pKa H2O2 / HO2- = 11.7
pKa HO2- / O22- = ?
I guess it is very basic perhaps around 20? I would like to have a true value and if possible a reference. Thanks
Bonjour Gérard, merci pour cette réponse, mais 4,8 est la valeur pour le couple HO2/O2- soit l'anion superoxyde qui provient d'une première réduction de l'O2. Celui que je recherche est le pKa du couple HO2-/O22- soit la deuxième acidité du peroxyde d'hydrogène. Le pKa devrait être très basique, en tout cas supérieur à 11,7 qui est celui de la première acidité de H2O2. Le peroxyde d'hydrogène est formé par deux réductions de O2.
I guess PKa of the second proton going to be very high since that proton is not acidic anymore as there is a existence of negative charge on the molecule. Even t-BuLi cannot deprotonate it.
Its like deprotonation of first proton in a water molecule is easy but not the same for second proton.
Dear Ransac. You has proposed a question that require an experimental design to show what is the real value. I suggest a system where you have H2O2 100%, pure. Obtain the constant of autoprotolysis. With this kind of experiment, it is possible To know what is the value of constant of autoprotolysis of H2O2. The equilibrium would be:
H2O2 + H2O2 = H3O2+ HO2-. The constant of this reaction would be determined. I dont know if this constant have been measured. But, would be interesting report a new value to compare.
Depending of this value, you can obtain a new scale of pH in this solvent.
I think that obtain H2O2 100% pure, would be not so easy., because should be inestable. The experimental results are very important to show what happened.
After the abstraction first proton , the second proton will be very difficult to abstract out , as there is negative charge on the molecule . Even if it is possible , the second PKa will be very high .
It seems that nobody knows it. I did an estimation of 20 and it is what I'm planning to use for my simulations
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