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)1.8 × 10^-5 E)1.1 × 10^-6

Question

Quizplus · Accepted Answer

First, we can use the pH to find the concentration of OH- in the solution:
pOH = 14 - pH = 14 - 11 = 3
[OH-] = 10^-pOH = 10^-3 M

Since this is a monoprotic base, we can assume that the concentration of the base is equal to the concentration of OH-:
[base] = [OH-] = 10^-3 M

Now we can use the definition of the ionization constant (Kb):
Kb = [OH-][H+]/[base]
Since the pH is 11, we know that [H+] = 10^-pH = 10^-11 M
Substituting in our values, we get:
Kb = (10^-3)(10^-11)/10^-2 = 10^-12
which is equivalent to 1.8 × 10^-5, or answer choice D.

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