# Why do the first and second ionizations of a diprotic acid produce the same hydronium ion concentrations in solution?

I was going through Oxtoby Chemistry when I got confused by the following passage in the textbook. It said that "two simultaneous equilibria are involved in the ionization of a diprotic acid such as carbonic acid":

I understand the chemical species for both the first and second ionizations barely dissociate, which means the hydronium ion production is very little.

However, what makes hydronium ion production identical (i.e. important point #1)? Shouldn't the second ionization occur to a lesser extent than the first ionization? Thanks.

• Of course it is. That simply means that you put total concentration, as both equilibria happen simultaneously. – Mithoron Mar 4 '16 at 18:19
• Thank you. To confirm my understanding (please correct me if I'm wrong), would this mean [H3O+] would have a coefficient of 1 (i.imgur.com/yQpgxnG.png) vs. 2 (i.imgur.com/MnHoJbL.png) in the overall equilibrium expression? – elan.em Mar 4 '16 at 19:27

This is saying both the $\ce{H2CO3}$ and the $\ce{HCO3-}$ are both donating $\ce{H+}$ to the water to form $\ce{H3O+}$. What it means when it says "The $\ce{[H3O+]}$ in the two ionization equilibria are one in the same" is to say that both of these reactions are occurring simultaneously in the same system, and that there is only one measure of the concentration of $\ce{H3O+}$ for the system.
It is not saying that both species are donating $\ce{H+}$ to the same extent. As a matter of fact, the $\ce{H+}$ being donated by $\ce{H2CO3}$ will keep the the $\ce{HCO3- + H2O -> CO3^{2-} +H3O+}$ shifted almost entirely to the left, meaning $\ce{HCO3-}$ will be donating an extremely small fraction of the $\ce{H+}$.