Thought Experiment on a new way to determine malt buffering (as yet an incomplete thought)

[Silver_Is_Money]

I'm currently thinking of a method to determine a malted or unmalted grains buffering capacity without any need for undertaking a series of titrations.

My thought is to mash two samples of the same malt/grain in DI (or alternately, very good distilled) water, mashing one the conventional way using 50 grams of sample within 150 mL of water, and holding at 150-155 degrees F. (65-68 degrees C.) for 60 minutes, and one using 50 grams of sample within 1,500 mL (1.5 L) of water, also holding it at 150-155 degrees F. (65-68 degrees C.) for 60 minutes.

The next step is to cool both samples to 77 degrees F. (25 degrees C.) and take their pH using all proper pH reading procedure and good practice in doing so.

At this juncture, my presumption is that if there is zero buffering capacity the grist mashed in 1,500 mL of DI water should exhibit a measured mash pH that is close to precisely 1.00 pH points higher than the sample mashed in 150 mL of water. But if there is some level of measurable buffering capacity associated with the malt/grain, then the pH for the sample mashed in 1,500 mL of water should exhibit a mash pH that is some measure less than 1.00 full pH points higher than for the sample mashed in 150 mL of water.

Provided that the degree of malt buffering is not so extremely high as to result in the 1,500 mL worts pH being measured at exactly the same as the 150 mL worts pH, the buffering capacity should thus be extractable from the difference seen between the two measured mash pH's.

I have not attempted this experiment as yet, but allow me to set up a purely hypothetical example case for our consideration:

For Example:
If Roast Barley at 300 Lovibond is mashed using 50 grams in 150 mL DI water and the resulting pH is found to be 4.70, then its molar concentration of Hydronium ions is nominally:

10^-4.7 = 0.000019953 moles/L

And if the very same Roast Barley, but this time for the case of 50 grams mashed in 1,500 mL of water comes out with a hypothetical mash pH of 5.3, then its molar concentration of Hydronium ions is nominally:

10^-5.3 = 0.000005012 moles/L

By ratio, 0.000019953/0.000005012 = 3.981

This means that the sample mashed in 150 mL of water is 3.981 times more acidic than the sample mashed in 1,500 mL of water (for this purely hypothetical case). [If there was zero buffering capacity the expected result (per my presumption above) should be a ratio of 10, meaning that the sample mashed in 150 mL of water should be fully 10-fold more acidic than the sample mashed in 1,500 mL of water.]

10/3.981 = 2.512

My question (intended to bring completion to this as yet incomplete thought experiment process) is:

Can a meaningful buffering capacity value for this hypothetical Roast Barley be extracted from this 2.512 ratio (or its inverse)?

 

[MildAvers]

As I've had the same inclination for quite some time, my initial reasoning is that buffering capacity of the grain would be more phosphate, sulfide and organic acid based and as such there would be little to no correlation.

 

[Gnomebrewer]

Maybe test a few grains that way and also titrate....see if the results correlate.

 

[ajdelange]

You will never understand how this works until you accept that where there is a pH change the protons lost by the malt are taken up by the water. As water has very small buffering relative to the malt it only takes a few protons to shift its pH appreciably and thus the malt must only give up way fewer resulting, as it has appreciable buffering, in a tiny pH shift. The pH of a mash made with a particular malt and deionized water is virtually independent of the amount of water.

As an example of this 50 grams of 600L chocolate mashed in 150 mL of DI water will come to pH 4.6995 (its DI mash pH) and the same amount of this malt mashed in 1.5 L of water will come to 4.7065 and mashed in 15L of water, to 4.7660. Water is a very weak base. This problem is solved the way all acid base problems are solved and has been stated in this forum dozens of times: ∆Qwater(∆pH) = ∆Qmalt(∆pH). Solve for ∆pH.

 

[Silver_Is_Money]

You will never understand how this works until you accept that where there is a pH change the protons lost by the malt are taken up by the water. As water has very small buffering relative to the malt it only takes a few protons to shift its pH appreciably and thus the malt must only give up way fewer resulting, as it has appreciable buffering, in a tiny pH shift. The pH of a mash made with a particular malt and deionized water is virtually independent of the amount of water.

As an example of this 50 grams of 600L chocolate mashed in 150 mL of DI water will come to pH 4.6995 (its DI mash pH) and the same amount of this malt mashed in 1.5 L of water will come to 4.7065 and mashed in 15L of water, to 4.7660. Water is a very weak base. This problem is solved the way all acid base problems are solved and has been stated in this forum dozens of times: ∆Qwater(∆pH) = ∆Qmalt(∆pH). Solve for ∆pH.

As @Gnomebrewer deftly stated above, the only way to find out if your information as seen above (or some other outcome) will be the reality of the matter will be to actually undertake this experiment for several malts and/or unmalted grains. Your presumption above is that malt buffering is indeed very high, but in a different thread discussing the buffering of mash water via measured additions of various quantities of weak acids and their salts you concluded for phosphoric acid and a salt of it (and the graph you had me seek out confirmed) that phosphate buffering is at its very weakest point within the span of roughly 4.2 to 5.8 pH, wherein the slope which represents phosphate buffering strength is almost non existent for pH's within that general range (which just happens to be right within the range of the DI_pH's of nearly all malted grains of any classification group or kiln/roast level).

 

[ajdelange]

I certainly don't want to discourage experimenting but it is helpful if you understand the theory enough to know what result the experiment is likely to produce. In some cases that understanding may change your approach to the experiment. If you stick your finger in a gas flame to see if it is hot you will get burned. If you add extra DI water to a malt mash it's pH won't change much. A major flaw in Bru'n water is that its mash pH estimates change dramatically if more DI water is added. I believe you were the guy that discovered this so at one time you apparently understood this fundamental fact.

I can only, again, encourage you to try to grasp how buffering works. The material is all available to you in this forum.

Re the particular comments:
1)Phosphate buffering is minimum between pK's (where mash pH tends to fall) but it is still, given the amount of phosphate in malt) much greater than the buffering of reasonable amounts of water. Thus, were phosphate the only buffer, adding more water wouldn't change the pH of the mixture much. Chemists refer to a change in a buffer's pH with additional added water as "dilution error". Water is an acid or a base (though a very weak one) and so do pull buffer pH a bit. Try diluting some pH 7 (that's a phosphate buffer) or pH 4 pH meter calibration solution with some DI water to get a handle on how much a phosphate buffer is effected by dilution.
2)Phosphate is not the only thing that buffers in malt. A malt's titration curve (smooth) looks very little like the phosphate titration curve (staircase).

 

[Silver_Is_Money]

A.J., I'm not denying that malts and grains exhibit inherent buffering. I'm merely questioning whether the quantified extent (or strength) of such buffering is at such a high level as to make a 10-fold, and perhaps even a 100-fold dilution show very little movement with respect to the originally measured DI_pH.

 

[ajdelange]

And I am repeating again that the simple tools for answering that question are before you and have been for years. It would seem that you are determined to ignore them.

What is the buffering of DI water between pH 7 (its intrinsic pH) and pH 5.4? 1000*10-5.4 = 0.004 mEq/kg. What is the buffering of a typical base malt between it's intrinsic pH of about 5.7 and pH 5.4? Typically 40*(5.4 - 5.7) = - 12 mEq/kg. That's 3014 times greater than water's.