posted by Anonymous .
As a technician in a large pharmaceutical research firm, you need to produce 400. mL of 1.00 M potassium phosphate solution of pH = 7.18. The pKa of H2PO4- is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M K2HPO4 will you need to make this solution?
Use the Henderson-Hasselbalch equation.
pH = pKa + log[(base)/(acid)]
7.18 = 7.21 + log (B/A)
B/A = 0.933
and the equation B+A = 1
Solve the two equation simultaneously for B and A. Then
B*400 = mL base
A*400 = mL acid
I get approximately, but you need to do it more accurately, 207 mL acid and 193 mL base.