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equivalence point

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Raiden2012

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If a diprotic acid is titrated with a strong base, what is the pH at the second equivalence point?
Below 7
7
Above 7
Not enough info
 

circulus vitios

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polycurve.gif


Separate the proton dissociations in your head. If you titrate to the first proton's equivalent point and you assume the first proton is a strong acid, the pH will be neutral at the equivalence point because it's a strong acid / strong base titration. If you assume the first proton is not a strong acid, the pH will be slightly basic at the equivalence point because it's effectively a weak acid / strong base titration.

Dissocation constants (Ka/pKa) are progressively weaker for progressive proton dissocations of any polyprotic acid. You can see the trends here: http://www2.ucdsb.on.ca/tiss/stretton/database/polyprotic_acids.htm

Remember from the first paragraph that the pH is at least 7, maybe higher depending on the acidity of the first proton. Titrating a less acidic species in a neutral or slightly basic solution with a strong base means the solution will definitely be basic (pH >7) at the equivalence point.
 
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sciencebooks

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If a diprotic acid is titrated with a strong base, what is the pH at the second equivalence point?
Below 7
7
Above 7
Not enough info

Above 7 tends to be the case with weak acids and strong bases. Even if the diprotic acid happened to be a strong acid (H2SO4), I believe the second equivalence point would occur at above 7. The poster above me offered another possible explanation. Is that the answer?
 

LoLCareerGoals

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I would say not enough info.
Basically the question we have to answer is H(X)- a strong or a weak acid? Where X can be several atoms. If it is strong, asnwer is 7, weak means above 7. Since we can't be expected to know every 2 proton acid in the universe and its HX- anion's acidity, I would go with not enough info.

Edit: Scratch all that. CV's explanation makes sense. I am ******ed.

Double edit: I would add to CV's explanation though:
1) We can bound the PH of the first titration to be >=7 (=7 if strong, >7 if weak, but it can bever be less then 7).
2) Second titration (even if first is =7) has to be at least a delta (small value) bigger than first, so even if first =7, then after another titration we are at 7+.

I don't think we should bring up specific acids here, this a logic/bounding question.
 

V5RED

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If you go to the first equivalence point it would be like you started with a solution of HA-.

If you go to the second equivalent point it would be like starting with all A2-.

Thinking logically, since we start with all A2- in water that would otherwise be neutral, our pH must be >7. A2- will attract protons from the autoionization of water and thus raise the pH above 7. A2- has no acidic protons to donate and is highly unlikely to be an electron acceptor, so it will not act as an acid. >7 must be the answer.
 
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shaboobly

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If you go to the first equivalence point it would be like you started with a solution of HA-.

If you go to the second equivalent point it would be like starting with all A2-.

Thinking logically, since we start with all A2- in water that would otherwise be neutral, our pH must be >7. A2- will attract protons from the autoionization of water and thus raise the pH above 7. A2- has no acidic protons to donate and is highly unlikely to be an electron acceptor, so it will not act as an acid. >7 must be the answer.

this. think of it in terms of the product you are making.. something with a -2 charge for instance is basic
 
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