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 $\mathrm{R2}$

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    $\begingroup$ 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. $\endgroup$
    – Poutnik
    Commented Jul 30, 2023 at 13:58
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    $\begingroup$ 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 $\ce{CO_2(g)}$. $\endgroup$
    – porphyrin
    Commented Jul 30, 2023 at 14:52
  • $\begingroup$ @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? $\endgroup$ Commented Jul 30, 2023 at 15:44

1 Answer 1


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


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

Such simplifications often lead to exact solutions, e.g. a quadratic equation, of an inexact, simplified task.


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