The pH of pure water at some temperature is 6.30. What is the value of Kw at this temperature?

Answer 1

Given that the pH = 6.3 , [H3O+] = 10^-6.3

2H2O ---> H3O+ + OH-

[H3O+] = [OH-]

Kw = [H3O+][OH-]/[H2O]^2

Since water is in huge excess and there occurs no appreciable change in water concentration we can ignore it.

Kw = [H3O+][OH-]

= 10^-6.3 x 10^-6.3

= 10^-12.6

pKw = 12.6

Answer 2

Kw is a constant value = 1.00x10-14
At pH = 6.30, the H+ concentration = 5.01x10-7 M and the OH- concentration = 2.00 x 10-8 M
Kw = (5.01x10-7) x (2.00x10-8) = 1.00x10-14