Given that kw for water is 2.4 x 10^-14 at 37oC, compute the PH of a neutral aqueous solution at 37 OC which is the normal human body temperature

Answer 1

Kw = [H⁺][OH^-]

2.4x10^-14 = [H⁺][OH^-] = (x)(x)

x² = 2.4x10^-14

x = 1.55x10^-7 M = [H⁺] = [OH^-]

pH = -log [H⁺] = -log 1.55x10^-7

pH = 6.81