Question

A common biological buffer is a “phosphate buffer,” which involves the equilibrium between H2PO4- and HPO4^2-, with a pKa at 25 Celcius of 6.86. Describe in words how you would prepare 500. mL of 0.250 M phosphate buffer at pH = 7.2, given appropriate glassware and 500. mL each of 0.250 M NaH2PO4 and 0.250 M Na₂HPO₄ at 25 Celcius. (Note: your explanation needs to include actual volumes. It will be helpful to know that the concentrations of each component must add to the buffer concentration. i.e. [A-] + [HA] = 0.250 M)

Answer 1

pH = pKa + log ([A-]/[AH])

We have pH = 7.2 , pKa = 6.86

and [A-] + [AH] = 0.250 M .....1

therefore,

7.2 = 6.86 + log ([A-]/[AH])

log ([A-]/[AH]) = 0.34
([A-]/[AH]) = 2.188
[A-] = 2.188 [HA]

from equation 1.

2.188 [HA] + [HA] = 0.25 M
[HA] = 0.25/3.188 = 0.078 M

[A-] = 0.250-0.078 = 0.172 M

We need H2PO4- 0.078 M of 500 mL solution. We have 0.250 M, so we need to calculate the volume we need to take.

M1V1= M2V2
0.078 M x 500 mL = 0.25 M x V2
V2 = 156 mL

Similarly, we need HPO42- 0.172 M of 500 ml solution. We have 0.250 M, so we need to calculate the volume we need to take.

M1V1= M2V2
0.172 M x 500 mL = 0.25 M x V2
V2 = 344 mL

Therefore, we need to mix 344 mL of 0.250M of NaH2PO4 and 156 mL of 0.250M Na2HPO4 to make (344 +156 = 500) 500. mL of 0.250 M phosphate buffer at pH = 7.2