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According to the below two titrations,

pH titration-1

pH titration-2

Image Reference

If we consider the reactions, at the first reaction (in first figure after adding 1.0), there is $\ce{HCO3-, NaCl,}$ and $\ce{H2O}$ at the first equivalence point. At the second one (after adding 1.0), there is $\ce{HCO3-,}$ and $\ce{H2O, Cl-}$ at the first equivalence point. $\ce{HCO3-}$ only affect to the $\mathrm{pH}$ of the medium. But,

  • According to the first figure, $\ce{HA- + H2O → H2A + OH-}$ it acts as a base
  • According to the second figure, $\ce{HA– + H2O → A2– + H3O+}$ it acts as a acid

I am doubtful on why do $\ce{HCO3-}$ perform as two types at these different situations? Do anything in the medium manipulate $\ce{HCO3-}$ to work as a base or an acid? How to describe these two scenarios? Generally I can't see a theoretical method for describe it.

I searched this in so many resources, but I couldn't find a satisfied answer. So, I hope a theoretical answer, for this problem, not a mathematical description.

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The problem is that both titration curves are wrong: at the first equivalence points, the $\mathrm{pH}$ does not equal $\mathrm{p}K_\mathrm{a1}$ and the $\mathrm{pOH}$ does not equal $\mathrm{p}K_\mathrm{b1}$. As well, the two provided dissociation constants for carbonate ion, in the OP's upper figure, are also wrong.

The figure below shows two titration curves: 1) for $\pu{10 mL}$ of $\pu{0.1 M}$ "carbonic acid", with $\pu{0.1 M}$ $\ce{NaOH}$ titrant (black rising curve) and 2) for $\pu{10 mL}$ of $\pu{0.1 M}$ sodium carbonate solution, with $\pu{0.1 M}$ $\ce{HCl}$ titrant (red falling curve):

Titration curves

The $\mathrm{pH}$ of the starting $\pu{0.1 M}$ carbonic acid is 3.67 and the $\mathrm{pH}$ of the starting $\pu{0.1 M}$ sodium carbonate is 11.65. At the first equivalence point, for both titration curves, the $\mathrm{pH}$ is approximately 8.34. The $\mathrm{p}K_\mathrm{a}$ values are as shown in the figure and are from (or derived from) the reference at the bottom.

The next two figures show Excel spreadsheet $\mathrm{pH}$ calculations. The first one shows that $\mathrm{pH} = 8.338$ at the first equivalence point, for the neutralization of sodium carbonate solution with $\ce{HCl}$:

Carbonate plus HCl

At the first equivalence point, the solution is $\pu{0.05 M}$ sodium bicarbonate plus $\pu{0.05 M}$ $\ce{NaCl}$. Ignoring ionic strength effects, the $\ce{NaCl}$ does nothing to the $\mathrm{pH}$.

The next one shows that $\mathrm{pH} = 8.339$ at the first equivalence point, for the neutralization of carbonic acid with $\ce{NaOH}$:

Carbonic acid plus NaOH

At the first equivalence point, the solution is $\pu{0.05 M}$ sodium bicarbonate. In both cases, the $\mathrm{pH}$ is approximately 8.34 and is consistent with this answer.

The two $\mathrm{p}K_\mathrm{a}$ values for carbonic acid are from D.G. Harris, In Quantitative Chemical Analysis, 7th Edition; Appendix G; W. H. Freeman and Company, ©2007. The two $\mathrm{p}K_\mathrm{b}$ values for carbonate ion are obtained from the two $\mathrm{p}K_\mathrm{a}$ values for carbonic acid via $\left(\mathrm{p}K_\mathrm{b1} = 14.00 - \mathrm{p}K_\mathrm{a2}\right)$ and $\left(\mathrm{p}K_\mathrm{b2} = 14.00 - \mathrm{p}K_\mathrm{a1}\right)$.

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    $\begingroup$ Your answer has described only half of my doubt. What about the titration of Na2CO3(Same concentration of your considered H2CO3) with HCl at the first equivalence point (after added 10ml of HC)l. Will the pH same to 8.34 as this?Or If it is not then what is the reason to having different pH values? $\endgroup$
    – Osal Thuduwage
    Commented Jul 6, 2020 at 15:24
  • $\begingroup$ I have shown the image reference in the question.(after edit). But while HCO3- acts as a base(according to the 8.34 pH value) how do HCO3- moreover react with NaOH? $\endgroup$
    – Osal Thuduwage
    Commented Jul 6, 2020 at 15:44
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    $\begingroup$ @MathewMahindaratne Thanks, it looks much better now! $\endgroup$
    – Ed V
    Commented Mar 16, 2021 at 14:50
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    $\begingroup$ @Ed V: It is the icing on the delicious cake! :-) $\endgroup$
    – Mathew Mahindaratne
    Commented Mar 16, 2021 at 14:53
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    $\begingroup$ The equation shown in the second spreadsheet is the same, with slightly different notation, as the equation here: chemistry.stackexchange.com/a/43422/79678. $\endgroup$
    – Ed V
    Commented Nov 16, 2021 at 23:51
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The (equivalence)points, which you have marked in the graphs, are where all of the acid present in the sample has been neutralized by the base added with the titrant –and vice versa.

But, pKa equals pH at the point where only half of the acid is neutralized. In other words: [HA] = [A-] -the concentration of acid equals the concentration of the corresponding base.

At these points, according to the Henderson-Hasselebalch equation, pH = pKa. They are called inflection points, and is where the titration curve is horizontal. See e.g. Wikipedia – titration curves

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    $\begingroup$ I think this is not the answer for which I have asked, I asked about the equivalence point which have every H2CO3 has converted into HCO3- and every Na2CO3 has converted into HCO3-(First equivalence point: Which has green color pointed pH=7.6(1) and pH=6.4(2) ). $\endgroup$
    – Osal Thuduwage
    Commented Jul 6, 2020 at 9:17

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