Water has a pH of 6.14 at 373K. Calculate the ionic product of water (Kw) at this temperature

Answer 1

From pH = 6.14, you can find the [H+] = [OH-]

[H+] = 1x10^-6.14 = 7.24x10^-7 M

Kw = [H+][OH-] = (7.24x10^-7)(7.24x10^-7)

Kw = 5.2x10^-13