Part A As a technician in a large pharmaceutical research firm, you need to prod

Question

Part A

As a technician in a large pharmaceutical research firm, you need to produce 300.mL of 1.00 M potassium dihydrogen phosphate buffer solution of pH = 7.04. 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?

Express your answer to three significant digits with the appropriate units

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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.

Part B

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

Express your answer numerically using two decimal places.

Explanation / Answer

PART A

6,83 = 7,21 + log (conc. HPO4)/(conc. H2PO4)
-0,38 = log (conc. HPO4)/(conc. H2PO4)
0,417 = (conc. HPO4)/(conc. H2PO4)
You also know that (conc. HPO4) + (conc. H2PO4) = 1,00 M which after rearraging gives (conc. HPO4) = 1,00 - (conc. H2PO4). This expression needs to be integrated into this: 0,417 = (conc. HPO4)/(conc. H2PO4) which gives you:
0,417 = (1,00 - (conc. H2PO4))/(conc. H2PO4)
After rearanging you will end up with:
conc. H2PO4 = 1,00/1,417 = 0,706 M
You have 2,00 L of 1,00 M H2PO4 (the stock solution). You need 100 mL of 0,706 M.
conc. stock x volume stock = conc. needed x volume needed
volume stock = (0,706 x 0,1)/1,00 = 70,6 mL

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See, there are really three possible answers to this question based on whether you're asked to account for acid-base buffering in the blood and renal compensation (how the kidneys react to changes in blood pH). So, I'll give you the three possible scenarios:

1. You aren't accounting for acid-base buffering or renal compensation; all you want to know is what the pH would be if neither of these occurred. Simply plug in the values you're given, it's that simple. pH = 6.1 + log (24/(0.03 x 23)) = 7.64.

2. You are accounting for acid-base buffering but not renal compensation. Simply, HCO3 levels in the blood change slightly in the same direction as the changes in CO2 levels because of the buffer system. For every decrease (or increase) in 10 mmHg of CO2 (normal value is 40mm Hg), HCO3 changes by ~2 units in the same direction. 40-23=17, so HCO3 would decrease by 3.4 units. 24-3.4=21.6. Use 21.6 instead of 24 for HCO3, and you get pH = 6.1 + log(21.6/(0.03 x 23)) = 7.59. This isn't a drastic change from 7.64 as you can see.

3. Within a few days, your kidneys will have excreted HCO3 into the urine to lower the HCO3 level to restore the pH back to 7.4 (it's never complete but it's better than nothing)...w/ renal compensation, HCO3 changes by 4 units for every change in 10 mmHg of CO2. In this case, you'd have a decrease in HCO3 by 6.8 units, so your new value for HCO3 would be 24-6.8=17.2. Use this value in the HH equation, and you get pH = 6.1 + log(17.2/(0.03 x 23)) = 7.49.

Keep in mind that small changes in blood pH have drastic effects. Blood is usually always maintained between 7.3 and 7.5. So the values shouldn't deviate too far from these. You shouldn't encounter a pH below 7.1 or above 7.8.

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