Hey, Boyago -- Part of the answer is that I keep a file of frequently asked chemistry and math questions that I've written answers to. A cheat sheet of sorts. This question about neutralizing acetic/citric/lactic acid has only come up .... oh maybe twenty bazillion times give or take a couple of thousand.
Another is that I have a chem and math background and like this kind of arcane geeky stuff. Here's your daily dose of algebra and chemistry:
Lactic acid + Sodium hydroxide => dissociated ionic species of the acid and base => Sodium lactate + water
C2H4OHCOOH + NaOH => C2H4OHCOO- + H+ + Na+ + OH- => C2H4OHCOONa + H2O
This chemical formula shows 1 molecule of lactic acid is needed to react with 1 molecule of sodium hydroxide.
Side note: This is a concept called the stoichiometric ratio, which is the number of moles of one substance needed to react exactly with the number of moles of another. So the stoichiometric ratio of this reaction is 1:1. The stoichiometric ratio of Sodium hydroxide with any soaping fat is 3:1 -- three molecules of NaOH are needed to react with 1 molecule of any soaping fat.
Side note: Another word to use in place of "molecule" is "mole" which is a more convenient measurement for us humans. A mole (shorthand: "mol") is based on Avogadro's constant, a standard measurement of 6.24 x 10^23 molecules. (Yes, I really do remember this off the top of my head.) Avogadro's number happens to be the precise count of atoms in 12 grams of Carbon-12, the isotope of carbon that weighs exactly 12 molecular units. (No I don't remember that, but Wikipedia is my friend.)
Molecular weight of Lactic acid: 90.08 grams / 1 mol (Source: Wikipedia or an old fashioned periodic table of the elements)
Molecular weight of Sodium hydroxide: 39.997 grams / 1 mol (ditto)
The stoichometric ratio is 1/1, but we don't normally measure in units of moles. We do use weight units, so I want to find a weight ratio that's means the same as the stoich ratio for this reaction.
It's nicer to write a ratio problem out properly by hand, but this is the only way I can show a ratio using this text editor:
(39.997 g NaOH / 1 mol) * (1 mol / 90.08 g lactic acid) = 0.4440 g NaOH / 1 g Lactic acid
Multiply top and bottom by 10 and you convert this ratio into something with easy-to-say numbers:
4.44 g NaOH / 10 g Lactic acid