Problem 61
Question
Hemoglobin (Hb) binds to both oxygen and carbon 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 \(\left[\mathrm{HbO}_{2}\right]=[\mathrm{HbCO}] ?\)
Step-by-Step Solution
Verified Answer
Answer: The partial pressure of carbon monoxide when the concentrations of HbO2 and HbCO are equal is 0.00105 atm.
1Step 1: Write the expression for the equilibrium constant (K)
Based on the given reaction, we can write the equilibrium constant K as follows:
$$K=\frac{[\mathrm{HbCO}][\mathrm{O}_{2}]}{[\mathrm{HbO}_{2}][\mathrm{CO}]}$$
2Step 2: Use the given information to set up the equation
We are given that the concentrations of HbO2 and HbCO are equal, so we can rewrite the K expression as:
$$K=\frac{[\mathrm{HbCO}][\mathrm{P}_{\mathrm{O}_{2}}]}{[\mathrm{HbO}_{2}][\mathrm{P}_{\mathrm{CO}}]}$$
Since the concentrations of HbO2 and HbCO are equal, we can replace both with a variable x.
$$K=\frac{x[\mathrm{P}_{\mathrm{O}_{2}}]}{x\mathrm{[\mathrm{P}_{\mathrm{CO}}]}}$$
3Step 3: Cancel out x
Now, since both the numerator and denominator have an x term, we can cancel them out:
$$K=\frac{\mathrm{P}_{\mathrm{O}_{2}}}{\mathrm{P}_{\mathrm{CO}}}$$
4Step 4: Solve for P_CO
To find P_CO, we can rearrange the equation to isolate the term:
$$\mathrm{P}_{\mathrm{CO}}=\frac{\mathrm{P}_{\mathrm{O}_{2}}}{K}$$
5Step 5: Substitute the given values
Now, we can substitute the given values for K and P_O2 into the equation:
$$\mathrm{P}_{\mathrm{CO}}=\frac{0.21\ \mathrm{atm}}{200}$$
6Step 6: Calculate P_CO
Finally, we can calculate the value of P_CO:
$$\mathrm{P}_{\mathrm{CO}}= 0.00105\ \mathrm{atm}$$
Therefore, the partial pressure of carbon monoxide when the concentrations of HbO2 and HbCO are equal is 0.00105 atm.
Key Concepts
Hemoglobin BindingPartial PressureChemical Equilibrium
Hemoglobin Binding
Hemoglobin is an essential protein found in red blood cells. It plays a crucial role in transporting oxygen from the lungs to tissues throughout the body. However, it is not exclusively bound to oxygen; it can also bind to other gases, such as carbon monoxide (CO). This dual binding capability of hemoglobin is interesting but can be dangerous.
- Oxygen (O2) and carbon monoxide both bind to the iron-containing part of the hemoglobin molecule.
- The binding of CO to hemoglobin forms carboxyhemoglobin (HbCO), which is a much more stable complex than oxyhemoglobin (HbO2).
- Because CO binds more strongly to hemoglobin than oxygen does, even small amounts of CO can prevent oxygen from binding, leading to tissue oxygen deprivation.
Partial Pressure
Partial pressure is a concept in chemistry that helps us understand the contribution each gas makes to the overall pressure of a mixture. It is an especially useful way to think about gases in terms of their interactions and equilibrium, such as in the reaction involving hemoglobin binding to oxygen and carbon monoxide.
- Each gas in a mixture exerts its own pressure as if it alone occupied the entire volume.
- This pressure is called the partial pressure, denoted as \(P\).
- In the chemical reaction described, the partial pressures of O2 and CO are considered to calculate the equilibrium.
Chemical Equilibrium
Chemical equilibrium is the state in which the concentrations of all reactants and products remain constant over time, in a closed system. With the hemoglobin binding reactions, understanding equilibrium helps explain how gases compete for binding sites on the protein.
- The reaction involving hemoglobin can reach an equilibrium where the rates of the forward and reverse reactions are equal.
- The equilibrium constant \( K \) quantifies the ratio of product concentrations to reactant concentrations at equilibrium.
- For the given reaction, \( K \) helps determine how likely the conversion of HbO2 and CO to HbCO and O2 is.
Other exercises in this chapter
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