I had a recent question on a test that asked what the products would be if sodium hydrogen carbonate were dissolved in water. I had a few candidate answers

  1. $\displaystyle\ce{NaHCO3 -> Na+ + HCO3-}$, but that one doesn't involve water at all
  2. $\displaystyle\ce{NaHCO3 + H2O -> Na+ + OH- + H2CO3}$, but doesn't $\ce{H2CO3}$ decompose into $\ce{H2O + CO2}$?
  3. $\displaystyle\ce{NaHCO3 + H2O -> Na+ + OH- + H2O + CO2}$, but that has water on both sides of the equation.

Part two of the question asked whether the solution would be acidic, basic, or neutral. What is the proper reaction for this?

  • $\begingroup$ once NaHCO3 is dissolved, the dissolution products are the Na+ ion and the HCO3- ion . According to wikipedia the HCO3- ion reacts with water to produce H2CO3 and a OH- ion, so the solution will be basic. The H2CO3 will decompose into water and carbon dioxide. So I guess that the most likely dissolution reaction will be the 3rd that you proposed $\endgroup$
    – user2117
    Feb 14, 2014 at 16:58

2 Answers 2


It's not as intuitive as it seems and your questions are all legitimate questions, but sometimes even good arguments can't be used as evidence in chemistry.

$\ce{NaHCO3 -> Na+ + HCO3-}$, but that one doesn't involve water at all

Look at this reaction: $$\ce{NaCl <=>Na+ +Cl-}$$

Even this reaction doesn't "involve" water in the schematics but is right, we assume that is a dissociation of a salt in water.

$\ce{NaHCO3 + H2O -> Na+ + OH- + H2CO3}$, but doesn't $\ce{H2CO3}$ decompose into $\ce{H2O + CO2}$?

Yes, but in fact this is an equilibrium: $$\ce{H2CO3 <=> H2O + CO2}$$

And then you have to take into account another important equilibrium many time ignored: $$\ce{CO2 (aq) <=> CO2 (g)}$$ So the amount of $\ce{CO2}$ present in the solution depends on by the type of system and the pressure exerted. Like a Coca-Cola bottle: if the system is closed, the $\ce{CO2}$ remains dissolved otherwise the equilibrium shift to the right and the $\ce{CO2}$ "escapes" until the rate between the $\ce{CO2}$ molecule dissolved and the $\ce{CO2}$ molecule escaped forms the solution is the same.

$\ce{NaHCO3 + H2O -> Na+ + OH- + H2O + CO2}$, but that has water on both sides of the equation.

Okay, it is not a net ionic equation but the stoichiometry is still right. Maybe water is used to indicate that we are talking about a reaction that takes place in water.

Which reaction is not right?

All the reactions here are right, I think maybe they put some formal inaccuracies to help you, but in fact they are all part of series of equilibria that normally occur and can be described as: $$\begin{multline} \ce{NaHCO3 + H2O <=>[A] Na+ + HCO3- + H2O <=>[B] Na+ + H2CO3 + OH-} \\ \ce{ <=>[C] Na+ + OH- + H2O + (CO2(aq)<=>[C_2]CO2 ^ (g))}\end{multline}$$

The question should be more specific asking for the ultimate products of the reaction because all these are products of the dissociation of sodium hydrogen carbonate and because are equilibria is probable that all the species are present in different amounts when the reaction reach the equilibrium. However, the ultimate products are described by the third answer. And this equation can give you a hint about the acidity of the solution. When $\ce{CO2}$ escape from the solution (equilibrium $\mathrm{C_2}$ goes to the right) the concentration of the manly acid species goes down but the hydroxides from the equilibrium $\mathrm{B}$ are still present so at the end the solution in an open system with an atmosphere with a low partial pressure of $\ce{CO2}$ will have a higher pH, will become mildly alkaline.


First, ionize the salt as you have shown. NaOH is a very strong base so Na+ must be a very weak acid. It is solvated. Being a singly-charged, rather large cation, it stops there. If it were $\ce{Al^{3+}}$, coordinated water would be acidic by charge withdrawal from that brutally small trication.

Bicarbonate is the salt of the first ionization of weak carbonic acid. It will then be a modestly weak base by hydrolysis, as you have shown (reversible). Bicarbonate solutions are weakly basic. If you add acid (even weak acetic acid) they fizz. Boiling the solution will drive out carbon dioxide.

Carbonate is the salt of the second ionization of weak carbonic acid. It is a strong base.

  • $\begingroup$ So what would be the net ionic equation? $\endgroup$
    – scrblnrd3
    Feb 14, 2014 at 16:49

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