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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?

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

2 Answers 2

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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 $\ce{Na2CO3}$

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There are two ways to react sodium carbonate (A) with hydrochloric acid (B):

  1. You can add A to B, or

  2. You can add B to A.

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.

enter image description here

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