Question

If the Kb of a weak base is 2.4 × 10-6, what is the pH of a 0.45 M solution of this base?

Answer 1

Kb = [OH-] [BH]/ [B]
we know that the concentration of B is 0.45, and that Kb = 2.4*10^-6. [OH-] = [BH], so
Kb = x^2/0.45-x
2.4*10^-6 = x^2/0.45(we can eliminate the -x on the bottom because 2.4*10^-6 is such a small number that subtracting x from 0.45 won't even make a significant difference to our answer)

thus, x = 0.00104, and [OH-] = 0.00104. we can then plug this into our pH and pOH equations:

pOH = -log [OH-], -log[0.00104] = 2.983
pOH + pH = 14, 14 - 2.983 = 11.016

pH = 11.086