So I wanted to know what the reaction between sodium hydroxide and carbon dioxide can be, and upon research I got 2 answers. The first one is

$$\ce{CO2 + NaOH(aq) -> NaHCO3(aq)}$$

and the second one is a two-step reaction, first with water then sodium hydroxide:

$$\ce{CO2 + H2O(l) -> H2CO3(aq)}, \\ \ce{H2CO3(aq) + NaOH(aq) -> NaHCO3(aq) + H2O(l)}$$

So my question is which one of these answers is right?

To me, I do not understand why the 2nd option takes into account water because really concentrated $\ce{NaOH}$ would/should be written just as $\ce{NaOH}$, which to me makes the 1st option the right one. Am I wrong?

  • 1
    $\begingroup$ Short answer is that you're right, but quite a bit can be told about carbonic acid and existence of H2CO3 molecule. $\endgroup$
    – Mithoron
    Aug 11, 2016 at 19:59
  • $\begingroup$ could you please give me the longer explanation? $\endgroup$
    – user510
    Aug 11, 2016 at 20:58
  • $\begingroup$ As Mithoron mentioned, H2CO3 is a unstable compound and was only isolated as a pure compound recently. You cannot obtain H2CO3 directly through chemical reaction. $\endgroup$ Aug 12, 2016 at 18:00
  • $\begingroup$ can you tell how much water will be in solution in the 1st equation? $\endgroup$
    – user510
    Aug 12, 2016 at 19:25

3 Answers 3


This question is a great opportunity to talk about state symbols, ionic bonding, and multi-step reactions.

Very often when we write an equation for a chemical reaction we only look at the starting material and the products, like when you wrote the first form: $$\ce{CO2 + NaOH (aq) -> NaHCO3 (aq)}$$

Multi-Step Reactions

Firstly, this form doesn't necessarily tell us what molecules are interacting with one another. Is one molecule of carbon dioxide colliding with one sodium hydroxide molecule to make a sodium bicarbonate molecule? In this case, no. Two different reactions happen in sequence.

  1. Carbon dioxide and water react to make carbonic acid. $$\ce{CO2(aq) + H2O(l) -> H2CO3(aq)}$$
  2. Carbonic acid reacts with hydroxide ions to make bicarbonate ions. $$\ce{H2CO3(aq) + NaOH(aq) -> NaHCO3(aq) + H2O(l)}$$

Reactions in Aqueous Solution

Secondly, the state symbols tell us important information about the reaction conditions. The (aq) state symbol tell us that the reaction is happening in aqueous conditions. That means the reactants are dissolved in water which can have a significant effect on how the reactants behave. Here the water reacts with carbon dioxide to form molecules of carbonic acid, $\ce{H2CO3}$, in solution which is an important intermediate in this reaction. $$\ce{CO2(aq) + H2O(l) -> H2CO3(aq)}$$

The aqueous conditions also mean that ionic compounds are separated into their constituent ions. So $\ce{NaOH(aq)}$ actually means a solution of separated positive sodium ions, $\ce{Na+}$, and negative hydroxide ions $\ce{OH-}$, which can each react on their own.

As you might notice, that means our solution contains molecules of carbonic acid and basic hydroxide ions. A neutralisation reaction occurs. The basic hydroxide ions remove a proton from the carbonic acid to make water and a bicarbonate ion. $$\ce{H2CO3(aq) + OH-(aq) -> HCO3-(aq) + H2O(l)}$$ Because the sodium ions don't react with anything, we can leave them out of the equation. This is similar to how we don't explicitly include water in the one step equation - because we would have the same amount of water on each side of the equation. If we add the sodium ions we get something equivalent to the equation for the second step. $$\ce{H2CO3(aq) + Na+(aq) + OH-(aq) -> Na+(aq) + HCO3-(aq) + H2O(l)}$$ In water, $\ce{NaOH(aq)}$ and $\ce{Na+(aq) + OH-(aq)}$ are the same because sodium salts dissociate into their ions in water.

