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4/19/2016
Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
The Danger of Antifreeze
• Each year, thousands of pets and wildlife
species die from consuming antifreeze
• Most brands of antifreeze contain ethylene
glycol
Chapter 16
Aqueous Ionic
Equilibrium
sweet taste
initial effect drunkenness
• Metabolized in the liver to glycolic acid
HOCH2COOH
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright  2011 Pearson Education, Inc.
Tro: Chemistry: A Molecular Approach, 2/e
Why Is Glycolic Acid Toxic?
• If present in high enough concentration in the
•
•
bloodstream, glycolic acid overwhelms the
buffering ability of the HCO3− in the blood,
causing the blood pH to drop
When the blood pH is low, its ability to carry O2
is compromised
acidosis
HbH+(aq) + O2(g)  HbO2(aq) + H+(aq)
One treatment is to give the patient ethyl
alcohol, which has a higher affinity for the
enzyme that catalyzes the metabolism of
ethylene glycol
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Making an Acid Buffer
glycolic acid
(aka a-hydroxyethanoic acid)
ethylene glycol
(aka 1,2–ethandiol)
2
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Buffers
• Buffers are solutions that resist changes in pH
when an acid or base is added
• They act by neutralizing acid or base that is
added to the buffered solution
• But just like everything else, there is a limit to
what they can do, and eventually the pH changes
• Many buffers are made by mixing a solution of a
weak acid with a solution of soluble salt
containing its conjugate base anion
blood has a mixture of H2CO3 and HCO3−
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How Acid Buffers Work:
Addition of Base
HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
• Buffers work by applying Le Châtelier’s Principle
to weak acid equilibrium
• Buffer solutions contain significant amounts of the
weak acid molecules, HA
• These molecules react with added base to
neutralize it
HA(aq) + OH−(aq) → A−(aq) + H2O(l)
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 you can also think of the H3O+ combining with the OH−
to make H2O; the H3O+ is then replaced by the shifting
equilibrium
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How Acid Buffers Work:
Addition of Acid
HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
How Buffers Work
H2O
new
A−
HA
HA
• The buffer solution also contains significant

A−−
+
•
H3O+
•
Added
HO−
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How Buffers Work
HA
8
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Common Ion Effect
HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
H2O
new
HA
amounts of the conjugate base anion, A−
These ions combine with added acid to make
more HA
H+(aq) + A−(aq) → HA(aq)
After the equilibrium shifts, the concentration
of H3O+ is kept constant
• Adding a salt containing the anion NaA, which

A−−
+
H3O+
•
is the conjugate base of the acid (the common
ion), shifts the position of equilibrium to the left
This causes the pH to be higher than the pH of
the acid solution
lowering the H3O+ ion concentration
Added
H3O+
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Common Ion Effect
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Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
write the reaction
for the acid with
water
HC2H3O2 + H2O  C2H3O2 + H3O+
construct an ICE
table for the
reaction
initial
change
equilibrium
enter the initial
concentrations –
assuming the
[H3O+] from
water is ≈ 0
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[HA]
[A−]
[H3O+]
0.100
0.100
≈0
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Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
HC2H3O2 + H2O  C2H3O2 + H3O+
initial
change
[HA]
[A−]
[H3O+]
0.100
0.100
0
x
+x
+x
equilibrium 0.100 x 0.100 + x
x
Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10−5
determine the
value of Ka
initial
because Ka is very
small, approximate change
the [HA]eq = [HA]init equilibrium
and [A−]eq = [A−]init ,
then solve for x
[HA]
[A−]
[H3O+]
0.100
0.100
≈0
−x
+x
+x
x
0.100x 0.100
0.100+x
0.100
substitute into the
equilibrium
constant
expression
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Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10−5
check if the
