Question

Excess protons in the blood decrease the amount of HCO − 3 and thus reduce the buffering capacity of blood. A rapid drop in pH could lead to death. Normal values for blood are pH = 7.4 , [ HCO − 3 ] = 24.0 mM , [ CO 2 ] = 1.2 mM . (a) If a patient has a blood pH = 7.03 and [CO2] = 1.2 mM, what is the [HCO3−] in the patient’s blood? The pKa of HCO3−= 6.1.


(b) Suggest a possible treatment for metabolic acidosis.


(c) Why might the suggestion for part (b) be of benefit to middle-distance runners?

Answer 1

Answer:

a) [HCO₃⁻] = 10,2 mM.

b) Sodium bicarbonate.

c) Yes.

Explanation:

a) The equilibrium of this reaction is:

CO₂ + H₂O ⇄ HCO₃⁻ + H⁺

Using Henderson-Hasselbalch equation:

pH = pka + log₁₀  $\frac{[HCO_{3}^-]}{[CO_2]}$

Replacing:

7,03 = 6,1 + log₁₀  $\frac{[HCO_{3}^-]}{[1,2 mM]}$

Thus, [HCO₃⁻] = 10,2 mM

b) A possible treatment of metabolic acidosis is with sodium bicarbonate. By Le Chateleir's principle the increasing of HCO₃⁻ will shift the equilibrium to the left decreasing thus, H⁺ concentration.

c) The shifting of the equilibrium to the left will increase CO₂ concentration producing in the body the need to increase breathing, increasing, thus, concentration of O₂ improving cardiac function in exercise.