Miscellaneous ramblings with regard to Buffering Capacity (BC) expressed as mEq/Kg_pH vs. as a log (base 10)

[Silver_Is_Money]

Note of disclaimer: All of the following is mere speculation on my part pending verification, and I welcome independent verification and/or comment by my peers and superiors. Being mere speculation it is certainly open to being incorrect thinking in part or in total, but I'm tossing my ramblings out there for constructive/destructive comment despite the potential to be way off base in my rambling, since discovering what doesn't work eventually narrows the possibilities to what does work. Be constructive, and don't shoot the messenger.

1st Rambling: In previous of my posts I have pondered and failed to understand why peer reviewed brewing scientists express Buffering Capacity (BC) in terms of logarithms rather than in terms of mEq/Kg_pH, and I was at a loss as to how to equate log based BC's to mEq/Kg_pH based BC's. Here are links to two examples of BC's expressed as logs:

https://www.onlinelibrary.wiley.com/doi/epdf/10.1002/jib.286https://www.onlinelibrary.wiley.com/doi/epdf/10.1002/jib.447
I've come to the conclusion that malt buffering factors expressed in mEq/Kg_pH terms as common among us amateurs can be converted to their scientific brewing community log values as follows:

Lets assume a hypothetical Pilsner base malt with a buffering capacity of 40 mEq/Kg_pH.

-log(1/40) = 1.6021

The BC of this malt as expressed in log notation is 1.6021

2nd Rambling: Converting BC's expressed in log notation into BC's in mEq/Kg_pH terms:

10^1.6021 = 40.00

3rd Rambling: As they correlate to pH's, logs represent molar concentrations of H+ ions in solution. And logs are meant to be summable. I believe that at present when two or more malts are combined to form a "grist", the overall grist buffer is determined via multiplying weight times the buffering capacity for each component and then summing and lastly dividing by overall grist weight. But logs are meant to be summed, and may potentially be the more correct way to determine the overall BC of the aggregate grist. I present this example:

Assume a grist consisting of 5 Kg. of 40 mEq/Kg_pH BC base malt and 0.5 Kg. of 70 mEq/Kg_pH dark roasted malt. Total grist weight = 5.5 Kg.

1st, the old/easy no-logs way:

5 x 40 = 200
0.5 x 70 = 35
(200+35)/5.5 = 42.73 mEq/Kg_pH for the overall aggregate grist buffer

2nd, the proposed log based method:

BC_log base malt = -log(1/40) = 1.6021
BC_log deep roasted malt = -log(1/70) = 1.8451

1.6021 x 5 = 8.0105
1.8451 x 0.5 = 0.92255
(8.0105 + 0.92255)/5.5 = 1.6242
10^1.6242 = 42.09 mEq/Kg_pH for the overall aggregate grist buffer

Via summing the logs of the BC's as opposed to the BC's themselves (both with respect to percentage of grist weight) the overall aggregate grist acidity "may" prove to be more precise. Your thoughts and comments are appreciated.

 

[Silver_Is_Money]

Continuing with our 2 malts from above:

Let's presume that for each malt we now have 1 Kg.
Let's presume the pHDI of the base malt to be 5.7 as a typical first guess
Let's presume the pHDI of the deep roast malt to be 4.7 as a typical first guess

For the base malt:
(5.7-5.4) = mEq's/(40 x 1)
mEq's = 12 mEq's of acid required to be added to move 1 Kg. of base malt to pH 5.4

For the deep roasted malt:
(4.7-5.4) = mEq's/(70 x 1)
mEq's = -49 mEq's required to move 1 Kg. of deep roasted malt to pH 5.4, or in other words 49 mEq's of base to be added

log BC base malt (from above) = 1.6021
log BC deep roast malt (from above) = 1.8451

1/(1.8451-1.6021) = 4.1152

49 mEq's/12 mEq's = 4.0833

4.1152 ~= 4.0833 (with the difference due to guessing both malts pHDI's?)

Kg. for Kg. our deep roasted malt is about 4.1 times more caustic (or basic) than our base malt is acidic with respect to a target of pH 5.4.

The advantage of knowing the log based BC for each malt appears to be that you no longer need to know either the malts pHDI or the malts BC in terms of mEq/Kg_pH, as log based BC's appear to potentially implicitly contain all of this information within a tidy single term. IMHO this is potentially a huge advantage, and is likely why modern brewing scientists use log based BC's!

