Like all equilibrium constants, Kw varies somewhat with temperature. Given that Kw is 7.51×10−13 at some temperature, compute

Question

the pH of a neutral aqueous solution at that temperature.

Answer 1

........H2O ==> H^+ + OH^-
I......liquid....0.....0
C......liquid....x......x
E.....liquid.....x......x

So Kw = (H^+)(OH^-)
Substitute and solve for H^+ and pH.