#2H_2O rightleftharpoonsH_3O^(+) + HO^-#
As you have correctly interpreted, this is a bond-breaking reaction, which requires substantial energy input to break the strong #O-H# bond, and the small equilibrium constant, #10^-14# (at #298*K#), reflects this strength. Now this reaction was exhaustively measured at a precise temperature of #298*K#. At higher temperatures, given a bond breaking reaction, we would expect that the equilibrium should lie farther over to the RHS, the product side, and thus #pH# should DECREASE at higher temperatures.
And thus at #100# #""^@C#, #pH=6.14#, as anticipated (this is almost a tenfold increase in the extent of reaction!). See the lower part of here.. Of course under this scenario, #pOH=6.14# necessarily.
Note that typically any equilibrium is a function of temperature. A standard temperature of #298*K# is assumed for simplicity.
