# How does sodium carbonate raise pH?

Sodium carbonate is used in pools to raise pH. From my understanding, it would dissociate in water like so:

$$\ce{Na2CO3 <=> 2 Na+ + CO3^2-}$$

To raise pH, doesn't it have to absorb hydrogen ion/s? If so then wouldn't it just form carbonic acid that would then just dissociate again?

• It's all about equlibrium. Try to write the corresponding equation(s) for hydrolysis and see what dissociates better, sodium hydroxide or hydrocarbonate. Also, the term "to absorb" looks weird in this context. Jul 14 '19 at 6:30
• This whole thing is silly. An aqueous solution of sodium carbonate has basic pH because carbonate ions hydrolize to yield hydroxide ions plus bicarbonate ions. Even just starting with sodium bicarbonate (baking soda), the pH is basic: around 8.3 for reasonable concentrations. So why is this bountied question at all?
– Ed V
Mar 25 '20 at 13:10
• @EdV In similar tune to \@Maurice, I equally wonder about the bounty set. The answer by \@LilBluey basically adds $K_b$'s, «comment like». I set a flag for the moderators, maybe then \@Haha Hahaha adds a comment why the bounty was set. Mar 25 '20 at 13:17
• @Buttonwood I think you can use the notification @hahaha to 'contact' the user setting the bounty. (It will not autofill though.) I believe this is possible because s..he is an editor of the post, see help center, but please don't put a gun to my head if that information is not accurate. You can only notify one (additional) person per comment though. Mar 25 '20 at 15:20
• @Martin-マーチン Well, meanwhile ... I identified comment (below chemistry.stackexchange.com/questions/127027/…) of @HahaHahah which offered an autofill where I asked him / her about the motivation for this bounty. We will see «what happenz». Thank you. Mar 25 '20 at 15:34

Sodium carbonate indeed dissociates in water : $$\ce{Na2CO3 <=> 2 Na+ + CO3^2-}$$ But that is just beginning, as in solutions with carbonates, bicarbonates or dissolved carbon dioxide happen multiple chemical equilibriums: ( See a lot of info at Carbonic acid on Wikipedia):

Carbonate anion partially hydrolyzes in water: $$\ce{CO3^2- + H+ <=> HCO3-}, \mathrm{p}K_\mathrm{a2}=10.33$$ or $$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$

Consumption of $$\ce{H+}$$ and production of $$\ce{OH-}$$ is complemented by water autodissociation $$\ce{H2O <=> H+ + OH-}$$ with $$\mathrm{p}K_\mathrm{w}=14$$

As a result, $$\mathrm{pH}$$ gets more alkalic.

Another hydrolytic step, that occurs significantly only in acidic and about neutral solutions is:

$$\ce{HCO3- + H+ <=> H2CO3}$$ respectively $$\ce{HCO3- + H2O <=> H2CO3 + OH-}$$ with $$\mathrm{p}K^{*}_\mathrm{a1}=6.35$$, $$\mathrm{p}K_\mathrm{a1}=3.6$$.

$$\mathrm{p}K^{*}_\mathrm{a1}$$ pretends all carbon dioxide converts to carbonic acid. $$\mathrm{p}K_\mathrm{a1}$$ considers the equilibrium of formation of carbonic acid.

And finally: $$\ce{H2CO3 <=> CO2(aq) + H2O}$$ $$\ce{CO3(aq) <=> CO2(g)}$$

You have the equilibrium in water: $$\ce{H2O <=> H+ + OH-}$$ The protons are removed from the equilibrium: $$\ce{H+ + CO3^2- <=> HCO3-}$$ Now you have more $$\ce{OH-}$$ in the solution which increases the pH.

• How can we do more than repeat what Deuti said in July '19 ? Could Haha Haha explain us what he wants and what he does not understand ? Where is the problem ? Mar 25 '20 at 12:49

The reaction system can be described as:

$$\ce{H2O <=> H+ + OH-}$$

$$\ce{H+ + CO3^{2-} <=> HCO3-}, \mathrm{p}K_\mathrm{a2}=10.33$$

Note, the net of the first two reactions imply a rise in pH in the presence of carbonate. Further, with time and carbon dioxide exposure:

$$\ce{H2O + CO2 <=> HCO3- + H+}$$

$$\ce{H2CO3 <=> H2O + CO2}$$

And, with the introduction of bicarbonate, the $$\ce{Na2CO3}{/}$$NaHCO3 equilibrium apparently progresses to a pKa of 10.33, which makes it a poor pH 9 buffer.

The reaction system, upon reaching equilibrium, has been succintly presented as:

$$\ce{H2CO3 <=> HCO3- + H+ <=> H2O + CO2(g)}$$

which could move to the right with warming.

[EDIT] A more quantitative (and useful) illustrative tool of the pH effect is given, in logarithmic form, as the Henderson-Hasselbalch Equation, see (4) in this source, presented below:

$$\ce{pH = pK + log([CO_2]/[HCO3^-])}$$

However, the above assumes an equilibrium situation, and in the case of sodium carbonate added to a swimming pool with continuing exposure to air containing carbon dioxide, in time perhaps, such an equilibrium could be established measuring at the same temperature.