# Why is CO2 more likely to dissolve in a basic solution than an acidic solution? [closed]

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?

• 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. – Poutnik Dec 7 '19 at 10:16

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.