A 0.0925 g sample of a monoprotic base (mm = 17.03 g/mol) was dissolved in water to produce 100 mL of solution with a pH = 11.00. What is the ionization constant of this base? A) 1.8 × 10^-6 B) 1.8 × 10^-11 C) 1.8 × 10^-21 D) 1.9 × 10^-5 E) 1.1 × 10^-6

Question

Quizplus · Accepted Answer

First, we can calculate the molarity of the base solution:
molarity = (0.0925 g / 17.03 g/mol) / (0.1 L) = 0.054 M

Next, we can use the pH to calculate the pOH of the solution:
pOH = 14 - 11 = 3

From the pOH, we can calculate the concentration of hydroxide ions:
[OH-] = 10^(-pOH) = 10^(-3) = 0.001 M

Since this is a monoprotic base, we know that its ionization reaction is:

B + H2O ↔ BH+ + OH-

Let x be the concentration of BH+ at equilibrium. Then, the equilibrium concentration of OH- will also be x since the base dissociates 1:1 with OH-. The initial concentration of the base (before any dissociation) is 0.054 M. Thus, the equilibrium concentration of the base will be (0.054 - x).

Using the equilibrium concentrations, we can write the expression for the ionization constant (Kb) of the base:

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

Kb = x(0.001) / (0.054 - x)

We can approximate (0.054 - x) to be equal to 0.054, since x is expected to be much smaller than 0.054. Thus:

Kb ≈ x(0.001) / (0.054)

Now, we need to solve for x. We know that the equilibrium concentration of OH- is also equal to 0.001 M. Thus:

Kb = (0.001)^2 / (0.054)

Kb = 1.85 × 10^-5

Therefore, the ionization constant of the base is 1.9 × 10^-5, which corresponds to choice (D).

A 00925 G Sample of a Monoprotic Base (Mm = 17