Question

The pH of human blood is steady at a value of approximately 7.4 owing to the following equilbrium reactions:

${\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\rightleftarrows }{\mathbf{H}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\rightleftarrows }{\mathbf{HCO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

Acids formed during normal celluar respiration react with the HCO₃^- to form carbonic acid, which is in equilibrium with CO₂(aq) and H₂O(l). During vigorous exercise, a person’s H₂CO₃ blood levels were 26.3 mM, whereas his CO₂ levels were 1.63 mM. On resting, the

H₂CO₃ levels declined to 24.9. What was the CO₂ blood level at rest?

Short answer

The CO₂ blood level at rest is 1.54 mM.

Step-by-step solution

Acid dissociation 

The reaction for the dissociation of acid is,

$\text{HA}\left(\text{aq}\right)\rightleftarrows {\text{H}}^{\text{+}}\left(\text{aq}\right){\text{+A}}^{\text{-}}\left(\text{aq}\right)$

The equilibrium constant expression for the above reaction is,

$\text{K=}\frac{\left[{\text{A}}^{\text{-}}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[\text{HA}\right]}$

Determination of concentration of CO2

The reaction is,

${\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{CO}}_3$

The equilibrium constant for the reaction is,

$\begin{array}{l}\mathrm{K}=\frac{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}{\left[{\mathrm{CO}}_2\right]}\\ \mathrm{K}=\frac{26.3\mathrm{mM}}{1.63\mathrm{mM}}\\ \mathrm{K}=16.1\end{array}$

The concentration of CO₂ is,

$\begin{array}{c}\left[{\mathrm{CO}}_2\right]=\frac{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}{\mathrm{K}}\\ =\frac{24.9\mathrm{mM}}{16.1}\\ =1.54\mathrm{mM}\end{array}$

Thus, the CO₂ blood level at rest is 1.54 mM.