Part A As a technician in a large pharmaceutical research fi

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 = 6.77. 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.

Part B)

If the normal physiological concentration of HCO3 is 24 mM, what is the pH of blood if PCO2 drops to 33.0 mmHg ?

Extra Info:

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 pKaof 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 CO2constricts arteries and reduces blood flow to the brain, causing dizziness or even fainting.

Solution

We use Henderson-Hasselbalch equation.

pH= pKA + log [Base] / [Acid]

6.79 = 7.21 + log [HPO4] / [H2PO4]

log [HPO4] / [H2PO4] = -0.42

[HPO4] / [H2PO4] = 0.380 / 1.00

But, we know that [HPO4] + [H2PO4] = 1.00 M

[HPO4] = 1.00 - [H2PO4]

0.380 = [1.00 - [H2PO4] ] / [H2PO4]

0.380 [H2PO4] = 1.00 - [H2PO4]

1.380 [H2PO4] = 1.00

[H2PO4] = 0.725 M

We have 2.00 L of 1.00 M KH2PO4 stock solution. We need 150.0 mL of 0.725 M

Thus, M1V1 = M2V2

(1.00)(V1) = (0.725) (150)

V1 = 108.75 mL = 109 mL

Thus, we need 109 mL of 1.00 M KH2PO4 is needed.