The titration of 0.02500 L of a diprotic acid solution with 0.1000 M NaOH requires 34.72 mL of titrant to reach the second equivalence point. The pH is 3.95 at the first equiva-lence point and 9.27 at the second equivalence point. If the acid solution contained 0.2015 g of the acid, what is the molar mass, pKa1, and pKa2 of the acid?

Titration is a quantitative analytical technique used to determine the concentration of a solute in a solution. In the context of a diprotic acid, there are two equivalence points during titration, where the amount of titrant added is stoichiometrically equivalent to the amount of acid present. The first equivalence point corresponds to the complete neutralization of the first proton, while the second equivalence point indicates the neutralization of the second proton.

The pKa value is a measure of the strength of an acid in solution, representing the negative logarithm of the acid dissociation constant (Ka). For diprotic acids, there are two pKa values corresponding to the two dissociation steps. The lower the pKa, the stronger the acid, as it indicates a greater tendency to donate protons. The pH at the equivalence points can be used to calculate the pKa values of the acid.

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To calculate the molar mass of the diprotic acid in this scenario, one can use the mass of the acid (0.2015 g) and the number of moles present at the equivalence point, which can be determined from the volume and concentration of the titrant used. This relationship allows for the determination of the molar mass based on the stoichiometry of the titration.