I didn't get a chance to check the math yesterday so here goes. I'll work through the problem manually which, while it may be a bit challenging to read should be informative. The math is what's behind some of the curves in Palmer's book so it may be interesting to readers from that POV.
We start with 100 mg of what we presume is pure Ca(OH)2. This has a GMW of 74.093 grams per millimole so we have 100/74.093 = 1.34966 millimoles of slaked lime each of which contains two milliequivalents of hydroxyl (OH-) ion thus we are going to neutralize 2*1.34966 = 2.69931 milliequivalents of hydroxyl ion and to do that we need the same amount of hydrogen ions:
H+ + OH- ---> H2O.
At any pH there are some hydroxyl and hydrogen ions floating around but at pH 7 they are in equal concentration of 10^-4 milliequivalent/L each and we can neglect their presence. So we specify that we are going to add acid until the lime is all dissolved and the pH is 7. The question becomes "how much phosphoric acid is required to deliver 2.69931 mEq of protons (hydrogen ions) at pH 7.
Phosphoric acid, H3PO4 sheds protons as pH increases. The first shedding is of the first proton: H3PO4 --> H+ + H2PO4-. The ratio of the number of acid molecules that shed this proton to the number that don't is 10^(pH - 2.123) = 10^(7 - 2.213) = 75335.6. Thus, at pH 7, the vast majority of acid molecules will have given up this first proton and been converted to H2PO4-. In turn H2PO4- sheds one of its protons: H2PO4- --> H+ + HPO4--. The ratio of the concentration of those that have shed the second proton (dibasic) to that of those that have retained it (monobasic) is 10^(ph - 7.214) = 10^(7 - 7.214) = 0.610942 i.e. a bit over half of the monobasic ions will have yielded up the second proton. Then the dibasic ions can give up their single protons to become phosphate: H2PO4-- --> H+ + PO4---. The relevant ratio here is 10^(pH - 12.44) = 10^(7 - 12.44) = 3.63078e-06. Very few of the dibasic ions emit their protons at pH 7.
To determine the number of protons from each millimole of phosphoric acid at pH 7 then we assume that at pH 7 there remain P millimoles of H3PO4, note that there would be 75335.6*P millimoles of monobasic phosphate ion and 75335.6*0.610942*P moles of dibasic phosphate and 75335.6*0.610942*3.63078e-06*P moles of phosphate. Each of these species contains 1 phosphorous so the total amount of phosphorous in the solution is P + 75335.6*P + 5335.6*0.610942*P + 75335.6*0.610942*3.63078e-06*P = P*(1 + 75335.6 + 75335.6*0.610942 + 75335.6*0.610942*3.63078e-06) = 121362*P. The fraction of the total which is still phosphoric acid is P/121362*P= 1/121362 = 8.23979e-06. The fraction which is monobasic is 75335.6 times this or 0.620749 and the fraction which is dibasic is 0.610942 times the fraction which is monobasic or 0.610942*0.0.620749 = 0.379242 and the fraction which is phosphate is 3.63078e-06 times the fraction which is dibasic which is an insignificant 1.37694e-06.
Given that the charge on phosphoric acid is 0, the charge on a monobasic ion is -1, the charge on dibasic phosphate is -2 and on phosphate is -3 the total charge on phosphate species at pH 7 is
0*8.23979e-06 - 1*0.620749 - 2*0.379242 - 3*1.37694e-06 = -1.37924 milliequivalents of charge per millimole of phosphate in the solution. Since all the phosphate come from the phosphoric acid we add the total charge is -1.37924 mEq per mmol of added phosporic acid. Since each negative charge comes from the loss of a proton it is clear that if we need 2.69931 mEq of protons we will have to add 2.69931/1.37924 = 1.9571 mmol of phosphoric acid. As the molecular weight of phosphoric acid is 98 mg/mmol we will need 98*1.9571 = 191.176 mg of phosphoric acid. If we use 10% w/w acid we'll need 10 times the amount of solution or 1912 mg i.e. 1.912 grams to get 191.2 mg of the actual acid and as the density of 10% phosphoric acid is about 1.050 g/ml we'll need 1.912/1.050 = 1.82 mL.