Adding equations

To see the equivalence between the multi-step form and the shorter one step form we can add together the equations for the two steps to get an idea of what is happening overall. $$\ce{CO2(aq) + H2O(l) -> H2CO3(aq)}\\+\\ \ce{H2CO3(aq) + Na+(aq) + OH-(aq) -> Na+(aq) + HCO3-(aq) + H2O(l)}\\ \implies \ce{CO2(aq) + H2O(l) + H2CO3(aq) + Na+(aq) + OH-(aq) ->\\ Na+(aq) + HCO3-(aq) + H2O(l) + H2CO3(aq)}$$ There's a lot going on in this equation, but we can make it more readable but taking away $\ce{H2O(l)}$ and $\ce{H2CO3(aq)}$ from both sides, just like any mathematical equation. $$\ce{CO2(aq) + Na+(aq) + OH-(aq) -> Na+(aq) + HCO3-(aq)}$$ In aqueous solution, sodium salts dissociate into their ions so we can write the ion pairs as the aqueous sodium salts. $$\ce{CO2 + NaOH (aq) -> NaHCO3 (aq)}$$

So we've gone full circle, from a single summary equation, to the two reactions involved, back to the summary equation once again.


No both are right.

If you add the latter equations, then you will get the former.

If sodium hydroxide solution was highly concentrated like you said, then both equations are wrong since sodium hydroxide solution will strip away the second hydronium as well forming carbonates.

All state notations with (aq) implies that it is dissolved in water. If it was pure NaOH, it would be in a solid state under room temperatures. It is also noteworthy to know that NaOH must be contained in a dry environment as it will spontaneously absorb water from the air.

  • $\begingroup$ oh okay but would the second one have access water molecules while 1 is just all the NaHCO3 in ion form. In the sense that even the 2nd one will have Na and HCO3 ions, but will it also have access water molecules? $\endgroup$
    – user510
    Aug 12, 2016 at 4:11
  • $\begingroup$ Both are aqueous solutions. As I've stated in the answer, state notations are important. Ionic compounds with sodium usually dissociate all the way in water, but to know how much an aqueous substrate dissociates, you need to know its equilibrium constant. $\endgroup$
    – Caprica
    Aug 12, 2016 at 4:17
  • $\begingroup$ okay, how would you calculate concentration of sodium hydroxide to use to make sure HCO3- remains in solution and not carbonate ions? $\endgroup$
    – user510
    Aug 12, 2016 at 13:00
  • $\begingroup$ You'd need to know the ratio of NaOH and CO2 dissolved in moles. $\endgroup$
    – Caprica
    Aug 12, 2016 at 13:13
  • $\begingroup$ Which one dominates would also depend on how you perform the experiment. Bubbling CO2 into concentrated NaOH would favour the carbonate more than adding a NaOH solution to water continuously saturated with CO2 $\endgroup$
    – Beerhunter
    Aug 15, 2016 at 11:41

The mechanism of reaction between the acidic oxide and the alkali depends on the concentration of the alkali solution.

(a) When the alkali ($\ce{NaOH}$) solution is very dilute ($\mathrm{pH} < 8$), carbon dioxide will first react with water to form carbonic acid ($\ce{H2CO3}$) slowly. The acid thus formed then reacts with the alkali to give sodium hydrogencarbonate ($\ce{NaHCO3}$).

$$\begin{align} \ce{CO2 + H2O &= H2CO3}\\ \ce{NaOH + H2CO3 &= NaHCO3 + H2O} \end{align}$$

(b) When the alkali solution is a fairly concentrated one ($\mathrm{pH} > 10$), carbon dioxide directly reacts with it to form the bicarbonate, which further reacts with the alkali to form sodium carbonate ($\ce{Na2CO3}$) as the main product by complete neutralization.

$$\begin{align} \ce{NaOH + CO2 &= NaHCO3}\\ \ce{NaHCO3 + NaOH &= Na2CO3 + H2O} \end{align}$$

Thus, only when the concentration of the alkali solution is quite low, the reaction proceeds via the formation of carbonic acid. But the acidic oxide is not completely neutralized in this case.


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