Question

If the Kb of a weak base is 9.0 × 10-6, what is the pH of a 0.12 M solution of this base?

Answer 1

Kb = [OH-] [BH]/ [B]
we know that the concentration of B is 0.12, and that Kb = 9 e-6. [OH-] = [BH], so
Kb = x^2/0.12-x
9 e-6 = x^2/0.12 (we can eliminate the -x on the bottom because 9e-6 is such a small number that subtracting x from 0.12 won't even make a significant difference to our answer)

thus, x = 0.00103923, and [OH-] = 0.00103923. we can then plug this into our pH and pOH equations:

pOH = -log [OH-], -log[0.00103923] = 2.9832
pOH + pH = 14, 14 - 2.9832= 11.0167

pH = 11.0167