If the Kw of water changes with temperature so that pH is lower at higher temperatures, and [H+]=[OH-] so that the new Kw=[H+]^2, then the...

Kw is the equilibrium constant for the self-dissociation of water .

It is the equilibrium constant for :

H2O =  H+   + OH-

Kw = [H+][OH-]

From the equation we see that there is a 1:1 mole ratio of H+ to OH- ions.

So [OH-] = [H+]  for pure water only

Thus Kw = [H+]^2

As the temperature is increased more water molecules acquire enough kinetic energy to dissociate ( break apart) as a...

Kw is the equilibrium constant for the self-dissociation of water .

It is the equilibrium constant for :

H2O =  H+   + OH-

Kw = [H+][OH-]

From the equation we see that there is a 1:1 mole ratio of H+ to OH- ions.

So [OH-] = [H+]  for pure water only

Thus Kw = [H+]^2

As the temperature is increased more water molecules acquire enough kinetic energy to dissociate ( break apart) as a result of increased intermolecular collisions associated with the increased temperature.

Thus at 25 degrees celsius Kw is 1.00 x 10^-14  and at 50 degrees celsius Kw is 5.476 x 10^-14

Now pKw = - log Kw = - log ([H+][OH-])

using the laws of logs we can show that

pKw = pH + pOH

but since [H+] = [OH-] for pure water

pH = pOH for pure water as well

Thus pKw = 2pH for pure water only

So at 25 degrees celsius , pKw = - log (1.00 x 10-14) =14

and thus 14 = 2pH  so pH = 7  at 25 degrees celsius

At 50 degrees celsius , pKw = - log (5.476 x 10^-14) = 13.26

So  13.26 = 2pH and thus pH = 6.63 at a pH of 50 degrees celsius.

Thus, the pH of pure water decreases from 7.00 at 25 degrees celsius to 6.63 at 50 degrees celsius.

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