Suppose you hada solution that was 10 g/L of citric acid. Citric acid has three titratable protons and a molecular weight of 192 g/mol. What would the solutions acidity be in equivalent g of acetic acid/100 mL? Confirm that it would be 9.39 g/L. Do the calculation in the prelab section of your lab notebook.

Helpful Info that helps for how to do this:

Malic acid has a molecular weight of 134 g/mol, and it has two protons that will react with NaOH. Acetic acid has a molecular weight of 60.1 g/mol, and just one titratable proton.

Suppose we titrate 25.00 ml of wine and find that it reacts with 19.31 mL of 0.1034 M NaOH. The titratable [H+] in the wine is VbaseMbase/Vacid = 19.31*0.1034/25.00 = 0.07987 M. The molar concentration of malic acid would be half this value, since each malic acid has two titratable protons. So the mass concentration would be 0.07987 M/2*134 g/mol = 5.35 g/L. That would be a pretty tart (acidic) wine!

Lets do the calculation using acetic acid. The molar concentration of acetic acid would be 0.07987 M, since it has just one acidic proton. So the mass concentration would be 0.07987 M*60.1 g/mol = 4.80 g/L. Not much different, because the molecular weight per acidic proton is about the same.