What is the resulting pH if 1/2 mL of 88% Lactic Acid is added to 5 Gal. of pure water?

[Silver_Is_Money]

What happens to the pH of 5 gallons of deionized, distilled, or very high quality RO water if 1/2 mL of 88% Lactic Acid is added to it:

Knowns
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Density of 88% Lactic Acid = 1.209 g/mL
Molecular Weight of Lactic Acid = 90.779 grams/mol
pKa of Lactic Acid = 3.86
5 gallons of water = 18.927 Liters

Lets go:
--------
0.5 mL x 1.209 g/mL x 0.88 = 0.53196 grams of lactic acid added
0.53196 g./90.779 g./mol = 0.00586 moles of lactic acid added
0.00586 moles/18.927 Liters = 0.00030961 Molar lactic acid solution

0.00030961 x 10^-3.86 = 0.000000043
X^2 = 0.000000043
X = 0.000206731 molar H3O+
-log(0.000206731) = 3.685 pH

Answer: 1/2 mL of 88% Lactic Acid added to 5 Gallons of distilled or quality RO water results in water with a pH of ~3.69 (presuming that I carried out the above math correctly)

 

[Silver_Is_Money]

The above method only works well for monoprotic acids (such as for Lactic Acid). Could someone show me how to properly solve for pH when polyprotic acids (such as phosphoric acid, which is tri-protic, for example) are involved?

Actually, when I apply the above method to phosphoric acid I come up with pH answers commensurate with various answers available via web searches. But I'm skeptical as to the validity of those internet search derived answers. Never trust anything just because you've seen it on the internet.

 

[Silver_Is_Money]

The weirdest thing is that if you apply the above method to 0.5 mL of 10% Phosphoric Acid added to 5 gallons of distilled water, the resulting pH (of 3.35) is lower than the pH for 0.5 mL of 88% Lactic Acid added. Intuitively that sounds quite strange, seeing as for mash or Wort pH modification the ballpark rule of thumb is that around 11 times more 10% Phosphoric Acid will be required as for 88% Lactic Acid. This is a huge mystery. Calling @ajdelange.

 

[Silver_Is_Money]

I believe I can see a potential reason as to why this monoprotic method works decently for phosphoric acid. The magnitude of Ka1 is ~115,000 times greater than the magnitude of Ka2 for phosphoric acid. This would not be the case for an acid like citric though, where the difference in magnitudes between the Ka's is only in the 10's.

 

[ajdelange]

I've probably posted the answer to this question a hundred times here over the years but as I haven't done it for a while it will be good for me to review it. The method is, as for all acid base problems, to find the pH which causes the charge on the H+ ions to just balance the sum of the charges on the OH- ions and acid anions. The method works equally well for strong and weak acids and for acids with any number of protons to yeild.

1)Guess what the pH might be. Using that pH:
1)Calculate r1 = 10^(pH - pK1) where pK1 is the first pK of the acid you are adding
2)Calculate r2 = 10^(pH - pK2) where pK1 is the second pK of the acid you are adding
3)Calculate r's for any other protons
4)Calculate f0 = 1/(1 + r1 + r1*r2 +...) for as many r's (protons) as you have
5)Calculate f1 = r1*f0
6)Calculate f2 = r2*f1. Continue for as many protons as you have
7)Calculate Q = f1 + 2*f2 + 3*f3 for as many protons as you have
8)Multiply Q by the number of millimoles of the acid added per liter of waterAdd 10^(pKw - pH)
9)Add 1000*10^(pKw - pH) in which pKw is the pK of water to Q. Q is the negative ion charge magnitude per liter
10)Compute P = 1000*10^-pH. This is the magnitude of the hydrogen ion charge per liter
11)Keeping in mind that INCREASING pH INCREASES Q and DECREASES P adjust the pH until P = Q

Obviously this is best done in a spreadsheet and trivially easily done if Excel's SOLVER is used to do the PQ matching. If doing this it is convenient to set the problem up for, say, 4 protons. Use a very high pK (+40) for protons that don't exist (e.g. the 4th proton of phosphoric acid) and a very low pK (-10) for strongly acidic protons of a strong acid (e.g. the first proton of sulfuric acid).

 

[Silver_Is_Money]

Welcome back A.J.! When you apply your method to 1/2 mL of 88% lactic acid added to 5 gallons of distilled water, what resulting pH do you see? Ditto for 1/2 mL of 10% phosphoric acid.