A practical application of the combined 'pHDI via Malt Color' and 'Buffering Coefficient via Malt Color' spreadsheets

[Silver_Is_Money]

And now for a practical application of the two most recent spreadsheets I placed in the 'Brewing Software' forum:

What if (for example) we had a grist comprised of 11 Lbs. of 1.8L Base malt and 2 Lbs. of 10L Caramel malt. And we are mashing it in DI water. What correction is needed whereby to get this mix of two barley malts to mash at pH 5.4?

Givens (using the spreadsheet values):

Base malt:
pHDI = 5.74
Buffering Capacity Coefficient= 37.88 BC
Kg.= 11/2.20462 = 4.99

Caramel Malt:
pHDI = 5.16
Buffering Capacity Coefficient = 47.67 BC
Kg. = 2/2.20462 = 0.907 Kg

How many mEq are required to move each malt to pH 5.4 individually?

mEq = Delta-pH x (BC x Kg.)

mEq_base_malt_pH 5.4 = (5.74-5.4) * (37.88*4.99)
mEq_base_malt_pH 5.4 = 64.27 mEq of Acid (acid since positive)

mEq_caramel_malt_pH 5.4 = (5.16-5.4) * (47.64*0.907)
mEq_caramel_malt_pH 5.4 = -10.37 mEq of Base (base since negative)

Combine the malts mEq's:
64.27 + -10.37 = 53.9 mEq's of Acid still needed

88% Lactic Acid has an acid strength of 11.451 mEq's/mL at pH 5.4

Therefore to mash this specific combination of malts at pH 5.4 in DI water requires:
53.9 mEq/11.451 mEq/mL = 4.71 mL of added 88% Lactic Acid

 

[Silver_Is_Money]

Further to this, lets now additionally assume BIAB, no sparge, in ~9 gallons (34L) of water with 65 ppm Calcium, 10 ppm Magnesium, and 75 ppm Alkalinity. How does this information change the mash conditions whereby to achieve 5.4 pH in the mash.

Per Kolbach, 3.5 mEq's of Calcium will react with the malt to release 1 mEq of acid from the malt phosphates.
Per Kolbach, 7 mEq's of Magnesium will react with the malt to release 1 mEq of acid from the malt phosphates.

Ca++ has a molecular weight of 40.078, and a valence of +2, for an Eq weight of 20.039.
Mg++ has a molecular weight of 24.305, and a valence of +2, for an Eq weight of 12.153.

For calcium:
34L x 65 mg/L = 2,210 mg
2,210/20.039 = 110.3
110.3/3.5 = +31.51 mEq's of released acid

For magnesium:
34L x 10 mg/L = 340 mg
340/12.153 = 28
28/7 = 4 mEq's of released acid

Total acid generated by minerals = 31.51 mEq + 4 mEq = 35.51 mEq

On to the waters Alkalinity:

34L x 75 mg/L = 2,550 mg CaCO3 equivalent
Molecular Weight of CaCO3 = 100.0869
Valence = 2
Equivalent Weight = 100.0869/2 = 50.04345

2,550/50.04345 = 50.96 mEq of base

Sum total of Acid from minerals and Base from Alkalinity = 50.96 - 35.51 = 15.45 mEq's of additional base vs. the example as above

In the example above we wound up with 53.9 mEq's of base to be acidified

Now we have 53.9 + 15.45 = 69.35 mEq of base to be acidified

69.35 mEq/11.451 mEq/mL of 88% Lactic Acid = 6.05 mL of 88% Lactic Acid to be added

 

[Silver_Is_Money]

With all of this information, anyone should now be able to construct their own mash pH assistant spreadsheet or software. The only things missing involve:

1) Was Kolbach correct in his presumptions for acid liberation via calcium and magnesium?
2) Are the various grist components buffering coefficients impacted by ones actual mill gap crush vs. lab pulverized? And if so, to what magnitude and in what direction (upward or downward)?

 

[Silver_Is_Money]

PS: In Mash Made Easy version 9.95, when the above parameters are entered, and both the Grist Buffer Multiplier and the Kolbach pH Shift Multiplier are set to 1.00, and the above hypothetical malt buffering and pHDI values are entered via MME's manual override provisions, the MME answer whereby to hit a mash pH of 5.40 is to add 6.05 mL of 88% Lactic Acid. Same as for post #2 above.