A 0.505 g sample of a monoprotic base (mm = 45.09 g/mol)was dissolved in water to produce 100.0 mL of solution with a pH = 11.84.What is the ionization constant of this base? A)4.3 × 10^-5 B)1.9 × 10^-23 C)4.3 × 10^-4 D)3.4 × 10^-1 E)1.3 × 10^-11

Question

Quizplus · Accepted Answer

First, we need to calculate the concentration of the base, which is given by:

C = m/MM = 0.505 g / 45.09 g/mol = 0.0112 mol/L

Since we know the pH of the solution, we can calculate the concentration of hydronium ions:

pH = -log[H3O+]

11.84 = -log[H3O+]
[H3O+] = 10^-11.84 = 1.48 × 10^-12 M

At equilibrium, the concentration of hydronium ions must be equal to the concentration of hydroxide ions:

[H3O+] = [OH^-]

Since the base is monoprotic, we can write its ionization reaction as:

B + H2O ⇌ BH+ + OH^-

The equilibrium constant for this reaction is:

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

At equilibrium, the concentration of BH+ is equal to the concentration of base that has ionized:

[BH+] = [B] - [OH^-]

Substituting the expressions for [BH+] and [OH^-] into the expression for Kb, and solving for Kb, we get:

Kb = ([B] - [OH^-]) [OH^-] / [B] = ([B]/[OH^-]) - 1

We already know the concentration of OH^- from the pH of the solution:

[OH^-] = 1.48 × 10^-12 M

Substituting that value and the value of [B] into the equation for Kb, we get:

Kb = (0.0112 mol/L) / (1.48 × 10^-12 M) - 1
Kb = 4.3 × 10^-4

Therefore, the answer is C) 4.3 × 10^-4.

A 0505 G Sample of a Monoprotic Base (Mm = 45