How much Mash pH prediction error does ignoring Mash water volume actually induce?

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Larry Sayre, Developer of 'Mash Made Easy'
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In another current thread there is a concern that Mash pH prediction should be measurably different with changes in mash thickness (mash water volume). This is in direct response to that question.

For the case of Mashing in 15L of Distilled water vs. Mashing in 30L of Distilled Water:

1) mEq's of Acid required to move 15 Liters of water from pH 7 to pH 5.40

10^-5.40 = 0.000003981 moles of H+ per Liter
0.000003981 x 1000 = 0.003981 mmoles H+ per liter
Charge for H+ = 1
Therefore 0.003981 mmoles/L = 0.003981 mEq/L
0.003981 mEq/L x 15L = 0.05972 mEq

It takes 0.05972 mEq's of Acid to move 15 Liters of Distilled Water from pH 7 to pH 5.40

88% Lactic Acid at pH 5.40 has an Acid Strength of 11.451 mEq/mL

Therefore:
0.05972 mEq ÷ 11.451 mEq/mL = 0.00522 mL of 88% Lactic Acid to move 15L of Distilled Water to pH 5.40

2) For brevity let's cut right to the chase and logically admit that it takes 0.05972 x 2 = 0.11944 mEq's of Acid to move 30L of distilled water to pH 5.40 pH

That is 0.00522 x 2 = 0.01044 mL of 88% Lactic Acid required to move 30L of Distilled Water to pH 5.40

3) Since MME is ignoring water volume in the Mash it is predicting an 88% Lactic Acid addition to be 0.00522 mL lower than is actually required for the case of Mashing in 15L of distilled water, and 0.01044 mL lower than is actually required for the case of Mashing in 30L of distilled water.

If we presume a 5 Kg. grist with a DI Mash pH of 5.70, and a Buffering Capacity (BC) of 35 mEq/Kg.pH, how far off will MME's Mash pH prediction be for the case of ignoring the acid required to acidify the water itself, both for the case of Mashing in 15L and 30L?

Let's calculate how many mEq's are required to move the grist alone (I.E., ignoring the mash water volume) from pH 5.70 to pH 5.40

(5.70 - 5.40) = mEq's/(35 x 5)
0.30 = mEq's/175
mEq's = 0.30 x 175
mEq's = 52.5
52.5 mEq's ÷ 11.451 mEq's/mL = 4.58475 mL of 88% Lactic Acid required to be added
(This is how MME would calculate it)

But for Mashing in 15L MME is 0.00522 mL short of 88% Lactic Acid, and 4.58475 + 0.00522 mL of 88% Lactic Acid is actually required. That equals 4.58997 mL as the correct 88% Lactic Acid addition.

And for Mashing in 30L MME is 0.01044 mL short of 88% Lactic Acid, and 4.58475 + 0.01044 mL of 88% Lactic Acid is actually required. That equals 4.59519 mL as the correct 88% Lactic Acid addition.

So we have MME telling us to add 4.58475 mL of 88% Lactic Acid vs. 4.58997 mL and 4.59519 mL respectively.

Ask yourself how accurately you can add 88% Lactic Acid as to mL decimal places.

Then ask what the maximum magnitude of MME's pH prediction error is:

The maximum error would be for the case of Mashing in 30 Liters, for which MME computes a deficit of 0.11944 mEq's as to its prediction output, as seen in #2 above.

Maximum_pH_error_MME = 0.11944/(35 x 5) = 0.0006825 pH points for the specific case of a 5Kg. grist with a pHDI of 5.70 and a BC of 35 mEq/Kg.pH

So by ignoring 30 Liters of water completely an output error of 0.0006825 pH is created. That would not likely even show up when rounding pH prediction to 2 decimal places.

Clearly in a buffered grist mash system the distilled water impact upon pH can be fully ignored, as can water volume doubling or halving as for Mash plus Sparge vs. No-Sparge.
 
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We said (within the other thread) that in a non buffered system doubling or halving acidic (or basic) water volume (or mash thickness) changed its pH by +/- 0.301 pH points.

In a 35 mEq/Kg.pH buffered system (of malts and grains evolving Wort into distilled water) the same degree of volume (or mash thickness) induced change is on the order of only 0.0007 pH points. This is the measure of the resistance to pH change that common grist buffering provides.
 
I shouldn't have jumped the gun in the 1st post above without first mentioning that:

Delta-pH_Mash = mEq's/(BC x Kg.)

Where:
1) Delta-pH_Mash = (pHDI - Mash pH Target)
and where: (pHDI = the mash pH as measured in deionized or distilled water)
2) KG. = Grist weight in Kilograms
3) mEq's = milliequivalents of added acid or base
4) BC = mEq's/Kg.Ph = the computed mEq's required to nominally move 1 Kg. of grist by 1 pH point for the specific case of the presumption of a linear titration.*

Trouble in River City:
* We know that Wort titrations are not factually linear. But if the BC is initially computed via careful titration to the very same pH as for the desired Mash pH target , then a functional linearity 'equivalent' is thereby achieved.