This is clearly an evolving miscellaneous ramble, and now there seems to be potential merit. All that remains is how to practically apply this knowledge. Again I call upon the assistance of my peers and superiors to help pick up this ball and run with it.

 

[Silver_Is_Money]

More miscellaneous rambling evolution:

We already know how to extract the implied mEq/Kg_pH from within the log based BC. The only question remaining is how to extract the implied pHDI from within the log based BC. That leaves us here so far:

Implied mEq/Kg_pH = 10^(log based BC)
Implied pHDI as derived from log based BC = ?

Lets move this train a step further down the track:

We know that the below formula is true for the mEq/Kg_pH based representation of what i will refer to as BC_conventional:
Delta_pH = mEq's/(BC_conventional x Kg_Grist)

But we now also know that:
BC_conventional = 10^(log based BC)

So therefore:
Delta_pH = mEq's/[10^(log based BC) x Kg_Grist]

And therefore:
mEq's = Delta_pH * [10^(log based BC) x Kg_Grist]

(Where Delta_pH = pHDI - pH_Target)
and (Where mEq's = the mEq's required to move malt or grist to target pH)

And lastly:

Implied pHDI = pH_Target + mEq's/[10^(log based BC) x Kg_Grist]

Lets test it:

Implied pHDI = 5.4 + 12/[10^1.6021 x 1]
Implied pHDI = 5.7

And lets test it again:

Implied pHDI = 5.4 + -49/[10^1.8451 x 1]
Implied pHDI = 4.7

Is that a wrap?

 

[Silver_Is_Money]

It appears that all one may need to know and retain (with respect to malts and unmalted grains, leaving water and water mineralization impacts aside for now) is an assigned and log based BC for each malt or grain (or perhaps, for utility) each malt/grain classification with respect to type, and in some instances (particularly caramel/crystal and Munich) also with respect to color (as either EBC or Lovibond). The BC's should perhaps (for further refinement and simplicity) be designated as BC+ and BC- whereby to classify malts/grains which are basic with respect to the chosen targeted ideal mash pH, and malts/grains which are acidic with respect to the desired target mash pH.

EDIT: But something is still missing. A mEq quantified malt acidity valuation appears to be a requirement for each malt/grain in addition to its log based BC, and that seemingly implies a requisite knowledge of pHDI. Or does it? Your assistance and comments are needed and welcomed. My miscellaneous ramblings seem to be running up against a brick wall.

 

[Silver_Is_Money]

Technically, in lieu of retaining a pHDI one could gather a list of malt/grain "acidities" to retain in conjunction with a log based BC. From the combination of log based BC and malt "acidity" a pHDI can easily be inferred whenever and if ever needed, as we have already seen accomplished.

The acidities for the two malts we have been looking at above would be '12' for the base malt, and '-49' for the deep roasted malt.

 

[Silver_Is_Money]

Of course, malt acidities can be played with as logs also. And this potentially opens new avenues to be explored.

-log(1/12) = 1.07918 = the log of our hypothetical base malts mEq/Kg_pH requirement for added acidity whereby to hit 5.4 pH

-log(1/49) = 1.6902 = the log of our deep roasted malts mEq/Kg_pH requirement for added causticity whereby to hit pH 5.4

1.6902 - 1.07918 = 0.61102

10^0.61102 = 4.0833 (where have we seen this value before?)

49/12 = 4.0833 (indicating that 1 Kg of our deep roast is 4.083 times more acidic with respect to 5.4 pH than our base malt is caustic)

 

[Big Monk]

My miscellaneous ramblings seem to be running up against a brick wall.

Well Larry, the topic is very advanced. The whole concept of estimating pH with this level of detail is only really understood by the 1% of the 1%. Frankly, nothing beats actual measurements of the malts used in these scenarios and frankly, that's a ton of work for even the most devoted brewers.

Sometimes it's better to just use the best approximations you can and adjust. I'm not sure this topic really advances any of the charge conservation topic as much as confuses some of it, i.e. an advanced topic within an advanced topic is liable to make peoples eyes glazed over.

I always like reading this stuff, however, even if I am not really around all that much anymore.

 

[Silver_Is_Money]

... I'm not sure this topic really advances any of the charge conservation topic as much as confuses some of it, i.e. an advanced topic within an advanced topic is liable to make peoples eyes glazed over.

