pOH and pH of an acid solution

Question

A 0.660 M solution of a weak base (B:) is made. The Kb for this acid is 2.17⋅10^−9. 

How do I find the pH and pOH of a solution given Kb and M of the solution? I'm getting this question wrong.

Answer 1

The first thing you do is find the pOH using the Kb and concentration (0.66 M)

Kb = [BH+][OH-]/[B]

Kb • [B] = [BH+][OH-]

(2.17•10^-9)(0.66) = [BH+][OH-]

1.432•10^-9 = [BH+][OH-] The trick here is to remember that [BH+] and [OH-] will be equal. So in reality [BH+]•[OH-] = [OH-]^2 = [BH+]^2. All three are equivalent statements, so you need to take the square root to find [OH-].

[OH-] = sqrt(1.432•10^-9) = 3.784 • 10^-5

From here pOH = -log[OH-] = 4.422

Since pOH + pH = 14; therefore pH= 14-4.422 = 9.578