0
$\begingroup$

I'm having trouble thinking through this problem. I know that when $\ce{CO2}$ dissolves in water it can release $\ce{HCO3-}$ and $\ce{H+}$ and cause a drop in pH. But can I use this to justify why $\ce{CO2}$ (g) might dissolve more easily in a basic solution than an acidic one? Is the water acting as the base?

$\endgroup$
  • $\begingroup$ Hydroxide ions keep concentration of dissolved carbon dioxide low, converting it to bicarbonate and eventually carbonate. Therefore while alkalic enough, the solution does not gets saturated. $\endgroup$ – Poutnik Dec 7 '19 at 10:16
2
$\begingroup$

A simple way to think about this is as follows (assuming we are talking about an aqueous solution).

The dissolution of $\ce{CO2}$ in water is followed by three chemical reactions:

$$ \begin{align} \ce{CO2 + H2O &<=> H2CO3}\label{rxn:R1}\tag{R1}\\ \ce{H2CO3 &<=> HCO3- + H+}\label{rxn:R2}\tag{R2}\\ \ce{HCO3- &<=> CO3^2- + H+}\label{rxn:R3}\tag{R3} \end{align} $$

Now the more basic the solution becomes, the further reactions \eqref{rxn:R2} and \eqref{rxn:R3} (\eqref{rxn:R3} becomes significant at high pH) lie to the right as hydroxide consumes $\ce{H+}$. As reaction \eqref{rxn:R2} depletes $\ce{H2CO3}$, reaction \eqref{rxn:R1} is also forced to the right. The net result is more $\ce{CO2}$ residing in solution (dissolved) as the solution becomes more basic.

$\endgroup$
  • $\begingroup$ The carbonic acid equilibrium lies massively to the left, but will be driven to the right by the hydrogencarbonate formation. However, you also have the hydroxide acting as a direct nucleophile on carbon dioxide which cannot be discounted. $\endgroup$ – Beerhunter Dec 7 '19 at 8:42
  • $\begingroup$ There is need to extend the equilibrium chain by the carbonate formation in alkalic solutions. $\endgroup$ – Poutnik Dec 7 '19 at 10:08
  • $\begingroup$ The forward part of reaction (1) is slow, with the rate constant 23 s^-1 $\endgroup$ – Poutnik Dec 7 '19 at 10:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.