A.) You need to produce a buffer solution that has a pH of 5.24. You already hav

Question

A.) You need to produce a buffer solution that has a pH of 5.24. You already have a solution that contains 10. mmol(millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74.

B.) As a technician in a large pharmaceutical research firm, you need to produce 450.mL of a potassium dihydrogen phosphate buffer solution of pH = 6.89. 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.)

C.)If the normal physiological concentration of HCO3? is 24 mM, what is the pH of blood if PCO2 drops to 28.0mmHg ?

Explanation / Answer

a) pH of the buffer 5.24

5.24 = 4.74 + log (acetate/10)

log (acetate/10) = 0.5

[acetate] = 31.62 mmol

b) Using Handerson-Hasselbalch equation:

pH = pKa + log [HPO42-]/[H2PO4 -]

6.89 = 7.21 + log [HPO42-]/[H2PO4-]

[HPO42-]/[H2PO4-] = 0.478

[HPO42-] = 0.478 x[H2PO4-]

Total moles of phosphate = final volume x total conc. of phosphate

= 450/1000 x 1 = 0.45 mole

So,

[HPO42-] + [H2PO4-] = 0.45

0.478 [H2PO4-] + [H2PO4-] = 0.45

[H2PO4-] = 0.304

So volume of [H2PO4-] required is 0.304 L = 304 ml

C) using Handerson-Hasselbalch equation:

pH = pKa + log[HCO3](0.030)(PCO2)
pH = 6.1 + log[24 mM](0.030)(28.0 mmHg)
pH = 6.1 + log(20.16)
pH = 6.1 + 1.3045
pH = 7.4045

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