The pH of a 2.1 *10^-³ M solution of a weak base is 9.87. Calculate K_b for this base.

Question

Quizplus · Accepted Answer

First, we need to determine the pOH of the solution:

pH + pOH = 14

pOH = 14 - 9.87 = 4.13

Next, we can use the equation for the ionization of a weak base:

B + H2O ⇌ BH+ + OH-

The equilibrium expression for this reaction is:

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

We know that [OH-] = 10^(-pOH) and [B] = 2.1 M.

To determine [BH+], we need to use the fact that the solution is primarily composed of the weak base and its conjugate acid:

BH+ + OH- ⇌ B + H2O

Kw = [BH+][OH-] = 1.0 x 10^-14

At equilibrium, we can assume that [BH+] ≈ [OH-], so we can substitute this into the expression for Kw:

Kw = [OH-]^2 = [BH+]^2

[BH+] = [OH-] = 10^(-4.13) = 6.4 x 10^-5 M

Now we can plug in the values for [BH+], [OH-], and [B] into the expression for Kb:

Kb = ([BH+][OH-])/[B] = (6.4 x 10^-5)(6.4 x 10^-5)/(2.1) = 2.6 x 10^-10

Therefore, the best choice is A: 2.6 x 10^-10.

The PH of a 2