Chapter 15 Notes 1

5. weak bases: more fun stuff!equilibria & Kb

B(aq)+H2O(l) BH1+ (aq)+OH1-(aq)

BOHBH

Kb

11

examples of N-bases: the proton attaches to the N lone pair

Chapter 15 Notes 2

questions and calculations:1. Ammonia has Kb=1.8x10-5; pyridine has Kb=1.7x10-9; methyl amine has Kb=4.4x10-4. Thus, given 0.010 M solutions of each,

a. pyridine is the strongest base so it has the highest pH.b. pyridine is the weakest base so it has the highest pH.c. methyl amine is the strongest base so it has the lowest pH.d. methyl amine is the strongest base so it has the highest pH.

2. Calculate the pH and the concentration of all equilibrium species in a 0.57 M solution of methylamine.

Chapter 15 Notes 3

questions and calculations:3. Consider a solution of sodium acetate•Write down all the predominant species (those species present prior to any reaction).

•Look for any possible direct reaction between these species

predominant species: Na1+, C2H3O21-, H2O

none in this case

Chapter 15 Notes 4

•Write down any reactions, including the equilibrium expression and constant, that lead to secondary species, such as the autoionization of water

H2OH1++OH1- KW=1x10-14

secondary species: H1+, OH1-

•Look for reactions between the predominant and secondary species, and write the equations and the equilibrium expression and constant

H1++C2H3O21-HC2H3O2 1/Ka=1/(1.8x10-5)

Chapter 15 Notes 5

•Add the two reactions together to get the net reaction; multiply the equilibrium expressions and constants together to the the net equilibrium expression/constant.

H2OH1++OH1- KW=1x10-14

H1++C2H3O21-HC2H3O2 1/Ka=1/(1.8x10-5)

C2H3O21-+H2OHC2H3O2+OH1- K=KWx(1/Ka)=

5.56x10-5

in general: Kw=KaxKb

Chapter 15 Notes 6

questions and calculations:

4. Calculate the pH of a 0.10 M solution of sodium phenolate, NaOC6H5; phenol is a weak acid with Ka=1.3x10-10.