The Henderson-Hasselbalch equation relates the pH of a buffer solution to the p

Question

The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa 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 150. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.86. 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.)

Part B.

Carbon dioxide (CO2) and bicarbonate (HCO3) 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=pKa+log[HCO3](0.030)(PCO2)

where [HCO3] is given in millimoles/liter and the arterial blood partial pressure of CO2 is given in mmHg. The pKa of carbonic acid is 6.1. Hyperventilation causes a physiological state in which the concentration of CO2 in the bloodstream drops. The drop in the partial pressure of CO2 constricts arteries and reduces blood flow to the brain, causing dizziness or even fainting.

If the normal physiological concentration of HCO3 is 24 mM, what is the pH of blood if PCO2drops to 22.0 mmHg ?

Explanation / Answer

Part A : pH = pKa + log(base/acid)

let x amount of KH2PO4 is added

6.86 = 7.21 + log(150x/150 - x)

67 - 0.44x = 150x

x = 0.445 mmol

Volume of 1.00 M KH2PO4 required = 0.445 mmol/1 M = 0.445 L of solution required

Part B - Using the given equation,

pH = pKa + log([HCO3-](0.030)(pCO2)

with,

[HCO3-] = 0.024 M

pCO2 = 22 mmHg = 0.029 atm

we get,

pH = 6.1 + log(0.024 x 0.030 x 0.029) = 1.42

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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