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 =-log_(10)[H_3O^+]# #=# #7#, at #298*K#.

But this relates to the autoprotolysis reaction:

#2H_2O rightleftharpoons H_3O^+ + 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, #H_3O^+# 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 #H_3O^+#. Therefore it seems that this analysis is not too offbase. If #H_3O^+# has increased, how does #HO^-# evolve?

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