Question

As a technician in a large pharmaceutical research firm, you need to produce 450. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.93. The pKa of H2PO4^- is 7.21. You have the following supplies: 2.00 mL of 1.00 M KH2PO4 stock solution, 1.50 m L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.) Express your answer to three significant digits with the appropriate units.

Answer 1

Let a = the volume of KH2PO4

b = volume of K2HPO4 needed in liters

Moles of KH2PO4 = volume x concentration = a x 1.00 = a mol

Moles of K2HPO4 = volume x concentration = b x 1.00 = b mol

Henderson-Hasselbalch equation:

pH = pKa + log([K2HPO4/[KH2PO4]])

6.93 = 7.21 + log(b/a)

b/a = 0.5248 => b = 0.5248 a

Total moles of phophate = final volume x total concentration of phosphate

= 450/1000 x 1.00 = 0.450 mol

Thus a + b = 0.450

a + 0.5348 a = 0.450

a = 0.295

Volume of KH2PO4 needed = a = 0.295 L

volume of  KH2PO4 needed = 295 mL