PLEASE HELP ASAP!!

Question

The student started part B with a 0.2024 g sample of an unknown diprotic acid. The volume of NaOH needed to neutralize the acid was determined to be 32.1 mL. What is the molar mass of the diprotic acid?

Molarity of NaOH= 0.105 M

Answer 1

The molarity and volume of NaOH allow anyone to determine how many moles of NaOH is used to neutralize the same number of moles of hydrated protons H3O+.

Moles of NaOH = (0.1.05 moles NaOH / liter of solution)(0.0321 liters of solution) = 3.37x10^-3 moles NaOH used

The NaOH is used to neutralize both protons on a diprotic acid H2A, so the stoichiometry of the reaction

2 NaOH (aq) + H2A = 2 HOH + 2 Na+ + A-2

offers a stoichiometric mole ratio of (1 moles H2A / 2 moles NaOH).

Moles of H2A = (3.37x10^-3 moles of NaOH)(1 moles H2A / 2 moles NaOH) = 1.69x10^-3 moles H2A in the sample

The molar mass of the diprotic acid is the ratio of the number of grams to the number of moles in that number of grams, which is what is determined by the titration.

(0.2024 grams H2A)/(1.69x10^-3 moles H2A) = 120. grams H2A /mole H2A