As a technician in a large pharmaceutical research firm, you need to produce 400
ID: 776278 • Letter: A
Question
As a technician in a large pharmaceutical research firm, you need to produce 400.mL of 1.00 Mpotassium dihydrogen phosphate buffer solution of pH = 6.82. 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?'
B) If the normal physiological concentration of HCO3? is 24 mM, what is the pH of blood if PCO2drops to 22.0mmHg ?
Explanation / Answer
Let a be the volume of KH2PO4 and b be the volume of K2HPO4 needed in liters
Moles of KH2PO4 = volume x concentration = a x 1.00 = a mol
Moles of K2HPO4 = volume x concentration = b x 1.00 = b mol
Henderson-Hasselbalch equation:
pH = pKa + log([K2HPO4/[KH2PO4]])
= pKa + log(moles of K2HPO4/moles of KH2PO4) since final volume (= 400 mL) is the same for both
7.21 = 6.82 + log(b/a)
log(b/a) = 0.39
b/a = 100.07 = 2.455 => b = 2.455a
Total moles of phophate = final volume x total concentration of phosphate
= 400/1000 x 1.00 = 0.4 mol
Thus a + b = 0.4
a + 2.455a = 0.4
a = 0.1157
Volume of KH2PO4 needed = a = 0.1157 L = 115.7 mL
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