6. Assume that the normal blood buffer contains 0.00080 M carbonic acid and 0.00

Question

6. Assume that the normal blood buffer contains 0.00080 M carbonic acid and 0.0085 M hydrogen carbonate; the pKa 6.35 for carbonic acid and the volume of blood in the body is 7.00 L. The blood pH, due to disruption, is now 7.20. What is the ratio of LHCO3 ]/LH2CO3] now that the blood has been challenged? How many moles of hydrogen carbonate must be added to the blood to bring the carbonic acid/hydrogen carbonate ratio back to a normal pH 7.4? How many milliliters of 0.10 M hydrogen carbonate must be added to the blood?

Explanation / Answer

pH= pKa+ log[HCO3-] /[H2CO3]

when pH= 7.2

7.2=6.35+ log [HCO3-]/[H2CO3]

0.85= log [HCO3-]/ [H2CO3]

7.079= [HCO3-]/[H2CO3]

for pH= 7.4

7.4-6.35= log [HCO3-]/[H2CO3]

[HCO3-]/[H2CO3]= 101.05= 11.22

moles of H2CO3= moles of HCO3- = 0.0085*7=0.0595

when x moles of H2CO3 is added, x moles of HCO3- gets added

So at equilibrium [H2CO3]= 0.0595-x and [HCO3-]= 0.0595+x

since its [HCO3-]/[H2CO3] mole ratio and concentration ratio remain the same.

(0.0595+x)/0.0595-x)= 11.22

0.0595+x =11.22*0.0595-11.22x

12.22x= 11.22*0.0595-0.0595 , x =0.8635M

Volume of H2CO3 to be added = 0.8635/0.1 = 8.635 L = 8635mL

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