The ionization constant of a weak base is Kb=6.0×10−7. What is the pH of an aqueous solution that contains 5.00×10−2M of this base?

Answer 1

Let B represent the weak base.

B + H₂O ==> BH⁺ + OH^-

Kb = [BH⁺][OH^-] / [B]

6.0x10^-7 = (x)(x) / 5.00x10^-2 - x and assume x is small relative to 5.00x10^-2 and ignore it

x² = 3.0x10^-8

x = 1.73x10^-4 (which is less than 1% of 5x10^-2, so above assumption was valid)

This is the [OH-], so ...

pOH = -log 1.73x10^-4 = 3.76

pH = 14 - 3.76

pH = 10.2