the ionization constant for water (kw) is 2.9 x 10^ -14 at 40 degrees C. What is the pOH of water at this temperature?

Answer 1

pOH = -log [OH^-], so we have to find the [OH-] and then take the negative log.

H₂O <==> H⁺ + OH^-

Kw = [H⁺][OH^-]

2.9x10^-14 = (x)(x)

x² = 2.9x10^-14

x = [H⁺] = [OH^-] = 1.70x10^-7 M

pOH = -log 1.70x10^-7

pOH = 6.77