Carbonic acid. H_2CO_3. and bicarbonate. HCO_3^- are present in the blood and pl
ID: 974256 • Letter: C
Question
Carbonic acid. H_2CO_3. and bicarbonate. HCO_3^- are present in the blood and play a major role in regulating blood pH, which is maintained within a narrow range of 7.35 to 7.45 in a healthy person, Acidosis results if the PH level drops below 7.35. alkalosis if the pH is raised above 7.45. A patient suffers from alkalosis as a result of an overdose of aspirin tablets. In the emergency room, the physician measures her blood ph to be equal to 7.55. Determine the ratio of HCO_3^- to H_2CO_3 in the patient's blood at this pH. How does this ratio compare to the ratio of HCO_3^- to H_2CO_3 in normal blood? (Use an average value of 740 for the pH of normal blood.) Can the H_2CO_3/HCO_3^- system work effectively as a buffer in this patient under these conditions? Explain, using concepts developed in this activity. The patient described in O41 has a blood bicarbonate concentration of 18 mM two hours after being brought to the emergency room. Normal blood bicarbonate concentrations are about 24 mM. Calculate the concentration of H_2CO_3 in the patient's blood and compare this value to the concentration of H_2CO_3 found in a normal persons blood Can the H_2CO_3HCO_3^- system work effectively as a buffer in this patient under these conditions? Explain, using concepts developed in this activity.Explanation / Answer
41. We treat the carbonic acid/bicarbonate buffer from a chemist’s point of view and employ Henderson-Hasselbalch equation. We write the dissociation of carbonic acid as
H2CO3 <======> H+ + HCO3-
The equilibrium constant, also called the acid dissociation constant, Ka is
Ka = [H+][HCO3-]/[H2CO3]
Taking logarithm on both sides,
log10Ka = log10{[H+][HCO3-]/[H2CO3]} = log10[H+] + log10[HCO3-]/[H2CO3]
or, - pKa = - pH + log10[HCO3-]/[H2CO3] (since pKa = - log10Ka and pH = -log10[H+])
or, pH = pKa + log10[HCO3-]/[H2CO3]……(1)
The pH of the blood in the patient is 7.55 and the pKa of carbonic acid is 6.35. Therefore,
7.55 = 6.35 + log10[HCO3-]/[H2CO3]
or, 1.2 = log10[HCO3-]/[H2CO3]
or, [HCO3-]/[H2CO3] = 101.2 = 15.85/1 (since we are taking a ratio)
In a healthy person, the blood pH = 7.40. Hence, substituting,
7.40 = 6.35 + log10[HCO3-]/[H2CO3]
or, 1.05 = log10[HCO3-]/[H2CO3]
or, [HCO3-]/[H2CO3] = 101.05 = 11.22/1 (again a ratio)
The bicarbonate concentration in the patient is too high for the carbonic acid/bicarbonate buffer to function effectively.