Please no pictures I deficiency the counter-argument typed quenched, thank you.
The Kw of soak varies with atmosphere. Consider the pH of soak at 46⁰C with a Kw = 1.219 x 10-14. Show total calculations ascititious to an counter-argument.
first we should consider the H3O+ concentration.
H2O + H2O ————–> H3O+ + OH–
as we know
[H3O+] [OH–] = Kw
[H3O+] = [OH–]
[H3O+]2 = Kw = 1.219 x 10-14
[H3O+] = 1.10 x 10-7 M
pH = – log [H3O+] = – log [1.1 x 10–7]
pH = 6.96