Why do we assume that Acid Malt drops mash pH by 0.1 points per % Acid Malt added to the grist?

[Silver_Is_Money]

The easy answer is that Weyermann told us that's what it will do, so that's what it will do (case closed). But they also mentioned that it is merely a simplified gereralization (or ballpark rule of thumb).

Let's look at an example:

Givens chosen for our example:
1) Grist weight is 6 Kg.
2) Grists aggregate (overall/blended) buffering factor is 32 mEq/Kg.pH (at 5.4 pH).
3) Our lot of Acid Malt has a measured (via titration with NaOH) acidity to pH 5.4 of 320 mEq/Kg

Time to get this train rolling:

6Kg x 1% = 0.06 Kg of Acid Malt to be added (for a presumed pH shift of 0.10 points)
320 mEq/Kg x 0.06 Kg = 19.2 mEq of acidity contributed by the acid malt addition

pH_Shift = mEq/(buffering_capacity x Kg_Grist)
pH_Shift = 19.2/(32 x 6)
pH Shift = 0.10 pH points (downward due to adding acidity)

It worked!

But there's a catch: The ideal case witnessed above is defined by the buffering capacity factor for the grist weighing in at 32, and the target pH of 5.4, and acid malt with 320 mEq of acidity to pH 5.4. All of these values are subject to degrees of variability.

 

[Die_Beerery]

Edit...
Yes you are correct.

 

[Qhrumphf]

I wouldn't understate the importance of "this lot of acid malt" either. While I've found it relatively consistent (though far less so than just using straight acid), some lots have been well outside the norm to either side.

Just personal experience.

 

[Silver_Is_Money]

And why (at pH 5.4) can we make the "effective" case that Acid Malt is 3% Lactic Acid by weight?

88% Lactic Acid has a density of 1.206. Therefore 1 mL of Lactic Acid at 88% concentration has:

1.206 g/mL x 0.88 = 1.06128 grams of pure lactic acid within each mL

MW Lactic Acid = 90.078 g./mol

1,000 x 1.06128 g/mL / 90.078 mg./mmol = 11.7818 mEq/mL (when it's H+ ion is 100% dissociated)

But at pH 5.4 the H+ ion in Lactic Acid is only 97.197% dissociated.

11.7818 x 0.97197 = 11.4516 mEq/mL H+ (acid strength) for 88% Lactic Acid

Now back to Acid Malt:

1 Kg = 1,000 g.

1,000 g. x 3% Lactic Acid = 30 grams 'effective' of pure Lactic Acid in 1 Kg. of Acid Malt

We know that 1 mL of 88% Lactic Acid contains 1.06128 grams of pure Lactic Acid.

30 g. / 1.06128 g/mL = 28.2678 mL of 88% Lactic Acid are 'effectively' contained within every Kg. of Acid Malt.

28.2678 mL x 11.4516 mEq/mL (at pH 5.4) = 323.7 mEq of H+ (acid strength)

In the first post of this thread the assumption was made that 1 Kg of acid Malt had an acid strength at pH 5.4 of 320 mEq. We have just shown that if Acid Malt 'effectively' contains 3% Lactic Acid by weight its pH 5.4 acid strength is 323.7 mEq/Kg

(323.7/320 x 100) - 100 = 1.156% error in our assumption.

So we can 'effectively' (though not definitively) conclude within only a small margin of error that for most practical purposes when one is shooting for a pH of 5.4 that 'typical, or nominal' Acid Malt is 3% by weight Lactic Acid.

Or alternately, if we de-facto accept 3% Lactic Acid in 1 Kg. of Acid Malt, we can proclaim that Acid Malts acid strength is indeed 323.7 mEq/Kg at pH 5.4 (as opposed to our initial assumption of 320 mEq/Kg).

 

[Silver_Is_Money]

Lets see how this works with mashing a base malt.

Lets assume a SMaSH grist with 5 Kg. of Pilsner malt.
Lets further assume its DI_pH is 5.8, and we are targeting 5.4 pH in the mash.
We will further presume distilled water only for the mash
Lastly we will assume that its buffering factor is 32 mEq/Kg.pH (at 5.4 pH).


