Ch. 16: Aqueous Ionic
Equilibrium
Dr. Namphol Sinkaset
Chem 201: General Chemistry II
I. Chapter Outline
I.
II.
III.
IV.
V.
Introduction
Buffers
Titrations and pH Curves
Solubility Equilibria and Ksp
Complex Ion Equilibria
I. Last Aspects of Equilibria
• In this chapter, we cover some final
topics concerning equilibria.
• Buffers are designed to take advantage
of Le Châtelier’s Principle.
• Solubility can be reexamined from an
equilibrium point of view.
• Complex ions are introduced and their
formation explained using equilibrium
ideas.
II. pH Resistive Solutions
• A solution that resists changes in pH by
neutralizing added acid or base is called
a buffer.
• Many biological processes can only
occur within a narrow pH range.
• In humans, the pH of blood is tightly
regulated between 7.36 and 7.42.
II. Creating a Buffer
• To resist changes in
pH, any added acid
or base needs to be
neutralized.
• This can be
achieved by using a
conjugate acid/base
pair.
II. How the Buffer Works
• For a buffer comprised of the conjugate
acid/base pair of acetic acid/acetate:
 OH-(aq) + CH3COOH(aq)  H2O(l) + CH3COO-(aq)
 H+(aq) + CH3COO-(aq)  CH3COOH(aq)
• As long as we don’t add too much OH- or
H+, the buffer solution will not change pH
drastically.
II. Calculating the pH of Buffers
• The pH of a buffer can be calculated by
approaching the problem as an
equilibrium in which there are two initial
concentrations.
• Since an acid and its conjugate base
are both in solution, problem can be
solved from a Ka or Kb point of view.
II. Sample Problem
• A solution was prepared in which
[CH3COONa] = 0.11 M and [CH3COOH]
= 0.090 M. What is the pH of the
solution? Note that the Ka for acetic
acid is 1.8 x 10-5.
II. Sample Problem
• A student dissolves 0.12 mole NH3 and
0.095 mole NH4Cl in 250 mL of water.
What’s the pH of this buffer solution?
Note that the Kb for ammonia is 1.8 x
10-5.
II. A Special Buffer Equation
• Buffers are used so widely that an
equation has been developed for it.
• This equation can be used to perform
pH calculations of buffers.
• More importantly, it can be used to
calculate how to make solutions
buffered around a specific pH.
II. Henderson-Hasselbalch Eqn.
II. The H-H Equation
• The Henderson-Hasselbalch equation works
for a buffer comprised of conjugate acid/base
pairs.
• It works provided the “x is small”
approximation is valid.
II. Sample Problem
• A student wants to make a solution
buffered at a pH of 3.90 using formic
acid and sodium formate. If the Ka for
formic acid is 1.8 x 10-4, what ratio of
HCOOH to HCOONa is needed for the
buffer?
II. Sample Problem
• A researcher is preparing an acetate
buffer. She begins by making 100.0 mL
of a 0.010 M CH3COOH solution. How
many grams of CH3COONa does she
need to add to make the pH of the
buffer 5.10 if the Ka for acetic acid is 1.8
x 10-5?
II. Upsetting the Buffer
• A buffer resists changes to pH, but it is
not immune to change.
• Adding strong acid or strong base will
result in small changes of pH.
• We calculate changes in pH of a buffer
by first finding the stoichiometric change
and then performing an equilibrium
calculation.
II. Illustrative Problem
• A 1.00 L buffer containing 1.00 M
CH3COOH and 1.00 M CH3COONa has
a pH of 4.74. It is known that a reaction
carried out in this buffer will generate
0.15 mole H+. If the pH must not
change by more than 0.2 pH units, will
this buffer be adequate?
II. Stoichiometric Calculation
• When the H+ is formed, it will react
stoichiometrically with the base.
• We set up a different type of table to find new
equilibrium concentrations.
• We use ≈0.00 mol because [H+] is negligible.
H+
Initial
Change
Final
0.15 mol
-0.15 mol
≈0.00 mol
+
CH3COO-
1.00 mol
-0.15 mol
0.85 mol
 CH3COOH
1.00 mol
+0.15 mol
1.15 mol
II. Equilibrium Calculation
• We use the new buffer concentrations in an
equilibrium calculation to find the new pH.
• Again, use ≈0.00 M because [H3O+] is negligible.
CH3COOH + H2O  H3O+ + CH3COOInitial
Change
Equil.
1.15 M
----
≈0.00 M
0.85 M
-x
----
+x
+x
1.15 – x
----
x
0.85 + x
Solve to get [H3O+] = 2.44 x 10-5 M, and pH = 4.613.
II. How the Buffer Works
II. A Simplification for Buffers
• In buffer problems, #
of moles can be
used in place of
concentration.
• This can be done
because all
components are in
the same solution,
and hence have the
same volume.
II. Sample Problem
• Calculate the pH when 10.0 mL of 1.00
M NaOH is added to a 1.0-L buffer
containing 0.100 mole CH3COOH and
0.100 mole CH3COONa. Note that the
Ka for acetic acid is 1.8 x 10-5.
II. Buffers Using a Weak Base
• Up until now, we’ve been creating
buffers with a weak acid and it’s
conjugate base.
• Can also make a buffer from a weak
base and its conjugate acid.
• Henderson-Hasselbalch still applies, but
we need to get Ka of the conjugate acid.
II. pKa/pKb Relationship
II. Sample Problem
• Calculate the pH of a 1.0-L buffer that is
0.50 M in NH3 and 0.20 M in NH4Cl
after 30.0 mL of 1.0 M HCl is added.
Note that the Kb for NH3 is 1.8 x 10-5.
II. Making Effective Buffers
• There are a few parameters to keep in
mind when making a buffer:
 The relative [ ]’s of acid and conjugate base
should not differ by more than factor of 10.
 The higher the actual [ ]’s of acid and
conjugate base, the more effective the
buffer.
 The effective range for a buffer system it +/1 pH unit on either side of pKa.
II. Buffer Capacity
• Buffer capacity is defined as the amount
of acid or base that can be added to a
buffer without destroying its
effectiveness.
• A buffer is destroyed when either the
acid or conjugate base is used up.
• Buffer capacity increases w/ higher
concentrations of buffer components.