I've conducted an experiment where a solution of $\pu{4.2 M}$ $\ce{NaOH}$ was left exposed to atmospheric $\ce{CO2}$ for a period of time and titrated using potentiometric titration and potassium hydrogen phthalate ($\ce{KHP}$) to investigate the degradation of $\ce{NaOH}$ over time. I was under the impression that the reaction for such a high pH solution would look like this:

$$\ce{2NaOH(aq) + CO2 <=> Na2CO3(aq) + H2O} $$

But the results I have got don't fit this. For example the rate of change of $[\ce{NaOH}]$ and $[\ce{Na2CO3}]$ should be a 2:1 ratio but I got 0.849:1. This is a significant difference. The data I have seem accurate and the titration curves lined up well on each run, I don't think this is a significant error on my part so I'm now wondering if instead of the above reaction, $\ce{NaHCO3}$ was being formed:

$$\ce{NaOH(aq) + CO2(g) <=> NaHCO3(aq)}$$

This would fit the ratio I got much better but I thought this wasn't possible at a pH of 10 or above? Here are my titration and Concentration vs time graphs. I had previously reasoned that the fact I was getting only two equivalence points was due to the third step being a weak neutralization with a pH of about 7 that would barely show but now it seems like it could be because $\ce{NaHCO3}$ is the reacting not $\ce{Na2CO3}$.

enter image description here

Graph of Concentration vs time

Any help working out what is happening here would be great...

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    – Buttonwood
    Aug 10, 2021 at 13:37
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    $\begingroup$ With the initial pH, it is obvious the $\ce{HCO3-}$ content must be almost negligible. But in the section between both steeper segments, $\ce{CO3^2-}$ is being converted to $\ce{HCO3-}$. On the plateou near 105 mL, the ratio $\ce{CO3^2-/HCO3-}$ is 1:1 - carbonate/bicarbonate pH buffer in action. $\endgroup$
    – Poutnik
    Aug 10, 2021 at 13:50
  • $\begingroup$ @Buttonwood Thanks for that, I haven't seen that formatting before so cheers. $\endgroup$ Aug 10, 2021 at 13:52
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    $\begingroup$ You add acid, right ? Does an acid convert carbonate ( in excess ) to bicarbonate ? $$\ce{NaOH + Na2CO3 ->[acid] Na2CO3 ->[acid]NaHCO3 + Na2CO3 ->[acid]NaHCO3 ->[acid]NaHCO3 + CO2 ->[acid] CO2}$$ $\endgroup$
    – Poutnik
    Aug 10, 2021 at 14:02
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    $\begingroup$ You do know about the concept of acido-basic equilibrium and acid dissociation constants, don't you ? $K_\mathrm{a2}=\frac{\ce{[H+][CO3^2-]}}{\ce{[HCO3-]}}$ $\endgroup$
    – Poutnik
    Aug 10, 2021 at 14:07

1 Answer 1


Real experiments and their outcomes are more interesting than expected so don't call it a trouble. First think of a couple of scenarios.

a) If you just leave a pellets of solid NaOH in open atmosphere. It would absorb water and form a protective coating of insoluble sodium carbonate. If you analyze the solution, that would be still be very high in NaOH conc.

b) If you leave concentrated NaOH (~ 50% w/w) in open atmosphere, you start to see a white crust and white powder settling to the bottom. Here again sodium carbonate is insoluble and acts like a protective coating. If you analyze the solution, you will see most of the NaOH is still intact.

c) Your NaOH solution is a relatively low concentration but its concentration is not changing as fast as you were expecting. The reason that you have a heterogeneous reaction. You have trace concentration of a gas reacting with quiet (unstirred) solution of NaOH. So two steps are involved:

(i) Absorption of carbon dioxide from a very dilute mixture (air) in NaOH. (ii) Irreversible reaction of carbon dioxide and NaOH to form sodium carbonate.

All of this is mainly happening at the air-solution interface. Your bulk is still protected by very slow diffusion. So these two steps will not follow your expectations.

People have worked out the expected rates of absorption of carbon dioxide in NaOH. For example in Absorption of Carbon Dioxide into Aqueous Sodium Hydroxide and Sodium Carbonate-Bicarbonate Solutions, The Chemical Engineering Journal, 11(1976), 131-141.

Then the absorption of solute gas $\mathrm{A}$ is accompanied by an irreversible chemical reaction, the rate of absorption $N_{\mathrm{A}}$ is represented by $$ N_{\mathrm{A}}=\beta\left(2A_{i} \sqrt{D_{\mathrm{A}} / \pi t}\right) $$ where $A_{\mathrm{i}}$ and $D_{\mathrm{A}}$ are the interfacial concentration and the liquid-phase diffusivity of the solute gas, respectively, $t$ is the exposure time and $\beta$ is the reaction factor.

Another, point is that I did not mention sodium bicarbonate anywhere. The reason is that your NaOH concentration is pretty high, as long as you do have NaOH, sodium bicarbonate will not form. So seven days are not enough. Wait for months perhaps!

If you were stirring everything for 7 days, then the rate of carbon dioxide will be faster but still you are limited by diffusion of the gas into the solution.

  • $\begingroup$ Thanks for the answer, if i understand you correctly youre saying that the rates will change based on many factors which i totally agree with - shouldnt the rate that the $NaOH$ decreases at still be related to the rate that the $Na_2CO_3$ increases at? Regardless of how much reacts, if i can work out how much $NaOH$ is gone i should be able to work out how much $Na_2CO_3$ should be gone and the rates should follow the same ratio? $\endgroup$ Aug 10, 2021 at 14:43
  • $\begingroup$ Also i think i possibly didnt phrase my question well, my problem is not that the concentration is changing slowly, it is that it is changing out of sync with the reactions products and therefore violating the stoichiometric relationship. $\endgroup$ Aug 10, 2021 at 14:59
  • $\begingroup$ The simple rates of reactions apply when the reaction were homogeneous. Your bulk is still NaOH, but sodium carbonate is being formed is only at the surface. $\endgroup$
    – AChem
    Aug 10, 2021 at 14:59
  • $\begingroup$ Regarding the stoichiometry issue, do you stir the solution before sampling for titration? $\endgroup$
    – AChem
    Aug 10, 2021 at 15:00
  • $\begingroup$ And regardless, if the analysis is done right, your mass balance must remain correct. Forget about the rates. Think about the moles of products and reactants only. $\endgroup$
    – AChem
    Aug 10, 2021 at 15:03

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