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 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.

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.