approximation
is valid by
seeing if x <
5% of
[HC2H3O2]init
initial
change
equilibrium
[HA]
[A−]
[H3O+]
0.100
0.100
≈0
−x
+x
0.100
0.100
+x
x
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Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
substitute x
into the
equilibrium
concentration
definitions
and solve
x = 1.8 x 10−5
initial
change
equilibrium
[HA]
[A−]
[H3O+]
0.100
0.100
≈0
−x
+x
+x
0.100x 0.100
0.100
+ x 1.8E-5
x
0.100
x = 1.8 x 10−5
the approximation is valid
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Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
substitute [H3O+]
into the formula
for pH and solve
initial
change
equilibrium
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[HA]
[A−]
[H3O+]
0.100
0.100
≈0
−x
0.100
+x
0.100
+x
1.8E−5
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Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
check by
substituting the
equilibrium
concentrations
back into the
equilibrium
constant
expression and
comparing the
calculated Ka to the
given Ka
Ka for HC2H3O2 = 1.8 x 10−5
initial
change
equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
[HA]
[A−]
[H3O+]
0.100
0.100
≈0
−x
+x
+x
0.100
0.100
1.8E−5
the values
match
18
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Practice − What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
Practice − What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
HF + H2O  F + H3O+
write the
reaction for the
acid with water
construct an
ICE table for
the reaction
enter the initial
concentrations
– assuming the
[H3O+] from
water is ≈ 0
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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
HF + H2O  F + H3O+
initial
change
[HA]
[A−]
[H3O+]
0.14
0.071
0
x
+x
+x
equilibrium 0.14 x 0.071 + x
x
initial
change
equilibrium
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[HA]
0.14
[A−] [H3O+]
0.071 ≈ 0
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20
Practice – What is the pH of a buffer that is
0.14 M and 0.071 M KF?
Ka pK
fora HF
for =
HF7.0
= 3.15
x 10−4
determine the
value of Ka
because Ka is
very small,
approximate the
[HA]eq = [HA]init
and [A−]eq = [A−]init
solve for x
initial
change
equilibrium
[HA]
[A−]
[H3O+]
0.14
0.071
≈0
−x
+x
0.14
x
0.012
0.100+x
0.071
+x
x
substitute into
the equilibrium
constant
expression
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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
Ka for HF = 7.0 x 10−4
check if the
approximation
is valid by
seeing if
x < 5% of
[HC2H3O2]init
initial
change
equilibrium
x = 1.4 x
[HA]
[A−]
[H3O+]
0.14
0.071
≈0
−x
+x
+x
0.14
0.071
x
10−3
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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
substitute x into
the equilibrium
concentration
definitions and
solve
initial
change
equilibrium
[HA]
[A−]
[H3O+]
0.14
0.071
≈0
−x
+x
+x
0.14x 0.071
0.072
0.14
+ x 1.4E-3
x
x = 1.4 x 10−3
the approximation is valid
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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
substitute
[H3O+] into the
formula for pH
and solve
initial
change
equilibrium
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[HA]
[A−]
[H3O+]
0.14
0.071
≈0
−x
0.14
+x
0.072
+x
1.4E−3
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Henderson-Hasselbalch Equation
• Calculating the pH of a buffer solution can be
•
Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
check by
substituting the
equilibrium
concentrations
back into the
equilibrium
constant
expression and
comparing the
calculated Ka to
the given Ka
Ka for HF = 7.0 x 10−4
initial
change
equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
[HA]
[A−]
[H3O+]
0.14
0.071
≈0
−x
+x
+x
0.14
0.072
1.4E−3
the values are
close enough
26
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Deriving the Henderson-Hasselbalch
Equation
simplified by using an equation derived from
the Ka expression called the HendersonHasselbalch Equation
The equation calculates the pH of a buffer
from the pKa and initial concentrations of the
weak acid and salt of the conjugate base
as long as the “x is small” approximation is valid
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Example 16.2: What is the pH of a buffer that
is 0.050 M HC7H5O2 and 0.150 M NaC7H5O2?
