As a technician in a large pharmaceutical research firm, you need to produce 450. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.95. 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.
A) How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.)
PART B:
If the normal physiological concentration of HCO3− is 24 mM, what is the pH of blood if PCO2 drops to 28.0 mmHg ?
Part A)
pH = 6.95
pKa = 7.21
total volume = 450 mL
volume of KH2PO4 = x
volume of K2HPO4 = 450 - x
pH = pKa + log [K2HPO4 / KH2PO4]
6.95 = 7.21 + log [450 - x / x]
450 - x / x = 0.5495
450 - x = 0.5495 x
x = 290
volume of KH2PO4 = 290. mL
Part B)
[HCO3-] = 24 mM = 0.024 M
PCO2 = 28 mmHg = 0.03684 atm
pH = pKa + log ([HCO3-] / 0.03 x pCO2 )
= 6.1 + log [0.024 / 0.03 x 0.03684]
pH = 7.44
As a technician in a large pharmaceutical research firm, you need to produce 450. mL of...