# Equivalence point of titration of Sodium carbonate

The pH at the equivalence point of titration of $$\ce{Na2CO3}$$ solution with $$\ce{HCl}$$ is around 3.7, as shown in this titration curve:

At this point, the following reaction is completed:
$$\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}$$ But considering the fact that $$\ce{NaCl}$$ is a neutral salt, shouldn't the pH be closer to 7? The $$\ce{H2CO3}$$ formed also decomposes to form $$\ce{H2O + CO2}$$, and shouldn't affect the pH either. Why then is the equivalence point at such a low pH value?

• Contrary to what you say, CO2 in water does affect the pH. You need a little bit more HCl to obtain the equivalence point. – Maurice Nov 27 '20 at 14:34
• @Maurice, Wouldn't the equivalence point be the point where the solution has been completely neutralized, and thus would only contain $\ce{NaCl + H2CO3}$? Also doesn't the $\ce{CO2}$ formed escape, and not cause a significant decrease in pH? – jerrythealpaca Nov 27 '20 at 15:40
• The equivalence point is not the point where the solution contains $\ce{NaCl}$ and that's all. No. The equivalent point is when the solution contains $\ce{NaCl}$ plus the small amount of $\ce{CO2}$ which remains dissolved. And this amount makes the solution a bit acidic. So the pH at the equivalence point is not $7$, but a bit less. – Maurice Nov 27 '20 at 17:25
• @Maurice, You are mixing the concept of "end-point" and "equivalence point". At the end-point there is slight excess of the acid or base. At the equivalence pt, stoichiometric amounts have reacted. – M. Farooq Nov 28 '20 at 0:45
• @M. Farooq. OK. You are right. I should have rewritten my text. But this does not change too much my message. – Maurice Nov 28 '20 at 9:56

When the $$Na_2CO_3$$ is completely neutralized by the $$HCl$$, the solution will be saturated with $$CO_2$$, so the pH will be lower than 7 - think of carbonated beverages. Although the $$CO_2$$ bubbles out, not all of it bubbles out - not even if stirred. If you had stopped the titration at pH = 7, the pH reduction from 11 would have been partly from $$HCl$$ and partly from $$H_2CO_3$$.
• Okay, so the low pH at the equivalence point can be attributed to just the dissolved $\ce{CO2}$, correct? – jerrythealpaca Nov 27 '20 at 16:02