# Reaction between sodium carbonate and hydrochloric acid

The 2 steps to the reaction between $$\ce{Na2CO3}$$ and $$\ce{HCl}$$ are:

1. $$\ce{Na2CO3 + HCl -> NaCl + NaHCO3}$$
2. $$\ce{NaHCO3 + HCl -> NaCl + CO2 + H2O}$$

If $$\ce{Na2CO3}$$ is in excess and we add limited amount of $$\ce{HCl}$$ (e.g. 2:1), why is it that after the addition of limited $$\ce{HCl}$$ , we still have $$\ce{NaHCO3}$$ present together with $$\ce{Na2CO3}$$?

What I was thinking is that as $$\ce{NaHCO3}$$ is formed, the $$\ce{HCl}$$ in the solution will react with it through step 2 to form $$\ce{CO2}$$ and $$\ce{H2O}$$. So, half of the $$\ce{HCl}$$ will react with $$\ce{Na2CO3}$$ and the other half will react with $$\ce{NaHCO3}$$. And so, if this happens there will only be $$\ce{Na2CO3}$$ present at the end. (since all the $$\ce{NaHCO3}$$ that is formed from step 1 will be reacted with $$\ce{HCl}$$ upon formation)

Why wouldn't this be the case? Why must step 2 only occur after step 1 is completed?

• Remember that carbonic acid is a weak acid and there are acido-basic equilibrii between the respective forms. Jun 3, 2021 at 5:05
• Also what you described isn't what normally goes under the term mechanism, or at least I think so. Jun 3, 2021 at 6:22

You are right. When adding a small amount of $$\ce{HCl}$$ to a solution of $$\ce{Na2CO3}$$ you produce some $$\ce{NaHCO3}$$. When adding more $$\ce{HCl}$$, you cannot avoid that some of this $$\ce{NaHCO3}$$ does react with the newly created $$\ce{NaHCO3}$$ producing some $$\ce{CO2}$$. But this $$\ce{CO2}$$ will immediately react with $$\ce{Na2CO3}$$ which is still in excess according to : $$\ce{CO2 + Na2CO3 + H2O -> 2 NaHCO3}$$ So it looks as if some more $$\ce{NaHCO3}$$ has simply been created by adding $$\ce{HCl}$$ to $$\ce{Na2CO3}$$
If the ratios differ from 1:1, you can get different reaction products by adding them differently. For instance, if you dump (A to B) a solution of 0.1 mole of Na$$_2$$CO$$_3$$ into a solution of 0.1 mole of HCl at a moderate rate, with stirring, the products will be 0.05 moles of CO$$_2$$ fizzing off instantly, plus 0.1 moles of NaCl, plus 0.05 moles of unreacted Na$$_2$$CO$$_3$$. You won't find any NaHCO$$_3$$ because your reaction Step 2 occurs rapidly enough to go to completion before all the Na$$_2$$CO$$_3$$ has been added. You have Steps 1 and 2 occurring for half of the Na$$_2$$CO$$_3$$, leaving half of the Na$$_2$$CO$$_3$$ unreacted. This scheme conforms to your supposition, but is not elegant chemical technique.
On the other hand, if you add, slowly, by using a tube to add (B to A) 0.1 mole of Na$$_2$$CO$$_3$$ at the bottom of a solution of 0.1 mole of HCl, something like the picture, you should be able to get a solution of 0.1 mole of NaHCO$$_3$$ plus 0.1 mole of NaCl. All the CO$$_2$$ will have been trapped and no bubbling out will have occurred. This is not as easy as it looks, and requires a lot of patience, but is a clear demonstration of your reaction Step 1.