3
$\begingroup$

I just need an explanation of why it is a strong acid in its first ionization and a weak acid its second ionization.

It becomes $\ce{HSO_4-}$. From what I have learned about oxoacids, the bond strength determines the strength of the acid. So why does it become harder for the molecule to ionize the second time after it loses a hydrogen atom. Just can't think of how it will affect the strength of the bond connecting oxygen and hydrogen.

|improve this question
$\endgroup$
  • $\begingroup$ After the second ionization there is no more oxygen hydrogen bond. $\endgroup$ – Dissenter Sep 28 '14 at 0:46
  • $\begingroup$ If you have enough energy to break one bond why can't you break two? $\endgroup$ – most venerable sir Sep 28 '14 at 0:47
  • $\begingroup$ Because ionization energies are additive. $\endgroup$ – Dissenter Sep 28 '14 at 0:48
  • $\begingroup$ If I were you I'd step away from analyzing bond strength and instead look at this the other way around. Consider the stability of the products. $\endgroup$ – Dissenter Sep 28 '14 at 0:49
  • 4
    $\begingroup$ There are a few ways to think about this, but consider how hard it would be to take $\ce{H+}$ from anionic $\ce{HSO4-}$. Clearly there's an electrostatic attraction which makes removing the second proton more difficult than the first. (There are other interpretations as well.) $\endgroup$ – Geoff Hutchison Sep 28 '14 at 1:56
1
$\begingroup$

The answer goes back to Pauling. pKa of oxoacies of general formula XOm(OH)n is independent of the central atom and the number of hydroxy groups. It only depends on the number of oxo groups, because that's the atoms the negative charge can be distributed over. The difference between successive ionizations is about 5 pKa units. That's because you are removing a proton from a charged species, and work has to be expended to do that.

|improve this answer
$\endgroup$
1
$\begingroup$

From what I have learned about oxoacids, the bond strength determines the strength of the acid.

Yes. But that is not the whole picture. You also have to consider what else is floating around in the solution.

The reaction of the species (I assume aqueous solution) is given below: $$ \ce{H2SO4 <=> HSO4- + H+ <=> SO4^{2-} + 2H+} $$

As you can clearly see, the concentration of the protons is raised when going from left to right. There is a natural equilibrium that pushes backwards.

However, as you correctly point out, the bond stability (or "strength") directly correlates to the stability of the acid species. It might well be (and I'd even bet on it) that the bond strength of $\ce{H-SO4-}$ is higher than $\ce{H-HSO4}$ (note that the bonds indicated here are not the real bonds as they occur but merely a representation) but this requires an in-depth (possibly computational) analysis, as a quick search of the internet did not yield any useful results.

Edit: I did some small, very basic geometry optimization calculations with turbomole (on default settings, so def-SV(P) basis set for all atoms) to find out the bond length of the hydrogen to its oxygen. Turns out my hunch was right, the bond is shorter in $\ce{HSO4-}$ (95.07 pm) than in $\ce{H2SO4}$ (95.72 pm), indicating stronger bonding once the first proton went flying away.

|improve this answer
$\endgroup$
  • $\begingroup$ you might want to specify that the proton is solvated. $\endgroup$ – Dissenter Oct 7 '14 at 20:56
  • 1
    $\begingroup$ @Dissenter As are the anionic species. I chose not to for enhanced legibility. (Personally I think $(\mathrm{aq})$ is far too often used when the solvent condition is already known.) $\endgroup$ – tschoppi Oct 7 '14 at 21:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.