I always like reading this stuff, however, even if I am not really around all that much anymore.

I'm not sure that it advances anything either! But to me it seems as if there is at least the potential for advancement locked up within logarithms, else why would brewing industry level experts be utilizing them, as well as why is pH itself a logarithm based measure of acidity and/or causticity. That's why I've been clear about opening this up to those more mathematically advanced than myself who may have the ability to unlock the magic, if such exists. I clearly have not to date unlocked it, and perhaps I never will.

 

[Big Monk]

I'm not sure that it advances anything either! But to me it seems as if there is at least the potential for advancement locked up within logarithms, else why would brewing industry level experts be utilizing them, as well as why is pH itself a logarithm based measure of acidity and/or causticity. That's why I've been clear about opening this up to those more mathematically advanced than myself who may have the ability to unlock the magic, if such exists. I clearly have not to date unlocked it, and perhaps I never will.

I'll give it a look. One of the things I always struggled with when trying to make a hybrid model using color based proxies and charge conservation was the relationship between the color based acidities and buffering. Maybe this is something to factor in.

 

[Silver_Is_Money]

I've teased out another piece of the log based methodology puzzle as follows:

First, lets consolidate our logs established previously above in one place so we don't have to keep seeking for them.

1) For our hypothetical base malt with pHDI = 5.7 and mEq/Kg_pH = 12
log_BC = 1.6021
log mEq/Kg_pH = 1.07918 (to abbreviate, lets call it 'log_AV going forward, where AV = acid value)

2) For our hypothetical deep roast malt with pHDI = 4.7 and mEq/Kg_pH = 49
log_BC = 1.8451
log_mEq/Kg_pH = 1.6902 (again, to abbreviate, lets call it 'log_AV going forward, where AV = acid value)

New discovery:
log_Delta-pH = log_AV - log_BC

Test #1, base malt:
log_Delta-pH = 1.07918 - 1.6021
log_Delta-pH = -0.52292
10^-0.52292 = 0.299971504
Recall that 5.7 pHDI - 5.4 target = Delta_pH = 0.3
Bingo! (a match within trivial rounding error)

Test #2, deep roast malt:
log_Delta-pH = 1.6902 - 1.8451
log_Delta-pH = -0.1549
10^-0.1549 = 0.700003159
Recall that 5.4 - 4.7 pHDI = Delta_pH = 0.7
Bingo again! (a match within trivial rounding error)

 

[Silver_Is_Money]

I'm not quite sure what to make of this one yet. Is it another piece of the puzzle?

From above:
log_AV base malt = 1.07918
Log_AV deep roast = 1.6902

1.6902 - 1.07918 = 0.61102

1) For our hypothetical base malt:
5.7pHDI - 0.61102 = 5.08898

For grins:
1 - 0.61102 = 0.38898

2) For our hypothetical Deep Roasted malt:
4.7pHDI + 0.38898 = 5.08898

5.08898 = 5.08898
Coincidence? (to no less than 5 decimal places)

Or could it perhaps be that if 1 Kg. of each of these two malts are mashed together the resulting mash pH is going to be ~5.08898?

Hmmm???

For now I'm going with coincidence.....

 

[Silver_Is_Money]

Another piece of the puzzle:

If 1 Kg. of our base malt requires the addition of 12 mEq of acid whereby to reduce it from 5.7 pHDI to the target of 5.4 pH, then 2 Kg's. must require an addition of 24 mEq of acid, and 3 Kg's. must require an addition of 36 mEq of acid, etc... (wherein 12 mEq/Kg. x Kg. of our base malt = required acid mEq's to be added)

1) For 2 Kg's:
-log(1/24 mEq) = 1.38021
But Kg's can be converted to log format also, thus:
-log(1/2 Kg.) = 0.30103
And interestingly enough:
1.07918 + 0.30103 = 1.38021
1.38021 = 1.38021

2) For 3 Kg's:
-log(1/36 mEq) = 1.55630
But Kg's can be converted to log format also, thus:
-log(1/3 Kg.) = 0.47712
And interestingly enough:
1.07918 + 0.47712 = 1.55630
1.55630 = 1.55630

We had previously determined that 4.0833 Kg. of our base malt was the caustic offset equiv. of 1 Kg. of our acidic deep roast.

3) For 4.0833 Kg.'s:
-log(1/(12 mEq/Kg. x 4.08333 Kg.) = 1.690196
But Kg's can be converted to log format also, thus:
-log(1/4.0833 Kg.) = 0.61101
And interestingly enough:
1.07918 + 0.611014 = 1.690194
1.690196 ~= 1.690194 (with any difference merely due to rounding)

And lastly, we had previously established that Log_AV_Deep_Roast = 1.6902 (when rounded to 4 decimal places)

We have a match here for 4.08333... Kgs. of our base to 1 Kg. of our deep roasted when using logarithms.

 

[Silver_Is_Money]

Silly me, taking the negative logs of inverted values when all along I could have been taking the positive logs of straight-up values whereby to arrive at the very same log valuations.

log(40) = 1.0621 = the log of the buffering coefficient of our hypothetical base malt
log(70) = 1.8451 = the log of the buffering coefficient of our hypothetical deep roast malt.
log(12) = 1.07918 = the log of the mEq's of acid required to move 1 Kg. of our base malt from pHDI 5.7 to mash pH 5.4
log(49) = 1.6902 = the log of the mEq's of caustic required to move 1 Kg. of our deep roast malt from pHDI 4.7 to mash pH 5.4
etc...

Someone should clearly have reached across the internet and smacked me over this by now. This simplifies things greatly. Silly me! And to think that my long passed elderly Dad often told me: "Son, don't ever get old.". He was indeed wise.

 

[Silver_Is_Money]

Another small piece of the log based puzzle:

log(mEq's) = log(AV) = log_BC + log_Kg. + log_Delta-pH
(where Delta-pH = pHDI - pH_Target = 5.7 - 5.4 = 0.3 for the case of our base malt)

Example for 4.08333 Kg of our hypothetical Base Malt:
log(AV) = 1.6021 + log(4.08333 Kg.) + log(0.3)
Log(AV = 1.6021 + 0.6110145 + -0.522879
Log(AV) = 1.6902 (rounded)
10^1.6092 = 49 mEq

Recall that 1 Kg. of our deep roast malt also has a log(AV) value of 1.6902 (rounded)

Example for 1 Kg of our hypothetical Base Malt:
log(AV) = 1.6021 + log(1 Kg.) + log(0.3)
Log(AV = 1.6021 + 0 + -0.522879
Log(AV) = 1.0792 (rounded)
10^1.0792= 12 mEq

Note that the log of 1 Kg. of any malt = 0
log(1) = 0

So for the special case of 1 Kg. of any malt:
log(mEq's) = log(BC) + log(Delta-pH)
(I presume this should hold true for blends of malts summing to 1 Kg. total)

 

[Silver_Is_Money]

Another puzzle piece, this one being for the determination of an aggregate grists mEq's:

mEq_Grist = 10^(log_BC + log_Kg. + log_Delta-pH) (+ or -) 10^(log_BC + log_Kg. + log_Delta-pH) (+ or -) 10^(log_BC + log_Kg. + log_Delta-pH)........etc...(for each malt in the grist/recipe).
[where (+ or -) = '+' for malts requiring acid, and where (+ or -) = '-' for malts requiring caustic whereby to achieve the target mash pH]

1) Example using 2 Kg. of our hypothetical base malt and 1 Kg. of our hypothetical deep roast malt:
mEq_Grist = 10^(1.6021+log (2 Kg.)+log (0.3))−10^(1.8451+log (1 Kg.)+log (0.7))
mEq_Grist = -25

2) Example using 4.08333 Kg. of our hypothetical base malt and 1 Kg. of our hypothetical deep roast malt:
mEq_Grist = 10^(1.6021+log (4.08333 Kg.)+log (0.3))−10^(1.8451+log (1 Kg.)+log (0.7))
mEq_Grist = 0.00425*

*mEq here would be 'zero' if not for rounding.

NOTE: This puzzle piece isn't quite as neat and clean as I would like. I'll keep hammering away at it to discover a more elegant solution. Others are encouraged to assist.

 

[Silver_Is_Money]

Another coincidental fluke?

Base Malt for 1 Kg. : (1.60206 BC+log(1 Kg.)+log(0.3 Delta-pH)) = 1.07918
10^1.07918 = 12 = mEq's of acid required to hit pH 5.4

Base Malt for 2 Kg. : (1.60206 BC+log(2 Kg.)+log(0.3 Delta-pH)) = 1.38021
10^1.38021 = 24 = mEq's of acid required to hit pH 5.4

Base Malt for 3 Kg. : (1.60206 BC+log(3 Kg.)+log(0.3 Delta-pH)) = 1.55630
10^1.55630 = 36 = mEq's of acid required to hit pH 5.4

Base Malt for 4.08333 Kg. : (1.60206 BC+log(4.08333 Kg.)+log(0.3 Delta-pH)) = 1.6902
10^1.6902 = 49 = mEq's of acid required to hit pH 5.4
--------------------------------------------------------------------------
Deep Roast Malt for 1 Kg. : (1.8451 BC+log(1 Kg.)+log(0.7 Delta-pH)) = 1.6902
10^1.6902 = 49 = mEq's of caustic required to hit pH 5.4

It turns out that for the strange world of logarithms, just as for the case where multiplication and division are accomplished via addition and subtraction, addition and subtraction are accomplished via multiplication and division. Lets give it a try.

(1.60206+log(1 Kg.)+log(0.3 Delta-pH))/(1.8451+log(1 Kg.)+log(0.7 Delta-pH))= 0.638494

10^0.638494 = 4.35

Well the number 4.35 isn't going to get us anywhere, so lets try something different.

0.638494 - 1 = −0.361506
5.4 pH + -0.361506 = 5.04, and this appears to potentially be the mash pH when 1 Kg. of base malt and 1 Kg. of deep roast are mashed together.

So lets continue...

(1.60206+log(2 Kg.)+log(0.3 Delta-pH))/(1.8451+log(1 Kg.)+log(0.7 Delta-pH))= 0.816597

0.816597 - 1 = -0.183403
5.4 pH + -0.183403 = 5.22, and this appears to potentially be the mash pH when 2 Kg. of base malt and 1 Kg. of deep roast are mashed together.

(1.60206+log(3 Kg.)+log(0.3 Delta-pH))/(1.8451+log(1 Kg.)+log(0.7 Delta-pH))= 0.92078

0.92078 - 1 = -0.07922
5.4 pH + -0.0.7922 = 5.32, and this appears to potentially be the mash pH when 3 Kg. of base malt and 1 Kg. of deep roast are mashed together.

(1.60206+log(4.08333 Kg.)+log(0.3 Delta-pH))/(1.8451+log(1 Kg.)+log(0.7 Delta-pH))= 1

1 - 1 = 0
5.4 pH + 0 = 5.40, and this appears to be the mash pH when 4.08333 Kg. of base malt and 1 Kg. of deep roast are mashed together.

Coincidence? (or) Potentially on to something here?

 

[Silver_Is_Money]

An observation and query. There isn't a lot of 'free' peer reviewed brewing science documentation regarding BC expressed as logs, but for that which I've uncovered, not a single source expresses BC as a function of grist weight in KG, such as is done at the amateur level. The actual method used by brewing scientists seems to involve titrating sample volume at some normalized specific gravity or Plato post separating and filtering the Wort from the malt/unmalted grist which birthed it. In short, buffering coefficients in the brewing industry (mainly as it existed pre micro-breweries) are an expressed function of the Wort as isolated from the grist, whereas for us homebrewers it is an expressed function of the malt or unmalted grist itself. I wonder if there is a connection here to D, M. Riffe's observation that in order to make grist weight derived BC's function more in correlation to observation one must introduce a compensating (as in correcting) factor which he has initially quantified as 0.6. Hmmm???? Is the amateur level method of mEq/Kg_pH based BC somehow flawed in comparison to mEq/L_pH_SG based BC?

 

[Silver_Is_Money]

Expanding upon my above post #17:

Even peer reviewed literature which references buffering and buffering coefficients without mention of logs explains that they are being derived from titrating a specified volume of Wort at a certain specific gravity. Again, no connection to the initial weight of malt utilized in the mash comes into play in the formula.

One document simply defined the mL's of 0.1N titrant (meaning here, either a strong acid or strong base) required to move 1 Liter of 1.040 SG Wort 1 full unit of pH to be the malts buffering coefficient. Another document did likewise, but called for normalization to 1.060 SG for buffering coefficients. There appears to be a certain rogueness which I feel derives from a lack of industry standardization.