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 = 7.10. The pKaof 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.)

Answer 1

Given data,

pH = 6.76

pKa of H2PO4 = 7.21

Let 'a' be the volume of KH2PO4 and

Let 'b' be the volume of K2HPO4

Moles of KH2PO4 = volume x concentration

= a x 1.00

= a mol

Moles of K2HPO4 = volume x concentration

= b x 1.00

= b mol

According to Henderson-Hasselbach equation:

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

6.76 = 7.21 + log(b / a)

log(b / a) = - 0.45

b / a = 10^-0.08

= 0.3548

b = 0.3548 a

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

= 450 / 1000 x 1.00

= 0.450 mol

Thus a + b = 0.450

a + 0.3548 a = 0.450

1.354 a = 0.450

a = 0.332

Volume of KH2PO4 needed = 0.332 L or 332 mL