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As a technician in a large pharmaceutical research firm, you need to produce 200. mL of...

As a technician in a large pharmaceutical research firm, you need to produce 200. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.85. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4stock 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.)

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Answer #1

6.85 = 7,21 + log (conc. HPO4)/(conc. H2PO4)
-0.36 = log (conc. HPO4)/(conc. H2PO4)
0.436 = (conc. HPO4)/(conc. H2PO4)
You also know that (conc. HPO4) + (conc. H2PO4) = 1.00 M which after rearraging gives (conc. HPO4) = 1.00 - (conc. H2PO4). This expression needs to be integrated into this: 0.436= (conc. HPO4)/(conc. H2PO4) which gives you:
0.436 = (1.00 - (conc. H2PO4))/(conc. H2PO4)
After rearanging you will end up with:
conc. H2PO4 = 1.00/1.436 = 0.696 M
You have 2.00 L of 1.00 M H2PO4 (the stock solution). You need 100 mL of 0.696 M.
conc. stock x volume stock = conc. needed x volume needed
volume stock = (0.696 x 0.1)/1.00 = 69.6 mL

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