The Kw of water at 60 C is 9.5 x 10-14. What is the pH of a neutral solution at 60 C?

Question

The Kw of water at 60 C is 9.5 x 10-14. What is the pH of a neutral solution at 60 C?

Answer 1

√(9.5x10^-14)=3.08x10^-7

pH=-log[3.08x10^-7]=6.5

The [H+] and [OH-] in a neutral solution is equal.

Kw=[H+][OH-]

Answer 2

The answer shouldn't just be -log(9.5E-14).

Since Kw = Kb * Ka.  In a neutal solution Kb = Ka.  So we would take the sqrt(9.5E-14), obtaining 3.08E-7.

Now we take the -log(3.08E-7) = 6.51.

So 6.51 is the pH.