How should the autoprotolysis reaction of water evolve at HIGHER temperature than 298⋅K?

1 Answer
anor277
Apr 30, 2016

Consider the definition of pH; it is clearly a BOND BREAKING PHENOMENON

Explanation:

By definition, pH=−log10[H3O+] = 7, at 298⋅K.

But this relates to the autoprotolysis reaction:

2H2O⇌H3O++HO−

Since this is a bond-breaking reaction, we would expect that autoprotolysis would become more facile at HIGHER temperature. (Why?). Thus the equilibrium should shift to the right.

Should the equilibrium shift rightwards, H3O+ should reasonably increase. Given the stated definition of pH, the value should thus decrease.

But as chemists, as physical scientists, we should seek out the data that inform our argument. From the interwebz, I found that at 333⋅K, pH=6.54. That is pH has decreased corresponding to an increase in H3O+. Therefore it seems that this analysis is not too offbase. If H3O+ has increased, how does HO− evolve?

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