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
assume the [HA] and HC7H5O2 + H2O  C7H5O2 + H3O+
[A−] equilibrium
Ka for HC7H5O2 = 6.5 x 10−5
concentrations are
the same as the
initial
substitute into the
HendersonHasselbalch
Equation
check the “x is small”
approximation
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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
find the pKa from the
given Ka
• The Henderson-Hasselbalch equation is
HF + H2O  F + H3O+
generally good enough when the “x is small”
approximation is applicable
• Generally, the “x is small” approximation will work
when both of the following are true:
assume the [HA]
and [A−] equilibrium
concentrations are
the same as the
initial
a) the initial concentrations of acid and salt are not very
dilute
b) the Ka is fairly small
substitute into the
HendersonHasselbalch
equation
• For most problems, this means that the initial
acid and salt concentrations should be over 100
to 1000x larger than the value of Ka
check the “x is
small”
approximation
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How Much Does the pH of a Buffer
Change When an Acid or Base Is Added?
• Though buffers do resist change in pH when acid
•
or base is added to them, their pH does change
Calculating the new pH after adding acid or base
requires breaking the problem into two parts
1. a stoichiometry calculation for the reaction of the
added chemical with one of the ingredients of the
buffer to reduce its initial concentration and increase
the concentration of the other
 added acid reacts with the A− to make more HA
 added base reacts with the HA to make more A−
2. an equilibrium calculation of [H3O+] using the new
initial values of [HA] and [A−]
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Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
fill in the table
– tracking the
changes in the
number of
moles for each
component
Do I Use the Full Equilibrium Analysis or
the Henderson-Hasselbalch Equation?
HC2H3O2 + OH−  C2H3O2 + H2O
HA
mols before
mols added
mols after
A−
0.100 0.100
─
─
0.090 0.110
OH−
≈0
0.010
≈0
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Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
If the added
chemical is a
base, write a
reaction for OH−
with HA. If the
added chemical
is an acid, write a
reaction for H3O+
with A−.
construct a
stoichiometry
table for the
reaction
HC2H3O2 + OH−  C2H3O2 + H2O
HA
mols before
mols added
mols after
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A−
0.100 0.100
─
─
OH−
0
0.010
Copyright  2011 Pearson Education, Inc.
Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
If the added
HC2H3O2 + OH−  C2H3O2 + H2O
chemical is a base,
write a reaction for
HA
A– OH−
OH− with HA. If the
added chemical is
mols before 0.100 0.100 0.010
an acid, write a
reaction for it with A−. mols change
construct a
stoichiometry table
for the reaction
mols end
new molarity
enter the initial
number of moles for
each
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Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
HC2H3O2 + OH−  C2H3O2 + H2O
using the added
chemical as the
limiting reactant,
determine how the
moles of the other
chemicals change
A−
HA
add the change to
the initial number of
moles to find the
moles after reaction
OH−
mols before
mols change −0.010 +0.010 −0.010
0
mols end
0.090 0.110
0
new molarity 0.090 0.110
0.100
0.100
0.010
divide by the liters
of solution to find
the new molarities
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Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
HC2H3O2 + H2O  C2H3O2 + H3O+
initial
change
[HA]
[A−]
[H3O+]
0.090
0.110
≈0
x
+x
+x
equilibrium 0.090 x 0.110 + x
x
Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
write the reaction
for the acid with
water
construct an ICE
table for the
reaction
enter the initial
concentrations –
assuming the
[H3O+] from water
is ≈ 0, and using
the new
molarities of the
[HA] and [A−]
HC2H3O2 + H2O  C2H3O2 + H3O+
initial
change
equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
[HA]
[A−]
[H3O+]
0.090
0.110
≈0
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38
Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10−5
determine the
value of Ka
[HA]
[A−]
[H3O+]
because Ka is very initial
0.100
0.100
≈0
small, approximate
change
−x
+x
+x
the [HA]eq = [HA]init
and [A−]eq = [A−]init
0.090x 0.110
0.110+x
equilibrium 0.090
x
solve for x
substitute into
the equilibrium
constant
expression
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Copyright  2011 Pearson Education, Inc.
Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
check if the
approximation
is valid by
seeing if
x < 5% of
[HC2H3O2]init
Ka for HC2H3O2 = 1.8 x 10−5
initial
change
equilibrium
[HA]
[A−]
[H3O+]
0.090
0.110
≈0
−x
+x
0.090
0.110
+x
x
Tro: Chemistry: A Molecular Approach, 2/e
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40
Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
substitute x
into the
equilibrium
concentration
definitions
and solve
initial
change
equilibrium
[HA]
[A-]
[H3O+]
0.090
0.110
≈0
−x
+x
0.090x 0.110
0.110+ x
0.090
+x
1.5E-5
x
x = 1.47 x 10−5
x = 1.47 x 10−5
the approximation is valid
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Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
substitute
[H3O+] into
the formula
for pH and
solve
initial
change
equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
[HA]
[A−]
[H3O+]
0.090
0.110
≈0
−x
+x
+x
0.090
0.110
1.5E−5
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43
Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10−5
check by
substituting the
equilibrium
concentrations
back into the
equilibrium
constant
expression and
comparing the
calculated Ka to
the given Ka
initial
change
equilibrium
[HA]
[A−]
[H3O+]
0.090
0.110
≈0
−x
+x
+x
0.090
0.110
1.5E−5
the values
match
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Copyright  2011 Pearson Education, Inc.
Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
write the reaction for
the acid with water
HC2H3O2 + H2O  C2H3O2 + H3O+
construct an ICE
table for the reaction
or, by using the
Henderson-Hasselbalch
Equation
enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0,
and using the new
molarities of the [HA]
and [A−]
[HA]
[A−]
0.090
0.110
initial
+x
x
change
equilibrium 0.090 x 0.110 + x
[H3O+]
≈0
+x
x
fll in the ICE table
Copyright  2011 Pearson Education, Inc.
Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
find the pKa from
the given Ka
HC2H3O2 + H2O  C2H3O2 + H3O+
assume the [HA]
and [A−] equilibrium
concentrations are
the same as the
initial
initial
change
Ka for HC2H3O2 = 1.8 x 10−5
[HA]
[A−]
[H3O+]
0.090
0.110
≈0
−x
+x
+x
equilibrium 0.090
0.110
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Example 16.3: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
substitute into the
HendersonHasselbalch
equation
HC2H3O2 + H2O  C2H3O2 + H3O+
pKa for HC2H3O2 = 4.745
check the “x is
small”
approximation
x
47
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Example 16.3: Compare the effect on pH of adding
0.010 mol NaOH to a buffer that has 0.100 mol
HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L to
adding 0.010 mol NaOH to 1.00 L of pure water
HC2H3O2 + H2O  C2H3O2 + H3O+
Practice – What is the pH of a buffer that has
0.140 moles HF (pKa = 3.15) and 0.071 moles
KF in 1.00 L of solution when 0.020 moles of
HCl is added?
(The “x is small” approximation is valid)
pKa for HC2H3O2 = 4.745
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Practice – What is the pH of a buffer that has
0.140 moles HF and 0.071 moles KF in 1.00 L of
solution when 0.020 moles of HCl is added?
If the added
chemical is a
base, write a
reaction for OH−
with HA. If the
added chemical
is an acid, write a
reaction for H3O+
with A−.
construct a
stoichiometry
table for the
reaction
F− + H3O+  HF + H2O
mols before
mols added
mols after
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F−
H3O+
HF
0.071
0
0.140
–
0.020
–
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a buffer that has
0.140 moles HF and 0.071 moles KF in 1.00 L of
solution when 0.020 moles of HCl is added?
If the added chemical
is a base, write a
reaction for OH− with
HA. If the added
chemical is an acid,
write a reaction for
H3O+ with A−.
construct a
stoichiometry table
for the reaction
F− + H3O+  HF + H2O
mols before
F−
H3O+
HF
0.071
0.020
0.140
mols change
new molarity
enter the initial
number of moles for
each
50
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a buffer that has
0.140 moles HF and 0.071 moles KF in 1.00 L of
solution when 0.020 moles of HCl is added?
fill in the table
– tracking the
changes in the
number of
moles for each
component
F− + H3O+  HF + H2O
mols before
mols added
mols after
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F−
H3O+
HF
0.071
0
0.140
–
0.020
–
0.051
0
0.160
Copyright  2011 Pearson Education, Inc.