(5.8 DI_pH - 5.4 Target pH) = mEq of acid required / (32 x 5)
0.4 pH shift (downward) = mEq of acid required / 160
mEq of acid required = 160 x 0.4
mEq of acid required = 64 mEq

64 mEq/323.7 mEq/Kg (for acid Malt) = .1977 Kg. x 1000 = 197.7 grams acid malt required to be added to hit pH 5.4 (or ~7 Oz.)

(or)

64 mEq/11.4516 mEq/mL (for 88% lactic acid) = 5.6 mL of 88% Lactic Acid required to be added to hit pH 5.4

 

[Silver_Is_Money]

And how does our calculation above jive with Weyermann's Acid Malt ballpark method?

1% Acid Malt per 0.1 points of pH movement (downward) and 0.4 pH points of drop required = 4% Acid Malt in the grist.

5 Kg. grist X 4% Acid Malt = 0.2 Kg = 200 grams = ~ 7 ounces of Acid Malt.

And if we simply ballpark that ~35.5 grams of Acid Malt = 1 mL of 88% Lactic Acid, then:

200/35.5 ~= 5.6 mL of 88% Lactic Acid.

Effective "enough" software could easily be written to compute acid requirements simply by making nominal equivalence comparisons of various liquid acids pH 5.4 strengths to Acid Malt at pH 5.4 and applying the Weyermann ballpark method. All that is required is to settle upon mashing at exclusively 5.4 pH, and to measure the current mash pH and subtract 5.4 from it.

Another ballpark is to assume that 1 gram of baking soda is the opposite pH direction equivalent of roughly the same 35.5 grams of Acid Malt.

There, isn't that easy.

 

[bkboiler]

Good to see a sample calc! Thanks for this!

 

[Robert65]

Yes, I've tried to figure this out before, but this is very helpful. And usable, more to the point. Thanks.

 

[Vale71]

I wouldn't understate the importance of "this lot of acid malt" either. While I've found it relatively consistent (though far less so than just using straight acid), some lots have been well outside the norm to either side.

Just personal experience.

And that is why getting malt analysis sheets would be a great plus. Acid malt analysis sheets would also give the actual measured amount of lactic acid in weight percentage. Unfortunately this is often not possible if buying from a homebrew shop and we're stuck with ballpark figures...

 

[Silver_Is_Money]

And that is why getting malt analysis sheets would be a great plus. Acid malt analysis sheets would also give the actual measured amount of lactic acid in weight percentage. Unfortunately this is often not possible if buying from a homebrew shop and we're stuck with ballpark figures...

But the very best part is that if you are only using Weyermanns ballpark rule of thumb whereby to equate Acid Malt to Lactic (or any other) Acid (Phosphoric or CRS/AMS, etc...) at pH 5.4 as your target as per the above, you are literally not concerned at all with Acid Malt's "real world" wild swings in strength variability, as you are only using Acid Malt's "nominal" 320 mEq/mL (or 323.7 mEq/mL if you prefer) acid strength at pH 5.4 to correlate it to pH 5.4 targeted mL's of the various other acids (or to grams of baking soda or slaked lime, etc...) as nominal mEq equivalents. Using the 35.5 grams of "hypothetical ideal" Acid Malt = 1 mL of 88% Lactic Acid (as to mEq's at pH 5.4) as an example, if you are factually adding Lactic Acid (as opposed to adding a specific 'lot' of Acid Malt that may very well deviate measurably from its ideal of 320 - 323.7 mEq/mL) you are completely avoiding the inevitable variability of the actual mEq/Kg @ pH 5.4 strength of your specific 'lot' of Acid Malt.

