As a technician in a large pharmaceutical research firm, you need to produce 450. mL of...
As a technician in a large pharmaceutical research firm, you need to produce 450. mL of a potassium dihydrogen phosphate buffer solution of pH = 7.05. 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.)
Homework Answers
pH = 7.05
pKa = 7.21
pH = pKa + log [conjugate base / acid]
7.05 = 7.21 + log [HPO42- / H2PO4-]
[HPO42- / H2PO4-] = 0.692
[HPO42- + H2PO4-] = 0.450
[HPO42- / H2PO4-] = 0.692
[0.450 - x / x ] = 0.692
0.450 - x = 0.692 x
x = 0.266
volume of KH2PO4 = 266 mL
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