Practice – What is the pH of a buffer that has
0.140 moles HF and 0.071 moles KF in 1.00 L of
solution when 0.020 moles of HCl is added?
using the added
chemical as the
limiting reactant,
determine how the
moles of the other
chemicals change
add the change to
the initial number of
moles to find the
moles after reaction
mols after
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Tro: Chemistry: A Molecular Approach, 2/e
F− + H3O+  HF + H2O
F−
H3O+
HF
mols before
0.071
0.020
0.140
mols change
−0.020
−0.020 +0.020
mols after
0.051
0
0.160
new molarity
0.051
0
0.160
divide by the liters
of solution to find
the new molarities
53
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Practice – What is the pH of a buffer that has
0.140 moles HF and 0.071 moles KF in 1.00 L of
solution when 0.020 moles of HCl is added?
HF + H2O  F + H3O+
write the reaction for
the acid with water
construct an ICE
table.
assume the [HA] and
[A−] equilibrium
concentrations are
the same as the
initial
initial
change
[HF]
[F−]
[H3O+]
0.160
0.051
≈0
+x
+x
0.051
x
−x
equilibrium 0.160
Basic Buffers
B:(aq) + H2O(l)  H:B+(aq) + OH−(aq)
• Buffers can also be made by mixing a weak
base, (B:), with a soluble salt of its conjugate
acid, H:B+Cl−
H2O(l) + NH3 (aq)  NH4+(aq) + OH−(aq)
substitute into the
HendersonHasselbalch
equation
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Tro: Chemistry: A Molecular Approach, 2/e
Henderson-Hasselbalch Equation for
Basic Buffers
•
•
•
The Henderson-Hasselbalch equation is written for a
chemical reaction with a weak acid reactant and its
conjugate base as a product
The chemical equation of a basic buffer is written with a
weak base as a reactant and its conjugate acid as a
product
B: + H2O  H:B+ + OH−
To apply the Henderson-Hasselbalch equation, the
chemical equation of the basic buffer must be looked at like
an acid reaction
H:B+ + H2O  B: + H3O+

57
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• Just as there is a relationship between the Ka
of a weak acid and Kb of its conjugate base,
there is also a relationship between the pKa of
a weak acid and the pKb of its conjugate base
Ka  Kb = Kw = 1.0 x 10−14
−log(Ka  Kb) = −log(Kw) = 14
−log(Ka) + −log(Kb) = 14
pKa + pKb = 14
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Example 16.4: What is the pH of a buffer that is
0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl?
NH3 + H2O  NH4+ + OH−
find the pKa of the
conjugate acid (NH4+)
from the given Kb
•
substitute into the
HendersonHasselbalch equation
•
check the “x is small”
approximation
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Henderson-Hasselbalch Equation for
Basic Buffers
•
assume the [B] and
[HB+] equilibrium
concentrations are the
same as the initial
Copyright  2011 Pearson Education, Inc.
Relationship between pKa and pKb
this does not affect the concentrations, just the way we are looking
at the reaction
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The Henderson-Hasselbalch equation is written for a
chemical reaction with a weak acid reactant and its
conjugate base as a product
The chemical equation of a basic buffer is written
with a weak base as a reactant and its conjugate
acid as a product
B: + H2O  H:B+ + OH−
We can rewrite the Henderson-Hasselbalch equation
for the chemical equation of the basic buffer in terms
of pOH
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Example 16.4: What is the pH of a buffer that is
0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl?
