Question

# The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of its conjugate...

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 100. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.99. 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.

Volume of KH2PO4 needed = ______
PART B

The Henderson-Hasselbalch equation in medicine

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 PCO2 drops to 29.0 mmHg ?

Express your answer numerically using two decimal places.

pH = _______

Part A

pH = pKa + log( [K2HPO4] / KH2PO4] )

log( [K2HPO4] / KH2PO4] ) = pH - pKa

[K2HPO4] / KH2PO4] = 10^(pH - pKa)

Cb x Vb / ( Ca x Va ) = 10^(pH - pKa)

Ca = Cb . . . therefore :

Vb / Va = 10^(pH - pKa)

Vb / Va = 10^(7.33 - 7.21)

Vb / Va = 1.318

. . . and :

Va + Vb = 100

. . . then, after solve :

Va = 43.14 ml of KH2PO4 . . . and . . . Vb = 56.86 ml of K2HPO4

Volume of KH2PO4 needed = 43.14 ml

Part B

pH = pKa + log[HCO3−]/(0.030)(PCO2)
pH = 6.1 + log[24 mM]/(0.030)(28.0 mmHg)
pH = 6.1 + log(27.58)
pH = 6.1 + 1.44
pH = 7.54

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