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 100. mL of a potassium dihydrogen phosphate buffer solution of pH = 6.92. 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 = _________________   _____(units)_____

Part B

If the normal physiological concentration of HCO3 is 24 mM, what is the pH of blood if PCO2 drops to 20.0 mmHg ?

Express your answer numerically using two decimal places.

pH=______________________

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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