As stated in the chapter, carbon monoxide has a much higher affinity for hemoglobin than oxygen does. (a) Write the equilibrium constant expression \(\left(K_{\mathrm{c}}\right)\) for the following process: $$ \mathrm{CO}(g)+\mathrm{HbO}_{2}(a q) \rightleftharpoons \mathrm{O}_{2}(g)+\mathrm{HbCO}(a q) $$ where \(\mathrm{HbO}_{2}\) and \(\mathrm{HbCO}\) are oxygenated hemoglobin and carboxyhemoglobin, respectively. (b) The composition of a breath of air inhaled by a person smoking a cigarette is \(1.9 \times 10^{-6} \mathrm{~mol} / \mathrm{L} \mathrm{CO}\) and \(8.6 \times 10^{-3} \mathrm{~mol} / \mathrm{L} \mathrm{O}_{2} .\) Calculate the ratio of \([\mathrm{HbCO}]\) to \(\left[\mathrm{HbO}_{2}\right]\), given that \(K_{\mathrm{c}}\) is 212 at \(37^{\circ} \mathrm{C}\).

Step by step solution

01

Writing down the equilibrium constant expression for the given process

According to the rule of equilibrium constants for a reaction \(a A + b B \rightleftharpoons c C + d D\), the equilibrium constant \(K_c\) is given by \(K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\), where a,b,c,d are the stoichiometric coefficients and [A],[B],[C],[D] are the equilibrium concentrations of the species involved. Applying this rule to the given process \(\mathrm{CO}(g)+\mathrm{HbO}_{2}(a q) \rightleftharpoons \mathrm{O}_{2}(g)+\mathrm{HbCO}(a q)\), the equilibrium constant expression becomes \(K_c = \frac{[\mathrm{O}_2][\mathrm{HbCO}]}{[\mathrm{CO}][\mathrm{HbO}_2]}\)

02

Calculate the ratio of [HbCO] to [HbO2]

Substitute given values into the equilibrium constant expression: \(K_c = 212 = \frac{[\mathrm{O}_2][\mathrm{HbCO}]}{[\mathrm{CO}][\mathrm{HbO}_2]}\). We can rearrange this equation to find the ratio of [HbCO] to [HbO2] as \([\mathrm{HbCO}] / [\mathrm{HbO}_2] = \frac{K_c [\mathrm{CO}]}{[\mathrm{O}_2]} = \frac{212 \times 1.9 \times 10^{-6}}{8.6 \times 10^{-3}}\)

03

Simplification

After computation, the ratio of [HbCO] to [HbO2] becomes approximately 0.05

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