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Which make HCO3- to show two pH values at two scenarios?
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Marcos Esteban
Which make HCO3- to show two pH values at two scenarios?
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):
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 $\ce{HCl}$:
At the first equivalence point, the solution is $\pu{0.05 M}$ sodium bicarbonate plus $\pu{0.05 M}$$\ce{NaCl}$. Ignoring ionic strength effects, the $\ce{NaCl}$ does nothing to the $\mathrm{pH}$.
The next one shows that $\mathrm{pH} = 8.339$ at the first equivalence point, for the neutralization of carbonic acid with $\ce{NaOH}$:
At the first equivalence point, the solution is $\pu{0.05 M}$ sodium bicarbonate. In both cases, the $\mathrm{pH}$ is approximately 8.34 and is consistent with this answer.
The problem is that both titration curves are wrong: at the first equivalence points, the $\mathrm{pH}$ does not equal $\mathrm{p}K_\mathrm{a1}$ and the $\mathrm{pOH}$ does not equal $\mathrm{p}K_\mathrm{b1}$. As well, the two provided dissociation constants for carbonate ion, in the OP's upper figure, are also wrong.
The figure below shows two titration curves: 1) for $\pu{10 mL}$ of $\pu{0.1 M}$ "carbonic acid", with $\pu{0.1 M}$$\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}$$\ce{HCl}$ titrant (red falling curve):
The $\mathrm{pH}$ of the starting $\pu{0.1 M}$ carbonic acid is 3.67 and the $\mathrm{pH}$ of the starting $\pu{0.1 M}$ sodium carbonate is 11.65. At the first equivalence point, for both titration curves, the $\mathrm{pH}$ is approximately 8.34. The $\mathrm{p}K_\mathrm{a}$ values are as shown in the figure and are from (or derived from) the reference at the bottom.
The next two figures show Excel spreadsheet $\mathrm{pH}$ calculations. The first one shows that $\mathrm{pH} = 8.338$ at the first equivalence point, for the neutralization of sodium carbonate solution with $\ce{HCl}$:
At the first equivalence point, the solution is $\pu{0.05 M}$ sodium bicarbonate plus $\pu{0.05 M}$$\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 $\ce{NaOH}$:
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.
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?More
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) ).More
But while HCO3- acts as a base(according to the 8.34 pH value) how do HCO3- moreover react with NaOH? While there is no excess H+ in the medium?(8.34 shows there is excess OH-) What is that exception?More
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):
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 $\ce{HCl}$:
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 $\ce{NaOH}$:
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 $\left(\mathrm{p}K_\mathrm{b2} = 14.00 - \mathrm{p}K_\mathrm{a1}\right)$.
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):
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 $\ce{HCl}$:
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 $\ce{NaOH}$:
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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