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 = 7.08. 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.)

Answer 1

Henderson–Hasselbalch equation

assuming the concentration of buffer of 450 ml is 1 M,

pH = pKa + log([salt]/[acid])

7.08 = 7.21 +  log([salt]/[acid])

0.74 = ([salt]/[acid])

then

[salt]+ [acid] = 1

hence

[acid] =0.575 M

[salt]=0.425 M

for preparing 450 ml of total solution of KH2PO4

N1V1 =N2V2

450 X 0.575 = 1 X V2

hence V2 = 25.87 ml

similarly for K2HPO4 of 0.425M

N1V1 =N2V2

450 X 0.425 = 1 X V2

V2 = 19.12

hence all we have to do is to add 25.87 ml of 1 M KH2PO4 and 19.12 ml of 1M K2HPO4 and dilute it to 450 ml to get the resultant pH.