Question

As a technician in a large pharmaceutical research firm, you need to produce 150. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.78. The pKa of H2PO4− is 7.21.

You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 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 be the volume of KH2PO4 and b be the 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]])

= pKa + log(moles of K2HPO4/moles of KH2PO4) since final volume (= 450 mL) is the same for both

7.13 = 7.21 + log(b/a)

log(b/a) = -0.08

b/a = 10^-0.08 = 0.8318 => b = 0.8318a

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 + 0.8318a = 0.450

a = 0.246

Volume of KH2PO4 needed = a = 0.246 L = 246 mL