As a technician in a large pharmaceutical research firm, you need to produce 200. mL of a potassium dihydrogen phosphate buffer solution of pH = 7.07. 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.)
There is a typo error, we are provided with
2.00L of 1.00M KH2PO4 olution and 1.50L of 1.00M K2HPO4 solution.
Using H. H equation,
pH = pKa + log [anion] / [acid]
7.07= 7.21 + log [anion] / [acid]
-0.14=log [anion] / [acid]
or, 0.724435= [anion] / [acid]
Ratio of concentrations in a solution, (200 ml), is the same as the ratio of moles in that volume.
arbitrarially choosing such that one to be added "X" ml of it , & the other becomes that its been added "250-X" ml
0.724436= (250 ml - X ml)(1.00 M K2HPO4) / (X ml)(1.00 M KH2PO4)
0.724436 = (250 - X) / X
After solving X = 144.97 ml
Thus we mix 144.97 ml of 1.00 MKH2PO4 with 105.03 ml of 1.00 M K2HPO4 to make 250 ml Solution
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