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deleted647690

There was a question in uworld that presented a picture of a polyprotic acid titration curve. It showed the titration of H3PO4 by NaOH. According to this curve:

At the first equivalence point: [H2PO4 -] = [OH-]

At the second equivalence point: [HPO4 2-] = 1/2 [OH-]

At the third equivalence point: [PO4 3-] = 1/3[OH-]

I'm having trouble understand how they came up with these ratios. I know that, at the half equivalence point, the concentration of an acidic species in solution is = the concentration of its conjugate base.
Also, at an equivalence point, one equivalent of OH- has been added to fully neutralize the acidic species.

Therefore, I'd expect these ratios:

1st equivalence point: [H3po4] = [OH-]

2nd equivalence point: [H2PO4-] = [OH-], and therefore, [H3PO4] = 2*[OH-]

3rd equivalence point: [HPO42-] = [OH-], and therefore, [H3PO4] = 3*[OH-].


Here is a picture of a titration curve for H3PO4 for reference:
1586985409902.png
 
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deleted936470

[brackets] refer to concentration of the entire solution and not just the volume or moles added of each species, which is why I think the ratios seem a bit counter-intuitive here.

1st equivalence point: [H2PO4-] = [OH-] - in words, the solution's concentration of acid and base are equal at its 1st equivalent point.

2nd equivalence point: [HPO4-] = [OH-], and therefore, [HPO4-] = 1/2*[OH-] - that is, the solution's [concentration] of acid is equal to half of its concentration of base (that you just added to reach the 2nd equivalence point). The base 'dilutes' the acid in this sense.

3rd equivalence point: [PO43-] = [OH-], and therefore, [PO43-] = 1/3*[OH-] - that is, the solution's [concentration] of acid is equal to a third of its concentration of base (that you just added to reach the 3rd equivalence point). The base further 'dilutes' the acid in this step.

This means the acid concentration is first equal, then twice and then three times more dilute / less concentrated for each respective equivalence point.
 
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D

deleted647690

[brackets] refer to concentration of the entire solution and not just the volume or moles added of each species, which is why I think the ratios seem a bit counter-intuitive here.

1st equivalence point: [H3PO4] = [OH-] - in words, the solution's concentration of acid and base are equal at its 1st equivalent point.

2nd equivalence point: [H2PO4-] = [OH-], and therefore, [H3PO4] = 1/2*[OH-] - that is, the solution's [concentration] of acid is equal to half of its concentration of base (that you just added to reach the 2nd equivalence point). The base 'dilutes' the acid in this sense.

3rd equivalence point: [HPO42-] = [OH-], and therefore, [H3PO4] = 1/3*[OH-] - that is, the solution's [concentration] of acid is equal to a third of its concentration of base (that you just added to reach the 3rd equivalence point). The base further 'dilutes' the acid in this step.

This means the acid concentration is first equal, then twice and then three times more dilute / less concentrated for each respective equivalence point.

Ah, I see I made an error for what I typed out here

"
1st equivalence point: [H3po4] = [OH-]

2nd equivalence point: [H2PO4-] = [OH-], and therefore, [H3PO4] = 2*[OH-]

3rd equivalence point: [HPO42-] = [OH-], and therefore, [H3PO4] = 3*[OH-].
"

Should be 1/2 and 1/3, respectively, because we've added 2 and then 3 equivalents of titrant base. The base titrant is therefore twice and three times more concentrated than the original acid (H3PO4). However, compared to the acid at each equivalence (IE, whether you are looking at the removal of the 1st, 2nd, or 3rd proton, H3PO4, H2PO4-, or PO42-) you are adding one equialent of titrant base for each. But compared to the original acid, it gets progressively more concentrated.


However, this still leaves confusion based on what they wrote in the diagram. At each equivalence point, they wrote the conjugate base as being equal to the titrant base (IE, at second equivalence point, [HPO4 2-] = 1/2 [OH-], and at the third equivalence point [PO4 2 -] = 1/3 [OH-]. Shouldn't this instead be written as you've outlined above?
 
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deleted936470

Ah, I see I made an error for what I typed out here

"
1st equivalence point: [H3po4] = [OH-]

2nd equivalence point: [H2PO4-] = [OH-], and therefore, [H3PO4] = 2*[OH-]

3rd equivalence point: [HPO42-] = [OH-], and therefore, [H3PO4] = 3*[OH-].
"

Should be 1/2 and 1/3, respectively, because we've added 2 and then 3 equivalents of titrant base. The base titrant is therefore twice and three times more concentrated than the original acid (H3PO4). However, compared to the acid at each equivalence (IE, whether you are looking at the removal of the 1st, 2nd, or 3rd proton, H3PO4, H2PO4-, or PO42-) you are adding one equialent of titrant base for each. But compared to the original acid, it gets progressively more concentrated.


However, this still leaves confusion based on what they wrote in the diagram. At each equivalence point, they wrote the conjugate base as being equal to the titrant base (IE, at second equivalence point, [HPO4 2-] = 1/2 [OH-], and at the third equivalence point [PO4 2 -] = 1/3 [OH-]. Shouldn't this instead be written as you've outlined above?

Oops, I copied and pasted what you wrote and now realized that these species in the equations weren't deprotonated like in the figure. The figure is correct because those protons dissociate from the acid at each equivalence point (1-3) and what is shown is the dominant species for each equivalent step. Edited original post above.

Protonation/deprotonation is a function of pH change - a key concept in biochem.
 
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