Since 10^-pH = moles of acid (hydrogen ion concentration) in solution, I'm at a loss as to why summing up the individual grist components quantifiable moles of acid and then via negative log reconverting them back to an "overall" wort pH is to be considered a highly invalid approach.
Because it in no way represents the chemistry. If you put a mole of hydrochloric acid into a liter of water it completely dissociates and so there will be a mole of H+ (ions though the pH will be around -10). If you now add a mole of lye (NaOH) it will absorb all those protons, the hydrogen ion concentration will drop to 1E-7 and the pH will be 7. Using your scheme we would have (36*1 + 40*0)/76 = 0.473684 (36 is the grams per mole of HCl and 40 the grams per mole for NaOH) for your estimated hydrogen ion concentration and consequently -log(36/76) = 0.32 for your pH estimate. The actual pH would be 7. The pH of a solution of 40 grams of NaOH in a gallon of water is about -14. Thus the weighted pH average is (-36*10 + 40*14)/76 = 2.63 which is, while not quite so ridiculous, is almost equally so.
If you dissolve 192.12 grams (1 mole) of citric acid in 5 gallons of water the pH will be 2.23 (the DI pH). If you now add 100 grams of lye the pH will be 6.42. Your method would give us (192.12*10^-2.23 + 100*10^-14)/292.12 = 0.00387268 for the H+ ion concentration and -log((192.12*10^-2.23 + 100*10^-14)/292.12) =2.41
So we have seen that your method does not work with strong acid and strong base nor with weak acid and strong base. Why would we expect it to work with weak acids and bases (which is what malts are or, rather, a mix of weak acids/bases)? So let's put a mole of citric acid and a mole of succinic acid in 5 gal of DI water. As we have seen the DI pH for the citric acid is 2.2280. For the succinic acid it is 2.7454. The molecular weight of succinic acid is 118.09 g. You would estimate the pH of the mix as
•print -log((118.09*10^-2.7454 + 192.12*10^-2.2280)/(118.09 + 192.12))
2.36173
But the weighted averages of the pH's is
•print (118.09*2.7454 + 192.12*2.2280)/(118.09 + 192.12)
2.42496
And the weighted harmonic average is
•print 1/( (118.09/2.7454 + 192.12/2.2280)/(118.09 + 192.12))
2.4002
As the actual answer for such a mix is 2.2074 all these answers are wrong. But why are they closer in this case than in the other examples? Because if you mix things with DI pHs of, respectively, 2.23 and 2.74 you are going to get a pH between those values. They are only 0.5 pH apart. Thus if you simply randomly guess a number between 2.23 and 2.74 you can't be more than 0.05 pH off.
pH itself is a logarithmic scale. Why then is it wrong to apply logs (exponents) to the solution and instead go with what is essentially a linearized solution?
Because the log of the sum of a set of numbers is not equal to the sum of the logs of those numbers as you can easily verify. One can approximately estimate the pH of a mixture of weak acids (malts) from the following formula.
This formula can be tied to the chemistry and I've done that so many times here that I won't get into it again. It's in the Stickies. You MUST understand that chemistry to understand how to estimate mash pH. Note that the estimate is not simply the sum of the pH's weighted by the malt masses but by the product of the mass and buffering (a's) of each malt. You MUST understrand buffering in order to be able to estimate mash pH. Now if you assume that all the malts have equal buffering then the a's cancel out and you have the 3rd expression above. That is the simple average of the pH's weighted by the masses. Your suggested method is represented by the rightmost expression. For validity it would have to equal the 3rd expression and it doesn't because the log of a sum doesn't equal the sum of the logs.
Now note that the average weighted pH (3rd expression) is an approximation to the first expression and that in turn is an approximation because the estimate it represents is the first step in the solution of a non linear equation from an initial estimate of pH = 0. And even that first step is in error because we are ignoring the effects of water alkalinity and any thing we have added to it and are assuming that the malt buffering is linear (which it isn't though in many malts it is close to being linear). Thus your proposed method is invalid for many reasons the easiest of which to understand is that log(1 + 2 + 3 + 4 + 5) = 1.17609 ≠ 2.07918 = log(1) + log(2) + log(3) +log(4 )+ log(5)