Question

Part A

As a technician in a large pharmaceutical research firm, you need to produce 150. mL of a potassium dihydrogen phosphate buffer solution of pH = 7.00. 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.) Express your answer to three significant digits with the appropriate units.

Part B

If the normal physiological concentration of HCO3− is 24 mM, what is the pH of blood if PCO2 drops to 24.0 mmHg ?

Express your answer numerically using two decimal places.

Answer 1

pH = pKa + log [anion] / [acid]
7 = 7.21 + log [anion] / [acid]
-0.21 =log [anion] / [acid]
10 to the x of both sides
0.6166 = [anion] / [acid]

the ratio of concentrations has to be 0.6166 , (to less sig figs, I know)...
but the ratio of concentrations in a solution, (150 mL), is the same as the ratio of moles in that volume,..so convert the mL of each acid and base added... into moles, where choose one to be where that added "X" millilitres of it , & the other becomes that added "150-X" ml
0.6166 = [M anion] / [M acid]

0.6166 = [M K₂HPO₄] / [M KH₂PO₄]

now multiply the molarity times the volume of each to get the moles of each ,

0.6166 = (150ml - X)(1.00 Molar K₂HPO₄) / (X ml)(1.00 Molar KH₂PO₄)

0.6166 = (150 - X) / (X)

0.6166 X = 150 - X

1.6166 X = 150 ml

X= 92.8 ml of 1.00 molar KH₂PO₄ "acid" is needed to make this solution.

Part B

pH = pKa + log [anion] / [acid]

pH = 6.1 + log (24/(0.03x24))

pH = 7.62