Question

derive the relation pkw=ph +p0h

Answer 1

Start with the expression for Kw:
Kw= 10-14 = [H3O+][OH-]
Take the negative log of both sides of the equation:
-log 10-14 = 14 = -log([H3O+][OH-])
Recall that log(ab) = log(a) + log(b)
14 = -log [H3O+] + -log [OH-]
Using pH = -log[H3O+] and pOH = -log[OH-] gives:
14 = pH + pOH

Answer 2

here is ur answer...................