Why is the OO-H dissociation is unlikely for the Caro's Acid? Let us consider the Caro's Acid: $\mathrm{H_2SO_5 \equiv SO_3H-OOH}$. 
It is widely known that the hydrogen tied with the $\mathrm{-OO-}$ group is pretty much unlikely to dissociate:
$$ \text {Terminal dissociation: }\mathrm{SO_3H - OOH \rightleftharpoons H^+ + [SO_3-OOH]^-} \\ 
   \text{An unlikely step: }\mathrm{[SO_3-OOH]^- \rightleftharpoons H^+ +[SO_3-OO]^{2-}} $$ 
The first time I encountered that fact I was presented a pretty inconclusive piece of argumentation: it was stated that $ \mathrm{-OO-} $ group is way more electronegative than $ \mathrm{-O-} $, which somehow brings about an obstacle for the dissociation. 

Unfortunately, it was a long time ago and I cannot remember the exact details. I stumbled upon this problem not so long ago. Having a better background in chemistry, I attempted to apply the Sanders' electronegativity balance theory, but the result was pretty inane in terms of our chemical reality: $\chi_{eq}(\mathrm{H_2SO_4}) \approx 2.89894; \chi_{eq}(\mathrm{H_2SO_5}) \approx 2.96803 $. In other words: $ \chi_{eq}(\mathrm{H_2SO_5}) &gt; \chi_{eq}(\mathrm{H_2SO_4}) $.

What am I doing wrong? Is it possible to tweak my attempt to use Sanders' theory? Is there any other approach to this problem?
One possibility is the formation of an intra-"molecular" hydrogen bond between the peroxy-bonded hydrogen and a third oxygen atom.  The structure shown in Wikipedia for the acid suggests this interaction.  The involved hydrogen is curled up into the five-membered ring formed by such a hydrogen bond, while the other hydrogen atom is freely exposed and ready to launch.  Potassium peroxymonosulfate corroborates shows only the non-peroxo hydrogen dissociated.

Hydrogen peroxide does not form this sort of intramolecular hydrogen bond, and as we expect from induction considerations the proton dissociates from this molecule more readily than from water.