Part A As a technician in a large pharmaceutical research firm, you need to prod

Question

Part A As a technician in a large pharmaceutical research firm, you need to produce 400.mL of 1.00 M a phosphate buffer solution of pH = 7.24. The pKa of H2PO4%u2212 is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution? Express your answer numerically in milliliters to three significant figures. Part A As a technician in a large pharmaceutical research firm, you need to produce 400.mL of 1.00 M a phosphate buffer solution of pH = 7.24. The pKa of H2PO4%u2212 is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution? Express your answer numerically in milliliters to three significant figures.

Explanation / Answer

Henderson%u2013Hasselbalch equation
pH = pKa + log [anion] / [acid]
7.24 = 7.21 + log [anion] / [acid]
0.03 =log [anion] / [acid]
10 to the x of both sides
1.07151930524 = [anion] / [acid]

The ratio of concentrations in a solution, (400 ml), is the same as the ratio of moles in that volume,

so I am going to convert the ml's of into moles, where I arbitrarially choose one to be where that I added "X" millilitres of it , & the other becomes that I added "400-X" ml

1.07151930524= [M anion] / [M acid]

1.07151930524= [M K2HPO4] / [M KH2PO4]

Multiply the molarity times the volume of each to get the moles of each

1.07151930524 = (400ml - X)(1.00 Molar K2HPO4) / (X ml)(1.00 Molar KH2PO4)

1.07151930524 = (400 - X) / (X)

of 1.00 molar KH2PO4 acid

Mix 193.09 of 1.00 Molar KH2PO4 with 216 ml of 1.00 Molar K2HPO4

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