Question

How many milliliters of 4.2 M HCl must be added to 3.1 L of 0.16 M K2HPO4 to prepare a pH = 7.49 buffer? Use the Acid-Base Table.

Answer 1

moles of HPO4-2 = 3.1 x 0.16 = 0.496

moles of HCl = 4.2 x V

HPO4-2 + H+ ----------------------> H2PO4-

0.496        4.2 V                             0

0.496-4.2V        0                                  4.2 V

pKa2 = 7.21

pH = pKa + log [HPO4-2 / H2PO4-]

7.49 = 7.21 + log (4.2 V / 0.496 -4.2 V)

1.905 = 4.2 V / 0.496 -4.2 V

V = 0.0774 L

V = 77.4 mL

volume of HCl = 77.4 mL