The #K_a# of carbonic acid is #4.3 xx 10^-7# #H_2CO_3 rightleftharpoons H^+ + HCO_3^-#. What does this mean for #H_2CO_3#?

1 Answer
Jun 14, 2017


It means that carbonic acid is a weak acid.


The acid dissociation constant, #K_a#, is a measure of the extent to which an acid ionizes in aqueous solution.

More specifically, the smaller the value of #K_a#, the lower the extent of the ionization. In other words, the lower the value of #K_a#, the more molecules of acid will not ionize, i.e. they will not donate a proton.

In your case, you know that the following equilibrium

#"H"_ 2"CO"_ (3(aq)) rightleftharpoons "H"_ ((aq))^(+) + "HCO"_ (3(aq))^(-)#


#K_a = (["H"^(+)] * ["HCO"_3^(-)])/(["H"_2"CO"_3]) = 4.3 * 10^(-7)#

The very small value you have for the acid dissociation constant tells you that the equilibrium lies to the left, which ultimately implies that carbonic acid is a weak acid because it does not ionize completely in aqueous solution.

In other words, a solution of carbonic acid will not contain significant concentrations of hydrogen ions (and, implicitly, of bicarbonate ions) when compared to the initial concentration of carbonic acid, hence why we say that carbonic acid is a weak acid.

Now, it's worth mentioning that carbonic acid is highly unstable in aqueous solution, which makes the following equilibrium

#"CO"_ (2(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 2"CO"_ (3(aq))#

a more accurate description of what's going on in solution.

You can thus say that you have

#"CO"_ (2(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 2"CO"_ (3(aq)) rightleftharpoons "H"_ ((aq))^(+) + "HCO"_ (3(aq))^(-)#