Attempting to predict the pH of 5 gallons of amazingly good distilled or perfectly deionized water with a TDS of zero when 1 drop of 88% Lactic Acid is added.

Knowns:

88% Lactic Acid is 11.78 Molar = 11.78 moles/L

Lactic Acid is 'effectively' monoprotic with respect to pH's below 7

False presumptions for the aid of initial simplicity:

Lactic Acid fully dissociates at pH 4-5(ish)

10 drops of 88% Lactic Acid = 1 mL

Our water is perfectly deionized

How many liters are there in one drop:

1/10 = mL = 0.1

0.1 mL/1000 = 0.0001 L

How many moles are there in one drop?

0.0001 L x 11.78 moles/L = 0.001178 moles

When this drop is added into 5 gallons of water, how many moles/L are there?

0.001178 moles / (5 Gal. x 3.7854 L/Gal.) = 0.000062239 moles/L

pH = -log(molar concentration) [for the case of monoprotic and full dissociation]

Therefore:

pH = -log(0.000062239) = 4.206 pH

However, we know that at pH 4.3 Lactic Acid is only 73.4% dissociated, so to improve upon our guess:

Actual molar H+ = 0.734 x 0.000062239 = 0.000045683 moles/L

pH = -log(0.000045683) = 4.34 pH

But we know that our RO or distilled water isn't likely to exhibit a TDS of a perfect zero, so in the real world our measured pH will be a smidge higher than 4.34 due to the buffering of some minuscule amount of remaining Alkalinity.

Call it pH 4.36 as our final guess, provided that 10 drops = 1 mL