I have wonders regarding the titration curve of the titration of sulphuric acid, or any diprotic acid, with a strong base. I am aware of the fact that a diprotic acid is protolysed in two steps but, how does this affect the titration curve?
1 Answers
The pKa values for sulfuric acid from Wikipedia are −3, and 1.99. So for all practical purposes the first ionization happens completely and would be unobservable with a pH electrode in aqueous solutions. So if you had a 0.1 molar solution of sulfuric acid it would almost look like you were titrating a 0.2N of $\ce{H^+X^-}$ where that acid had a pKa of 1.69 (pH at half NaOH used). This would be a strong acid too and the inflection point would be that for water at pH 7.
Note that the pKa 1.69 is assuming $\ce{H^+X^-}$ and is taken at 1/2 of the $\ce{Na^+OH^-}$ used. Knowing that $\ce{H2SO4}$ has two protons, the second pKa should be taken at 150% on the plot which would give the correct pKa value of 1.99.
If you had an acid with two lower pKa values then you'd see two pH transitions in the titrations. Look at the figure below. With two widely separated pKa values maleic acid shows two nice transitions. With two pkA values closer together the two transitions for malonic acid aren't as sharp. For succinic acid the knee at a volume of about 5 ml is a hint that there are two dissociations but you don't see two nice complete transitions like you do for maleic acid.
To go back to the sulfuric acid tritration, we'd really expect to see a knee at a low pH something like the titration curve we got for succinic acid. Such a knee isn't seen because a pH electrode just doesn't work at such high acid levels (pH range from -3 to 0).
PS - Thanks to the commenters for help to vastly improve the answer.
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1It's not true! Strong first dissotiation makes it even more visible that it's diprotic! Please don't add obviously wrong answers. – Mithoron Nov 05 '15 at 19:01
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I didn't do the math, but if I have 0.1 N $\ce{H2SO4}$ I'd expect about 0.1 moles $\ce{H^+}$ and $\ce{HSO^{-}_4}$ and very little $\ce{H2SO4}$ or $\ce{SO^{2-}_4}$. Right? – MaxW Nov 05 '15 at 19:09
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1Geez, if you titrate even stronger HCL curve is completely normal! Base is reacting especially with dissociated protons. @samueltober you shouldn't accept wrong answers. – Mithoron Nov 05 '15 at 19:12
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For HCl the transition from acidic to basic solution would happen at pH 7 which is indicative that the HCl was completely dissociated. If you titrate acetic acid you just get one break, not one for acetic acid and another for water. – MaxW Nov 05 '15 at 19:19
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Sorry but i did not know that this answer was wrong – Unknown Nov 05 '15 at 19:37
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I don't know that the answer is wrong. I still don't understand the objections. The one break for acetic acid happens at a pH lower than 7 indicating the dissociation constant of the acetic acid. – MaxW Nov 05 '15 at 19:39
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http://www.titrations.info/acid-base-titration-sulfuric-acid seems you're partially right, I misunderstood you; but it's one transition because of that the second proton is almost completely dissociated, if it was weakly dissociated there would be two breaks - edit it and I'll undo downvote – Mithoron Nov 05 '15 at 19:42
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That being said, I doubt your last sentence is right : indeed, the two pKas 4.5 and 2.5 are so close that we'd probably again observe only one jump. See for instance the titration curves of succinic acid. @MaxW – Hippalectryon Nov 05 '15 at 19:45
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1Ok, I did some major edits on the answer. Thanks for the help improving the clarity. – MaxW Nov 05 '15 at 20:43
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paH isn't here equal pKa, your value was ok – Mithoron Nov 05 '15 at 20:54
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@Mithoron - I assume you mean that the titration shows pKa of 1.69 whereas Wikipedia has 1.99 for pKa2? – MaxW Nov 05 '15 at 20:58
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Hmm, there's 0,1 M H2SO4 titrated (not N) so pH at half should be higher than 2 but maybe because it's activity not concentration, it's lower... dunno, but at 1 M (not 0,1) pKa should be equal pH, if I recollect right – Mithoron Nov 05 '15 at 21:10
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KNOWING that there are two H+ the second pKa would actually be at 150% on the graph shown, not 100%. – MaxW Nov 05 '15 at 21:26

