Question

Hemoglobin (Hb) hinds to both oxygen and carhon monoxide. When the carbon monoxide replaces the oxygen in an organism, the following reaction occurs: $$ \mathrm{HbO}_{2}(a q)+\mathrm{CO}(g) \rightleftharpoons \mathrm{HbCO}(a q)+\mathrm{O}_{2}(g) $$ At \(37^{\circ} \mathrm{C}, K\) is about 200 . When equal concentrations of \(\mathrm{HbO}_{2}\) and \(\mathrm{HbCO}\) are present, the effect of CO inhalation is fatal. Assuming \(\mathrm{P}_{\mathrm{O}_{2}}=0.21 \mathrm{~atm},\) what is \(\mathrm{P}_{\mathrm{CO}}\) when \(\left[\mathrm{HbO}_{2}\right]=\) \([\mathrm{HbCO}] ?\)

Short answer

Answer: The partial pressure of CO is approximately 0.00105 atm.

Step-by-step solution

Write the equilibrium constant expression

For the given reaction, we can write the equilibrium constant expression as follows: $$ K = \frac{[HbCO] \cdot P_{O_2}}{[HbO_2] \cdot P_{CO}} $$

Substitute the given values and conditions

We are given that \(K=200\), and we are asked to find \(P_{CO}\) when \([HbO_2]=[HbCO]\) and \(P_{O_2} = 0.21 atm\). Substitute these values and conditions into the equilibrium constant expression: $$ 200 = \frac{[HbCO] \cdot 0.21}{[HbO_2] \cdot P_{CO}} $$ Since \([HbO_2] = [HbCO]\), we can substitute \([HbO_2]\) in place of \([HbCO]\): $$ 200 = \frac{[HbO_2] \cdot 0.21}{[HbO_2] \cdot P_{CO}} $$

Solve for P_CO

Now, we can solve for \(P_{CO}\) by simplifying the equation: $$ P_{CO} = \frac{0.21}{200} $$ Calculating the value of \(P_{CO}\): $$ P_{CO} \approx 0.00105 \ atm $$

State the conclusion

When the concentrations of HbO2 and HbCO are equal, and the partial pressure of O2 is 0.21 atm, the partial pressure of CO that causes a fatal effect is approximately 0.00105 atm.