I think that's pretty close to what I said earlier. And it checks with my spreadsheet now so I don't think I made a math error. That, of course, still leaves a mystery.
We start with 100 mg of what we presume is pure Ca(OH)2. This has a GMW of 74.093 grams per millimole so we have 100/74.093 = 1.34966 millimoles of slaked lime each of which contains two milliequivalents of hydroxyl (OH-) ion thus we are going to neutralize 2*1.34966 = 2.69931 milliequivalents of hydroxyl ion and to do that we need the same amount of hydrogen ions:
H+ + OH- ---> H2O.
At any pH there are some hydroxyl and hydrogen ions floating around but at pH 7 they are in equal concentration of 10^-4 milliequivalent/L each and we can neglect their presence. So we specify that we are going to add acid until the lime is all dissolved and the pH is 7. The question becomes "how much phosphoric acid is required to deliver 2.69931 mEq of protons (hydrogen ions) at pH 7.
Phosphoric acid, H3PO4 sheds protons as pH increases. The first shedding is of the first proton: H3PO4 --> H+ + H2PO4-. The ratio of the number of acid molecules that shed this proton to the number that don't is 10^(pH - 2.123) = 10^(7 - 2.213) = 75335.6. Thus, at pH 7, the vast majority of acid molecules will have given up this first proton and been converted to H2PO4-. In turn H2PO4- sheds one of its protons: H2PO4- --> H+ + HPO4--. The ratio of the concentration of those that have shed the second proton (dibasic) to that of those that have retained it (monobasic) is 10^(ph - 7.214) = 10^(7 - 7.214) = 0.610942 i.e. a bit over half of the monobasic ions will have yielded up the second proton. Then the dibasic ions can give up their single protons to become phosphate: H2PO4-- --> H+ + PO4---. The relevant ratio here is 10^(pH - 12.44) = 10^(7 - 12.44) = 3.63078e-06. Very few of the dibasic ions emit their protons at pH 7.
To determine the number of protons from each millimole of phosphoric acid at pH 7 then we assume that at pH 7 there remain P millimoles of H3PO4, note that there would be 75335.6*P millimoles of monobasic phosphate ion and 75335.6*0.610942*P moles of dibasic phosphate and 75335.6*0.610942*3.63078e-06*P moles of phosphate. Each of these species contains 1 phosphorous so the total amount of phosphorous in the solution is P + 75335.6*P + 5335.6*0.610942*P + 75335.6*0.610942*3.63078e-06*P = P*(1 + 75335.6 + 75335.6*0.610942 + 75335.6*0.610942*3.63078e-06) = 121362*P. The fraction of the total which is still phosphoric acid is P/121362*P= 1/121362 = 8.23979e-06. The fraction which is monobasic is 75335.6 times this or 0.620749 and the fraction which is dibasic is 0.610942 times the fraction which is monobasic or 0.610942*0.0.620749 = 0.379242 and the fraction which is phosphate is 3.63078e-06 times the fraction which is dibasic which is an insignificant 1.37694e-06.
Given that the charge on phosphoric acid is 0, the charge on a monobasic ion is -1, the charge on dibasic phosphate is -2 and on phosphate is -3 the total charge on phosphate species at pH 7 is
0*8.23979e-06 - 1*0.620749 - 2*0.379242 - 3*1.37694e-06 = -1.37924 milliequivalents of charge per millimole of phosphate in the solution. Since all the phosphate come from the phosphoric acid we add the total charge is -1.37924 mEq per mmol of added phosporic acid. Since each negative charge comes from the loss of a proton it is clear that if we need 2.69931 mEq of protons we will have to add 2.69931/1.37924 = 1.9571 mmol of phosphoric acid. As the molecular weight of phosphoric acid is 98 mg/mmol we will need 98*1.9571 = 191.176 mg of phosphoric acid. If we use 10% w/w acid we'll need 10 times the amount of solution or 1912 mg i.e. 1.912 grams to get 191.2 mg of the actual acid and as the density of 10% phosphoric acid is about 1.050 g/ml we'll need 1.912/1.050 = 1.82 mL.
I think that's pretty close to what I said earlier. And it checks with my spreadsheet now so I don't think I made a math error. That, of course, still leaves a mystery.