Autoionization of water and pH 10 POINTS BEST ANSWER?

Question

At the temperature of the human body, 37D celsius, the value of Kw is 2.4 x 10E-14. calculate the H+, OH-, pH and pOH of pure water at this temperature. what is the relation between pOH, pH and Kw at this temperature? Is water neutral at this temperature?

Answer 1

pKw = - log (2.4 x 10^-14) = 13.62

H2O + H2O <===> H3O+ + OH-

Kw = [H3O+][OH-] = 2.4 x 10^-14

Let s = the ionization of water.

[H3O+] = s

[OH-] = s



(s)(s) = 2.4 x 10^-14

s = 1.55 x 10^-7

[H3O+] = [OH-] = 1.55 x 10^-7 M

pH = pOH =  -log (1.55 x 10^-7)  = 6.81

pH + pOH = 13.62