Example: We titrate 1 Kg of grist from a pHDI of 5.70 to a pH_Target of 5.40, and find that the mEq's of acid added is 10.5 mEq's. What is this grists (or malts) BC?

Delta-pH_Mash = mEq's/(BC x Kg.)
(5.70 - 5.40) = 10.5/(BC x 1)
0.30 = 10.5/BC
0.30(BC) = 10.5
BC = 35 (in units of mEq/Kg.pH) [The number of mEq's computed to move 1 Kg 1 pH point]

We can thus quasi-presume from this BC computation (by definition) that had we actually added 35 mEq's of acid to 1 Kg. instead of the 10.5 mEq's that we did add our resulting Wort pH would be measured at 4.70, or precisely 1 pH point lower than for pHDI. Yet (due to titration non-linearity) when we actually do add precisely 35 mEq's of acid to 1 Kg. of this same grist we are not very likely to subsequently measure the Wort pH to be precisely 4.70 as expected, albeit that since we only care about getting to pH 5.40 in the mash we can completely ignore any such error. But (and this is a big 'but') we can only ignore BC induced error for a BC computed initially via careful titration to pH 5.40 combined with subsequently targeting our Mash to hit the very same pH of 5.40.

The moral of the story is that if you are going to establish a mEq/Kg.pH units derived BC for each and every malt or grain or adjunct that exists today you had better not do so via titrating with an arbitrary fixed addition of mEq's of Acid or Base. Or via titrating to any pH other than 5.40. BC and targeted Mash pH are joined at the hip. They are in fact inseparable.

The major fly in the BC ointment is that many (if not factually most) of our historically available BC values as computed for us in the past have been computed via targeting to no specific target pH at all (via adding merely a fixed number of mEq's to each malt, etc... and accepting strict pH linearity in the face of factual non-linearity), or by targeting to pH 5.6, or 5.7, etc...(which is clearly not good for anyone subsequently targeting 5.40 pH in the mash). If any BC was not established to pH 5.40, then such a BC (as for values we find in the books) is assuredly going to be grossly in error, and such error subsequently finds its way into any mash pH prediction software which makes the presumption of BC_pH_5.40. Attempts whereby to adjust such deviant BC's to a standardized 5.40 pH are often dubious at best. And as a consequence, much existing BC valuation data for malts, grains, and adjuncts is thus highly questionable as to its viability as to achieving a mash pH target of 5.40.
 
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Interesting. Now apply this to a water with an alkalinity of say 200ppm as CaCO3 as opposed to distilled water.
 
Interesting. Now apply this to a water with an alkalinity of say 200ppm as CaCO3 as opposed to distilled water.

That's way too easy!

For 15 Liters:
200 mg/L ÷ 50.04345 mg/L = 3.99653 mEq/L
3.99653 mEq/L × 15L = 59.94795 mEq of Alkalinity (as CaCO3)

Remove 59.94795 mEq of Alkalinity and you are at pH ~4.30
Remove ~90% of 59.94795 mEq of Alkalinity and you are at ~pH 5.40

90% of 59.94795 mEq's = 53.9532 mEq's

Let's use 88% Lactic Acid, which has a strength of 11.451 mEq/mL at a pH of 5.40

53.9532 mEq ÷ 11.451 mEq/mL = 4.71 mL of 88% Lactic Acid to be added to bring the water to ~pH 5.40

And for 30 Liters:

4.71 mL x 30L/15L = 9.42 mL of 88% Lactic Acid to be added to bring the water to ~pH 5.40
 
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Agreed but what is the output error as calculated in your first post with an input water containing alkalinity as opposed to distilled water.
 
True, but what if the alkalinity isn't zeroed?

At pH ~5.40 it is not zeroed. That requires pH ~4.30.

HCO3-.png
 
What was meant by "accounted for" is that if the mash water alkalinity is sufficiently depressed to bring the mash water to 5.40 pH, and additional acid is added whereby to overcome the grist buffering and bring the Wort which is evolving from the grist to 5.40 pH, the alkalinity buffering is adequately accounted for. This is the essential basis for AJ deLange's sticky within this forum titled "The O Effective Alkalinity Method". The key lies within his use of the word "effective". Effective (such as for pH 5.40) does not imply definitive (as for pH 4.30).
 
Thank you. Very didactic.


But for the neophyte I am, where does the ~90% comes from ?
From the chart above (see where 0.10, or 10%, hits the HCO3- line at pH 5.4), and from this (which is the formula which draws the HCO3- curve seen on the chart):

Water's pH = 6.40 + log(mol fraction of HCO3- to H2CO3)

5.40 pH = 6.40 + log(mol fraction of HCO3- to H2CO3)

-1.00 = log(mol fraction of HCO3- to H2CO3)

10^-1.00 = mol fraction of HCO3- to H2CO3

mol fraction of HCO3- = 0.100 = 10.0%
mol fraction of H2CO3 = 1-0.100 = 90.0%

See also this thread:
https://www.homebrewtalk.com/thread...natural-water-via-its-ph.699663/#post-9254658
 
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