If a carbonic acid to bicarbonate reaction involves the release of H+, why does pH increase?

Question

OK, I'm an engineer that took Aqueous Geochem 6 years ago, so please forgive this basic (get it?) question. But why do the two reactions which start from dissolution of CO2 in water increase pH if they both release H+ ions? I.e., the conversion of carbonic acid to bicarbonate releases a hydrogen proton, and the conversion from bicarbonate to carbonate releases a hydrogen proton. Yet each reaction results in a higher pH. Specifically, why don't the additions of H+ decrease pH?

$\ce{H2CO3 + 2 H2O ⇌ HCO3- + H3O+ + H2O ⇌ CO3^{2-} + 2 H3O+}$

Thanks, and apologies for the (probably simplistic) question.

Mike

Thank you all for your help with this. I should clarify that what is really troubling me is the interpretation of the Bjerrum Plot for the carbonate system (aqueous geochem). Looking, for example, at that plot here:

https://en.wikipedia.org/wiki/Bjerrum_plot#Bjerrum_plot_equations_for_carbonate_system

For HCO3- to transition to CO32-, a H+ ion is released (going rightward in the plot), but that is associated with higher pH (less acidity). But if increasing H+ ions are the very definition of increasing acidity, why in the plot is that reaction associated with increasing pH?

Thanks again!

Mike

Answer 1

No, addition of carbon dioxide to water will never cause an increase in pH. Think of the simplest scenario. Take pure water and dissolve carbon dioxide in it. The pH will decrease from 7 to ~ 5. Recall rainwater is slightly acidic due to $\ce{CO2}$.

You are most likely confused about carbonate-bicarbonate buffering system in geological settings? See here

http://butane.chem.uiuc.edu/pshapley/environmental/l24/3.html

Answer 2

There is sure to be a more technical (and more pleasing to the physical chemist) answer to this, but as a heuristic, don't consider what the carbonic acid is doing to solution, consider how the solution is acting on the bicarbonate.

The plot your referencing describes mole fractions of diprotic acids in solutions relative to pH. This isn't the same as the pH of the carbonic by itself. Of course, carbonic acid in a cup of water decreases the pH of the solution, but instead consider an equilibrium condition in which a happy amount of bicarbonate and hydronium exist. We're not concerned with what happens if we add more carbonic acid, we only care about what happens to the bicarbonate if we vary the concentration of our hydronium.

In a solution rich with hydronium such that there are many free protons to act on bicarbonate, equilibrium will shift towards the production of carbonic acid. Thus, at a lower pH, carbonic acid is the dominant species, as we can clearly see by our mole fraction of the acid hitting a maximum at lower pH's.

In a solution quite poor in hydronium, such that there are very few protons to act on bicarbonate, equilibrium will shift to produce carbonate. This is because we perhaps have free hydroxide ions in solution, which are perfect homes for bicarbonate protons. Therefore, we see a higher mole fraction of our carbonate ion at higher pH's.

Again, the dissociation of carbonic acid, which is really just dissolved carbon dioxide, is in fact a proton-producing process. But the plot your describing is a relationship that describes which species are dominant at different concentrations of hydronium, not the concentration of hydronium (pH) as carbonic acid dissociates.

To directly address the equilibrium you've set shown us- the real bump on the road is the dissociation of the bicarbonate. Remember that in this situation, we must have a very hydronium poor solution, such that it's likely that we actually have free hydroxide ions. So, what happens to the 2 protons on the far right side of your formula? Well, they pair up with the hydroxide! The issue is that you can't see that, because again the equilibrium formula you have doesn't intuitively show that there is an entire solution that impacts the equilibrium. So where do we get these hydroxide ions from? Take a look at one of the answers below!

The Le Chatelier way of saying this is if we decrease the concentration of hydronium relative to the bicarbonate, equilibrium will shift to try to increase it again. But clearly the best way to do that isn't by producing more bicarbonate, because the stoichiometry shows the fastest way to get more hydronium is by producing more carbonate. For every one mole of hydronium relative to bicarbonate, we have two moles of hydronium relative to carbonate. So in a weird way, by increasing our pH to become hydronium deficient, equilibrium shifts towards the intuitively lower pH side of the reaction to try and counteract this and we get carbonate as a byproduct. This may be the source of your confusion.

"Ask not what your acid can do for your solution, but what your solution can do for your acid"
- JFpKa (buh-dum-tss)

Answer 3

It seems you have misconception that creation of bicarbonates and/or carbonates causes pH increase, because the plot has more of bicarbonate and carbonate at higher $\mathrm{pH}$ values. That is just a half of the truth.

The plot is about ratios of concentrations of respective carbonate forms in equilibrium.

There is equilibrium equation rewritten into logarithmic form, where square brackets are convention for molar concentration:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a1}}^{*} + \log {\frac {\ce{[HCO3-]}}{[\ce{CO2}]}}$$

When $\ce{CO2}$ is just dissolved in pure water with $\mathrm{pH}=7$, the right equation side does not match it, it is too low. So $\ce{CO2}$ converts to $\ce{HCO3-}$ until dropping left equation side and raising right equation side are equal, reaching equilibrium.

The similar equilibrium equation is for bicarbonate and carbonate:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a2}} + \log {\frac {\ce{[CO3^2-]}}{[\ce{HCO3-}]}}$$