As a technician in a large pharmaceutical research firm, you need to produce 150.mL of 1.00 M potassium dihydrogen phosphate buffer solution of pH = 6.79. 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?
Express your answer to three significant digits with the appropriate units.
We use Henderson-Hasselbalch equation.
pH= pKA + log [Base] / [Acid]
6.79 = 7.21 + log [HPO4] / [H2PO4]
log [HPO4] / [H2PO4] = -0.42
[HPO4] / [H2PO4] = 0.380 / 1.00
But, we know that [HPO4] + [H2PO4] = 1.00 M
[HPO4] = 1.00 - [H2PO4]
0.380 = [1.00 - [H2PO4] ] / [H2PO4]
0.380 [H2PO4] = 1.00 - [H2PO4]
1.380 [H2PO4] = 1.00
[H2PO4] = 0.725 M
We have 2.00 L of 1.00 M KH2PO4 stock solution. We need 150.0 mL of 0.725 M
Thus, M1V1 = M2V2
(1.00)(V1) = (0.725) (150)
V1 = 108.75 mL = 109 mL
Thus, we need 109 mL of 1.00 M KH2PO4 is needed.
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