Question

The binding of oxygen to hemoglobin in the blood involves the equilibrium reaction: Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction: HbO2(aq) + CO(aq)HbCO(aq) + O2(aq) a. If the normal pH of blood is controlled within a range of 7.35 to 7.45. If the pKw is 13.63, what is the normal range of [H+] and [OH- b. What would happen if oxygenated hemoglobin became too acidic? c. Use the reaction and associated equilibrium constants at body temperature to find the equilibrium constant for the reaction above if: Hb(aq) + O2(aq) ←→ Hb02(aq) Kc=1.8 Hb(aq)-CO(aq) ← → HbCO(aq) KC-306 Note that the value of K for the new equation is determined by combining both equiations like we did with Hess's Law in Gen Che d. Supposed that the air mixture becomes polluted with carbon monoxide at a level of 0.10%. Assuming the air contains 20.0% oxygen and that the oxygen and carbon monoxide ratios that dissolve in blood are identical to the ratios in the air, what is the ratio of HbCO to Hb02 in the bloodstream and what does that say about how much CO will interrupt the uptake of oxygen? (Hint: % is just a concentration!)

Answer 1

(a) [H^+] = 10^-p^H implies p^H = -log[H^+] = 10^-7.45 = 3.55 times 10^-8 m range of [H^+] = 3.55 times 10^-8 to 4.47 times 10^-8 m. [H^+] = 10^-7.35 = 4.47 times 10^-8 M p^OH + p^H = p^kw = 13.63 implies p^OH = 13.63 - 7.35 = 6.28 p^OH = 13.63 - 7.45 = 6.18 [OH^-] = 10^-6.28 = 5.25 times 10^-7 [OH^-] = 10^-6.18 = 6.61 times 10^-7 range of [OH^-] = 5.25 times 10^-7 to 6.61 times 10^-7 m (b) According to Le chatelier principle, if concentration of products are increased equilibrium will shift in the backward direction. HbH^exclus (aq) + O_2 (aq) rightleftarrow HbO_2 (aq.) + H^exclus (aq.) On increasing acidic strength of oxygenated hemoglobin, equilibrium will shift in backward direction i.e. to consume H^exclus ions. (c) Hb(aq.) + O_2 (aq.) rightleftarrow HbO_2 (aq.) k_c = 1.8 equation (1) Hb(aq.) + CO (aq.) rightleftarrow HbCO (aq.) k_c = 306 equation (2)