Carbon monoxide replaces oxygen in hemoglobin according to the reaction: HbO_2(a

Question

Carbon monoxide replaces oxygen in hemoglobin according to the reaction: HbO_2(aq) + CO(aq) HbCO(aq) + O2(aq) The equilibrium constants for the reactions below are: Hb(aq) + O_2(aq)o HbO_2 (aq) K=1.8 Hb(aq) + CO(aq) HbCO(aq) K 306 Compute the equilibrium constant for the reaction presented above. Suppose that air is polluted with CO to a level of 0.1 % (mole ratio) The air contains 20.0 % (mole ratio) oxygen. If the CO/O_2 m the blood is the same as that in air, what is the HbCO/HbO_2 ratio in the blood?

Explanation / Answer

(a)
Net reaction: HbO2(aq) + CO(aq) = HbCO(aq) + O2(aq)     
Individual equilibrium constants:

Reaction1: Hb(aq) + O2(aq) = HbO2(aq); K1 = 1.8
Reaction2: Hb(aq) + CO(aq) = HbCO(aq); K2 = 306

Reaction2 - Reaction1:

Reaction2: Hb(aq) + CO(aq) = HbCO(aq); K2
- Reaction1: HbO2(aq) = Hb(aq) + O2(aq); K' = 1/K1

CO(aq) + HbO2(aq) = HbCO(aq) + O2(aq); Keq = K2/K1 = 170

(b)
[CO(aq)]/[O2(aq)] = 0.1/20 = 0.005

CO(aq) + HbO2(aq) = HbCO(aq) + O2(aq); Keq = K2/K1 = 170

Keq = [HbCO(aq)]*[O2(aq)]/([CO(aq)]*[HbO2(aq)]) = 170

[HbCO]/[HbO2] = 170*[CO(aq)]/[O2(aq)] = 170*0.005 = 0.85

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