A phosphate buffer solution (25.00 mL sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The components of the buffer were sodium monohydrogen phosphate and sodium dihydrogen phosphate. The first endpoint occurred at a volume of 10.32 mL; the second occurred after an additional 18.62 mL was added, for a total volume of 28.94 mL. What was the approximate ratio of HPO₄^2- to H₂PO₄^-in the buffer?

Question

Quizplus · Accepted Answer

The first endpoint is the conversion of the dihydrogen phosphate ion to monohydrogen phosphate ion:

H2PO4- + H+ -> H3PO4,

so we can calculate the initial amount of H2PO4- from the volume of acid added up to the first endpoint:

0.1 M x 10.32 mL = 1.032 mmol H+

Since the H2PO4- and HPO42- are in a 1:1 ratio, this means there were 1.032 mmol of HPO42- initially as well.

At the second endpoint, all the HPO42- has reacted and the remaining acid is reacting with the H2PO4-:

H2PO4- + H+ -> H3PO4

so the volume of acid from the first to the second endpoint is proportional to the amount of H2PO4- present:

0.1 M x (18.62 mL - 10.32 mL) = 0.836 mmol H+

Therefore, the total amount of H2PO4- present was:

1.032 mmol + 0.836 mmol = 1.868 mmol

and the total amount of HPO42- present was:

1.032 mmol

The mole ratio of HPO42- to H2PO4- is therefore:

1.032 mmol / 1.868 mmol = 0.552

which is approximately 5:4 (option D).

A Phosphate Buffer Solution (25