As a technician in a large pharmaceutical research firm, you need to produce 300
ID: 487957 • Letter: A
Question
As a technician in a large pharmaceutical research firm, you need to produce 300. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.97. 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.)
As a technician in a large pharmaceutical research firm, you need to produce 300. mL of a potassium dihydrogen phosphate buffer solution of pH 6.97. The pKa of H2PO4 is 7.21. You have the following supplies: 2.00 L of 1.00 MKH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O How much 1.00 MKH2 POM will you need to make this solution? (Assume additive volumes.) Express your answer to three significant digits with the appropriate units. Volume of KH2PO4 needed .332 L Submit Hints My Answers Give Up Review Part incorrect, Try Again; 2 attempts remainingExplanation / Answer
PH = PKa + log[K2HPO4]/[KH2PO4]
PH = 6.97
PKa = 7.21
6.97 = 7.21 + log[K2HPO4]/[KH2PO4]
log[K2HPO4]/[KH2PO4] = 6.97-7.21
log[K2HPO4]/[KH2PO4] = -0.24
[K2HPO4]/[KH2PO4] = 10^-0.24
[K2HPO4]/[KH2PO4] = 0.575
[K2HPO4] = 0.575*[KH2PO4]
This means for every 1 moleK2HPO4 added 0.575 moles of KH2PO4
x + 0.575x = 300
x = 190.47
190.47ml of K2HPO4 added 109.52ml of KH2PO4
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