 

[Silver_Is_Money]

At specifically pH 5.4, and for the presumption of the "hypothetical ideal" Acid Malt being 323.7 mEq/Kg:

35.4 grams of Acid Malt = 1 mL of 88% Lactic Acid
31.7 grams of Acid Malt = 1 mL of 80% Lactic Acid
45.9 grams of Acid Malt = 1 mL of 85% Phosphoric Acid
37.9 grams of Acid Malt = 1 mL of 75% Phosphoric Acid
11.3 grams of Acid Malt = 1 mL of 30% Phosphoric Acid
3.37 grams of Acid Malt = 1 mL of 10% Phosphoric Acid
11.3 grams of Acid Malt = 1 mL of CRS/AMS
~30.9 grams of Acid Malt = 1 gram of Citric Acid
~-33.8 grams of Acid Malt = 1 gram of Baking Soda (moving pH in the opposite direction)
~-60.4 grams of Acid Malt = 1 gram of Ca(OH)2 (moving pH in the opposite direction, plus ballpark attempting to account for its simultaneous Ca++ downward pH shift impact)

NOTE: I have less overall confidence for the last three above with the ~'s. YMMV. As Ronald Reagan would say: "Trust, but verify" The very least confidence is for Ca(OH)2.

The proposed easy ballpark method:
Use Weyermann's ballpark formula to determine how many grams of "perfect world" Acid Malt must be added to the mash to move it to pH 5.4, then divide those calculated Acid Malt grams by one of the above "X Grams of Acid Malt =" values and use the corresponding acid to the right of the "=" sign instead of using Acid Malt. If the required Acid Malt grams come out to be negative, choose either Baking Soda or Ca(OH)2.

It would be very easy to write a mash pH assistant spreadsheet based upon this simple method. And it would be generally close enough. The real trick is determining the aggregate (or composite) grists DI_pH such that you can determine how much you need to shift its pH to hit the pH 5.4 target in order to compute the Acid Malt (or negative Acid Malt) required to accomplish the required shift.

 

[Qhrumphf]

But the very best part is that if you are only using Weyermanns ballpark rule of thumb whereby to equate Acid Malt to Lactic (or any other) Acid (Phosphoric or CRS/AMS, etc...) at pH 5.4 as your target as per the above, you are literally not concerned at all with Acid Malt's "real world" wild swings in strength variability, as you are only using Acid Malt's "nominal" 320 mEq/mL (or 323.7 mEq/mL if you prefer) acid strength at pH 5.4 to correlate it to pH 5.4 targeted mL's of the various other acids (or to grams of baking soda or slaked lime, etc...) as nominal mEq equivalents. Using the 35.5 grams of "hypothetical ideal" Acid Malt = 1 mL of 88% Lactic Acid (as to mEq's at pH 5.4) as an example, if you are factually adding Lactic Acid (as opposed to adding a specific 'lot' of Acid Malt that may very well deviate measurably from its ideal of 320 - 323.7 mEq/mL) you are completely avoiding the inevitable variability of the actual mEq/Kg @ pH 5.4 strength of your specific 'lot' of Acid Malt.

The lesson is, just use the acid and skip the acid malt entirely. It's cheaper too (at least in bulk it's a HUGE difference).

 

[Qhrumphf]

Or a guess a sauergut reactor if that's your thing.

 

[Silver_Is_Money]

I just averaged the mEq/Kg.pH (at 5.4 pH) buffering capacities of the grists from 25 recipes that I maintain, and the mean came to 33.57. The lowest (a very light colored Pils with a fair load of flaked corn in it) was 29.29, and the highest (a robust Russian Imperial Stout) was 35.84.

The average for the general Blonde Ale, Pilsner, Pils, Kolsch, and IPA color class of my recipes was 31.74
The average for recipes with 19 SRM or higher finished color was 34.10

I favor darker beers on the order of Stouts and Porters among my recipes, so my overall average of 33.57 reflects this.

All of this was derived for 'MME' with the buffer multiplier cell set at 0.70.

Does anyone have data or literature detailing what should be the average beers nominal buffering capacity? I seem to recall being told a couple years back that 34 might be a good ballpark average, but who knows? A buffer multiplier setting of 0.709 for MME would place the average buffer value for all of my recipes at 34. I believe Kolbach nominally presumed 32 though. In MME a buffer multiplier value of 0.667 would average all of my recipes to 32. But I would presume that Kolbach did most of his work on Pils/Pilsner/Kolsch color class beers, and for that class of grists I'm already at 31.74. Bumping the Pils/Pilsner/Kolsch color class beers to 32 would require a buffer multiplier setting of 0.706 for MME. So I think it will just stay set at 0.70 for now. With a setting of 0.7075 (splitting the difference between 0.706 and 0.709) as an option to hit ~32 for light beers and ~34 overall.