millimoles per liter of carbonate species (Ct) multiplied bu the charge (mEq/mmol) as a function of pH.

In my opinion something is wrong here, cause mEq/mmol must always be constant. If Ca 2+ has for example 2 Eq/mol that can't be changed, but only mol/L of Ca 2+. The only thing that should change with pH is concentration. I would like if somebody can explain me this better. I hope when I understand this, I can continue with this chapter.

Thanks again and cheers

 

[ajdelange]

The key is Fig. 22 which shows the charge on 1 mmol of carbo (which is mmol carbonic plus mmol bicarbonate plus mmol carbonate) as a function of pH. At high pH all the carbo converts to carbonate so the charge is -2 mEq/mmol. At pH 8.3 the distribution of species is 1% carbonic, 98% bicarbonate and 1% carbonate for a total of 2*.01 + 1*.98 = - 1.0 mEq/mmol. At low pH all the carbo converts to carbonic with 0 charge. In trying to manage mash pH you need to know how many protons are needed per liter of water to bring it to a desired pH. This is simply the difference in charge between the sample pH and the desired pH multiplied by the number of moles of carbo. That can be found from the sample alkalinity (which is he amount of acid required to bring the sample to a reference pH, 4.5, in the ISO method) and the initial pH. If the sample pH is 8.3 (with charge = -1) and the end point titration pH is 4.5 (with charge = -0.015) then alkalinity = carbo*(1 - 0.015) and carbo = alkalinity/(1 - 0.015) or you can use the chart below to figure it out, Just find the curved line closest to the intersection of alkalinity and pH - interpolating between the pair that surrounds the intersection is even better. I offered this to John but he decided not to put it in the book. His explanation is a bit difficult to follow.

So while it is true that the charge on Ca++ stays the same the charge on the anion of an acid or of an amphoteric ion of an acid does not. It's charge depends on how fully it is deprotonated and that depends on the pH.