Question

Buffers are common (and extremely important) in biological systems, including in human body. One of the most important biological buffers in human blood is the buffer established between carbonic acid and bicarbonate ion (H2CO3/HCO3–). At 25 oC, H2CO3 has a Ka1 of 4.45x10–7 and a Ka2 of 4.69x10–11.

a. Will Ka1 or Ka2 of H2CO3 be instrumental in determining blood pH? Briefly explain your reasoning.

b. Let’s pretend that there was an equal ratio of carbonic acid to bicarbonate (note that this ratio would not be sustainable to human life), what would be the pH of blood?

c. But wait, the pH of blood is regulated to be 7.45 (slightly basic pH)! What is the ratio of bicarbonate to carbonic acid concentration at this pH?

d. Metabolic acidosis is a condition in which the body produces excessive acid through metabolic processes. Consider the following clinical scenario: A 34-year-old man presents at the Mt. Nittany Medical Center emergency room with fatigue, nausea, headache, and low blood pressure. Upon further examination, it is found that his blood concentration of H2CO3 is 2.4 mmol/L, and the concentration of HCO3− in his blood is 22 mmol/L. What is the pH of his blood?

e. Excessive hyperventilation (rapid deep breathing) can lead to the opposite problem, respiratory alkalosis. Consider the following clinical scenario: A 23-year-old man with a coffee addiction, anxious about his move out to Steamboat Springs (altitude: 6,732 ft.), winds up at the Yampa Valley Medical Center after a lengthy bout of hyperventilation. Upon clinical examination, it is found that his blood pH is 7.53. If his blood concentration of HCO3– is 25 mmol/L, what is the concentration of carbonic acid?

Answer 1

a) The Ka1 will be instrumental in determing blood pH. Since [H+] coming from the Ka2 will be negligible in comparison of coming from Ka1

b)

pH = pKa1 + log[bicarbonate/carbonic acid]

pH = (7 - log(4.45)) + log(1) = 7 - 0.65 = 6.35

Hence the pH in this case will be equal to 6.35

c)

pH = pKa1 + log[bicarbonate/carbonic acid]

7.45 = 6.35 + log[bicarbonate/carbonic acid]

log[bicarbonate/carbonic acid] = 1.10

[bicarbonate]/[carbonic acid] = 10^(1.10) = 12.59

Hence the ratio of bicarbonate to carbonic acid at this pH will be 12.59

d)

pH = pKa1 + log[bicarbonate/carbonic acid]

pH = 6.35 + log(22/2.4) = 6.35 + 0.96 = 7.31

e)

pH = pKa1 + log[bicarbonate/carbonic acid]

7.53 = 6.35 + log[bicarbonate/carbonic acid]

[bicarbonate]/[carbonic acid] = 10^(1.18) = 15.13

[carbonic acid] = 25/15.13 = 1.65 mmol/L

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