3
$\begingroup$

It's well known that if you put carbonic acid $\ce{H2CO3}$ into water it will dehydrate into carbon dioxide $\ce{CO2}$ and water $\ce{H2O}$. Indeed this is how modern fizzy drinks generate their fizz. Taking this as a given what I don't understand is why doesn't sodium bicarbonate dissolved in water not slowly give off $\ce{CO2}$ via the following pathway:

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

This would lead to the solution of sodium bicarbonate slowly converting itself into a solution of sodium hydroxide. The first part of the reaction is reversible and is something that at equilibrium will have a non-zero amount of $\ce{H2CO3}$. In fact given that carbonic acid is a weak acid there will be a lot of $\ce{H2CO3}$ present at equilibrium. I also can't see how the presence of sodium ions in solution would inhibit the second reaction (since that is the only difference between this solution and a solution of pure carbonic acid which does effervesce) so according to my understanding we should be seeing a fizz .

However very clearly this bubbling does not happen in real life, as anyone can easily verify for themselves. So what inhibits it then?

$\endgroup$
5
  • $\begingroup$ Check all 4 parallel ongoing equilibrium processes and they eq. constants. CO2(g) <=> CO2(aq) <=> H2CO3(aq) <=> HCO3-(aq) <=> CO3^2-(aq). All of them are reversible. $\endgroup$
    – Poutnik
    Commented Feb 5 at 21:12
  • $\begingroup$ I agree with that. From my understanding these processes are the same in carbonic acid dissolved in water as they are with sodium bicarbonate dissolved in water. And in the former case for the level of partial pressure of CO2 in ambient air the equilibrium of this system lies very strongly towards CO2(g), however for the latter case it doesn't (as we don't see effervescence). My question can be rephrased as asking why there is a difference in equilibrium concentrations of CO2(g) between the two cases (as Na doesn't show up anywhere in the system you posted). $\endgroup$
    – Hadi Khan
    Commented Feb 5 at 21:20
  • 1
    $\begingroup$ Conditions are very different. pH is the key. $\endgroup$
    – Poutnik
    Commented Feb 5 at 21:27
  • 1
    $\begingroup$ Research the use of baking soda in baking, specifically soda bread; you will find at high temperatures bicarbonate does fizz. Aso the 400+ ppm of CO2 in the air raises the CO2 level in the water inhibiting the hydrolysis of HCO3- to give H2CO3 or CO2. Purging a bicarbonate solution with CO2 free air should eventually decompose all the bicarbonate. $\endgroup$
    – jimchmst
    Commented Feb 5 at 22:57
  • $\begingroup$ It certainly effervesces in hot enough water $\endgroup$ Commented Feb 6 at 17:24

1 Answer 1

7
$\begingroup$

When sodium bicarbonate is dissolved in water, it produces $\ce{HCO3-}$ ions. These $\ce{HCO3-}$ ions can react with acidic ions in water, and produce some $\ce{CO2}$, as described in your question. But the available amount of acidic ions in water is rather low, about $\pu{10^{-7}}$ M. It has two consequences. First, the concentration of dissolved $\ce{CO2}$ is probably in the same order of magnitude. At such low concentrations, $\ce{CO2}$ will stay dissolved, and the solution will never fizz. To get a fizz, the $\ce{CO2}$ concentration in water must be greater than $0.04$ M, according to Henry's law. Second, if some $\ce{NaHCO3}$ produces $\ce{CO2}$, it will also produce some $\ce{NaOH}$ in solution, and this $\ce{NaOH}$ will immediately react with all available $\ce{CO2}$ molecules around to produce $\ce{NaHCO3}$ backwards. This is why the $\ce{NaHCO3}$ solutions will never spontaneously produce any $\ce{CO2}$ bubbles.

$\endgroup$
6
  • $\begingroup$ HCO3- + H2O = H2CO3 + OH-; H2CO3 = H2O + CO2, Standard hydrolysis equation. H2O is the acid in highest concentration in pure water. Atmospheric CO2 is also involved [or not]. $\endgroup$
    – jimchmst
    Commented Feb 5 at 23:12
  • 2
    $\begingroup$ ... and as a kitchen experiment, adding a little vinegar to the water donates acidic ions, raising the concentration of $\ce{CO_2}$ enough to cause bubbling. $\endgroup$ Commented Feb 6 at 6:33
  • $\begingroup$ As a side-note, improperly stored samples of sodium hydroxide will react with moisture and carbon dioxide from the air to form the carbonate. $\endgroup$ Commented Feb 6 at 13:39
  • $\begingroup$ I was wondering how likely the direct decomposition $\ce{HCO3- <=>CO2 +OH-}$ in neutral water solution would be and ended up going down a quantum chemistry rabbit hole. It seems to be a topic of ongoing research, but AFAICT (from e.g. this recent paper) it may be a relevant route (maybe more than $\ce{HCO3- +H3O+<=>CO2 +2H2O}$ depending on pH), so it doesn't seem like water autodissociation is a necessary initial step. Of course, higher pH still drives the equilibrium left either way, so all this is really just mostly pointless quibbling. $\endgroup$ Commented Feb 6 at 16:32
  • $\begingroup$ (Also, AFAICT, apparently modern consensus is that $\ce{CO2 + H2O <=> H2CO3}$ doesn't happen in a single step in water, but goes through $\ce{HCO3-}$ first.) $\endgroup$ Commented Feb 6 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.