The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pK
ID: 1054722 • Letter: T
Question
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pK_a of its conjugate acid and the ratio of the concentrations of the conjugate base and acid. The equation is important in laboratory work that makes use of buffered solutions, in industrial processes where pH needs to be controlled, and in medicine, where understanding the Henderson-Hasselbalch equation is critical for the control of blood pH Part A As a technician in a large pharmaceutical research firm, you need to produce 300. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.99. The pK_a of H_2PO_4^- is 7.21. You have the following supplies: 2.00 L of 1.00 M KH_2PO_4 stock solution, 1.50 L of 1.00 M K_2HPO_4 stock solution, and a carboy of pure distilled H_2O. How much 1.00 M KH_2PO_4 will you need to make this solution? (Assume additive volumes.) Express your answer to three significant digits with the appropriate units. The Henderson-Hasselbalch equation in medicine Carbon dioxide (CO_2) and bicarbonate (HCO_3^-) concentrations in the bloodstream are physiologically controlled to keep blood pH constant at a normal value of 7.40. Physicians use the following modified form of the Henderson-Hasselbalch equation to track changes in blood pH: PH = pK_a + log [HCO_3^-]/(0.030)(P_CO_2) where [HCO_3^-] is given in millimoles/liter and the arterial blood partial pressure of CO_2 is given in mmHg. The pK_a of carbonic acid is 6.1. Hyperventilation causes a physiological state in which the concentration of CO_2 in the bloodstream drops. The drop in the partial pressure of CO_2 constricts arteries and reduces blood flow to the brain, causing dizziness or even fainting. Part B If the normal physiological concentration of HCO_3^- is 24 mM, what is the pH of blood if P_CO_2 drops to 35.0 mmHg ?Explanation / Answer
Part A:
Let a = volume of KH2PO4
b = 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)
6.99 = 7.21 + log(b/a)
b/a = 0.6026 => b = 0.6026 a
Total moles of phophate = final volume x total concentration of phosphate
= 350/1000 x 1.00
= 0.350 mol
Thus a + b = 0.350
a + 0.6025 a = 0.350
a = 0.218
Volume of KH2PO4 needed = a = 0.218 L = 218 mL
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