CHM112 Home Page Reference Materials Homework Assignments Lectures Exams & Quizzes Grades Study Aids URI Home Page


J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 15.52

Codeine, C18H21NO3, a commonly prescribed painkiller, is a weak base. A saturated aqueous solution contains 1.00 g codeine in 120 mL of solution and has a pH = 9.8. What is the Kb of codeine?

C18H21NO3(aq) + H2O(l)[C18H21NHO3]+(aq) + OH(aq) Kb = ?






Answer:

Treat this as any equilibrium problem: write the balanced chemical reaction, write the mass action expression for Kb, set up a table of concentrations, and then solve for the unknown variables using the given information. Assume 25 oC.


C18H21NO3(aq)
+
H2O(l)
[C18H21NHO3]+(aq)
+
OH(aq)
Initial
[B]ini
0
0
Change
– x
+ x
+ x
Equilibrium
[B]ini – x
x
x

[B]ini is found from the concentration information given:

molar mass of codeine = 18(12.01) + 21(1.01) + 1(14.01) + 3(16.00) = 299.4 g/mol

so the molar concentration of the saturated solution is:

[B]ini = 0.0278 M

The pH gives the hydronium ion concentration:

[H3O+] = 10–pH = 10–9.8 = 2×10–10 M

The hydroxide ion concentration is found from the hydronium ion concentration by:

[OH] = Kw/[ H3O+] = 1.0×10–14/2×10–10 = 5×10–5 M

Then,

x = [OH] = 5×10–5 M

[B]ini – x = 0.0278 – 5×10–5 = 0.0278 M

so Kb is found from: