# Equilibrium Solubility of CO2 in Aqueous Solution and its Dependence on H3O+ Concentration

The equilibrium solubility of $$\ce{CO2}$$ in an aqueous solution is given by three chemical reactions: \begin{align} \ce{CO2(g) &<=> CO2(aq)}\label{rxn:R1}\tag{R1}\\ \ce{CO2(aq) + H2O &<=> H2CO3(aq)}\label{rxn:R2}\tag{R2}\\ \ce{H2CO3(aq) + H2O &<=> HCO3-(aq) + H3O+(aq)}\label{rxn:R3}\tag{R3} \end{align} Considering a system $$E_1$$ at equilibrium with constant pressure and variable volume, an increase in the concentration of $$\ce{H3O+}$$ would shift the equilibrium towards $$\ce{CO2(g)}$$ and consequently result in a change in volume. Let's denote the new system with the altered equilibrium as $$E_2$$. The law of mass action for each reaction is defined as:

$$K_1 = \frac{\ce{[CO2(aq)]}}{\ce{[CO2(g)]}},\ K_2 = \frac{\ce{[H2CO3(aq)]}}{\ce{[CO2(aq)]}},\ K_3 = \frac{\ce{[HCO3-(aq)][H3O+(aq)]}}{\ce{[H2CO3(aq)]}}$$

Since $$K_1$$ is constant at a given temperature, and $$\ce{[CO2(g)]}$$ remains constant because the pressure is assumed to be constant, $$\ce{[CO2(aq)]}$$ is also constant. With the same argument, the concentration of $$\ce{H2CO3(aq)}$$ should be constant. However, the concentration of $$\ce{HCO3-(aq)}$$ in $$E_2$$ needs to be lower than in $$E_1$$ because the concentration of $$\ce{H3O+}$$ is higher in $$E_2$$, and $$K_3$$ is a constant. Therefore, the change in volume can be fully explained by the alteration in the concentration of $$\ce{HCO3-(aq)}$$.

Does that make sense? It seems quite strange to me that there is no change in concentration for the reactants in $$\mathrm{R1}$$ and $$\mathrm{R2}$$

• Concentrations of a given molecular entity must be the same in all equations, as it is the same system. // Be aware MathJax is preferred not to be used in CH SE Q Titles. Jul 30 at 13:58
• Add the equations or multiply the $K$'s, then as $\ce{H_3O^+}$ goes up $\ce{HCO_3^-}$ goes down because each $K$ is constant and so is $\ce{CO_2(g)}$. Jul 30 at 14:52
• @porphyrin Hello porphyrin. I was thinking the same but then I got the following. Lets sum them all and "magically" put some protons there so that the $\mathrm{pH}$ goes down. Then, $\ce{HCO3-}$ must go down. But it can only go down with the generation of $\ce{CO2(g)}$ (so the global reaction actually goes to the left), but $\ce{CO2(g)}$ is constant. What am I thinking wrong? Jul 30 at 15:44

General solution of such systems is solving a set of $$N$$ nonlinear equations for $$N$$ variables.

There are concenctrations of $$\ce{H+}$$, $$\ce{OH-}$$, $$\ce{CO2(aq)}$$, $$\ce{H2CO3(aq)}$$, $$\ce{HCO3-(aq)}$$, $$\ce{CO3^2-(aq)}$$ + $$p_{\ce{CO2}}$$, i.e. $$N=7$$.

• There is 1 equation for total $$\ce{CO2}$$ inventory
• There is 1 equation for charge neutrality.
• There are equilibrium equations:
• 1 equation for $$\ce{CO2}$$ dissolution
• 1 equation for $$\ce{H2CO3}$$ formation
• 2 equations for $$\ce{H2CO3}$$ dissociation
• 1 equation for water auto-ionization

So 7 variables and 7 independent equations.

With constant $$\ce{CO2(g)}$$ partial pressure, it can be simplified as $$[\ce{CO2(aq)}]$$ and $$[\ce{H2CO3(aq)}]$$ would be at equilibrium constant as well.

Exact solving of the equation set often lead to a cubic or more complicated equation for $$[\ce{H+}]$$. It may be impossible or challenging to resolve analytically.

But can be solved numerically e.g. by the MS Excel solver package, e.g. by minimalization of function:

$$z = \sum_{i=1}^N{\text{(EqLeftSide}_i - \text{EqRightSide}_i)^2}$$

The practical hint: Use for variables and parameters single-letter symbols, common in algebra, for easier manipulation.

Another way is simplification by evaluation of eventual strong inequalities.

E.g. $$[\ce{HCO3-}] \gg [\ce{CO3^2-}] \implies [\ce{HCO3-}] + [\ce{CO3^2-}] \simeq [\ce{HCO3-}]$$

or

$$|x| \ll 1 \implies \frac{1}{1 \pm x} \simeq 1 \mp x$$