NH3 + H2O  NH4+ + OH−
find the pKb if given Kb
Buffering Effectiveness
• A good buffer should be able to neutralize
moderate amounts of added acid or base
assume the [B] and
[HB+] equilibrium
concentrations are the
same as the initial
• However, there is a limit to how much can be
added before the pH changes significantly
• The buffering capacity is the amount of acid or
substitute into the
Henderson-Hasselbalch
equation base form,
find pOH
base a buffer can neutralize
• The buffering range is the pH range the buffer
can be effective
• The effectiveness of a buffer depends on two
check the “x is small”
approximation
factors (1) the relative amounts of acid and base,
and (2) the absolute concentrations of acid and
base
calculate pH from pOH
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Effect of Relative Amounts of Acid
and Conjugate Base
Effect of Absolute Concentrations of Acid
and Conjugate Base
A buffer is most effective with equal
concentrations of acid and base
A buffer is most effective when the
concentrations of acid and base are largest
Buffer 1
Buffer 2
0.100 mol HA & 0.100 mol A−
0.18 mol HA & 0.020 mol A−
Initial pH = 5.00 pK (HA) = 5.00 Initial pH = 4.05
Buffer 1
Buffer 2
0.50 mol HA & 0.50 mol A−
0.050 mol HA & 0.050 mol A−
Initial pH = 5.00 pK (HA) = 5.00 Initial pH = 5.00
a
HA + OH−  A + H2O
HA + OH−  A + H2O
a
after adding 0.010 mol −NaOH −
HA
A
OH
pH = 5.09
mols before
0.100
0.100
mols added
─
─
0.090
0.110
mols after
0
after adding 0.010 mol NaOH
HA= 4.25A−
OH−
pH
mols before
0.010 mols added
after adding 0.010
HA mol
A−NaOH
OH−
pH = 5.02
0.18
0.020
0
mols before
0.50
0.500
─
─
0.010
mols added
─
─
0.49
0.51
0.030 ≈ 0
≈ 0 mols afterCopyright0.17
 2011 Pearson Education, Inc.
Buffering Capacity
mols after
0
after addingHA
0.010 mol
A− NaOH
OH−
pH = 5.18
mols before
0.010 mols added
0.050
0.050
0
─
─
0.010
0.060 ≈ 0
≈ 0 mols afterCopyright0.040
 2011 Pearson Education, Inc.
Effectiveness of Buffers
• A buffer will be most effective when the
a concentrated
buffer can
neutralize
more added
acid or base
than a dilute
buffer
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[base]:[acid] = 1
equal concentrations of acid and base
• A buffer will be effective when
0.1 < [base]:[acid] < 10
• A buffer will be most effective when the [acid]
and the [base] are large
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Buffering Range
• We have said that a buffer will be effective when
0.1 < [base]:[acid] < 10
• Substituting into the Henderson-Hasselbalch
equation we can calculate the maximum and
minimum pH at which the buffer will be effective
Lowest pH
Highest pH
Example 16.5a: Which of the following acids
would be the best choice to combine with its
sodium salt to make a buffer with pH 4.25?
Chlorous acid, HClO2
pKa = 1.95
Nitrous acid, HNO2
pKa = 3.34
Formic acid, HCHO2
pKa = 3.74
Hypochlorous acid, HClOpKa = 7.54
Therefore, the effective pH range of a buffer is pKa ± 1
When choosing an acid to make a buffer, choose
one whose is pKa closest to the pH of the buffer
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Example 16.5b: What ratio of NaCHO2 : HCHO2
would be required to make a buffer with pH 4.25?
Formic acid, HCHO2, pKa = 3.74
The pKa of HCHO2 is closest to the desired
pH of the buffer, so it would give the most
effective buffering range
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Practice – What ratio of NaC7H5O2 : HC7H5O2
would be required to make a buffer with pH 3.75?
Benzoic acid, HC7H5O2, pKa = 4.19
To make a buffer with
pH 4.25, you would use
3.24 times as much
NaCHO2 as HCHO2
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Practice – What ratio of NaC7H5O2 : HC7H5O2
would be required to make a buffer with pH 3.75?
Benzoic acid, HC7H5O2, pKa = 4.19
to make a buffer with
pH 3.75, you would use
0.363 times as much
NaC7H5O2 as HC7H5O2
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Buffering Capacity
• Buffering capacity is the amount of acid or base
that can be added to a buffer without causing a
large change in pH
• The buffering capacity increases with increasing
absolute concentration of the buffer components
• As the [base]:[acid] ratio approaches 1, the ability
of the buffer to neutralize both added acid and
base improves
• Buffers that need to work mainly with added acid
generally have [base] > [acid]
• Buffers that need to work mainly with added base
generally have [acid] > [base]
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Titration
Titration
• In an acid-base titration, a solution of unknown
concentration (titrant) is slowly added to a
solution of known concentration from a burette
until the reaction is complete
 when the reaction is complete we have reached the
endpoint of the titration
• An indicator may be added to determine the
endpoint
 an indicator is a chemical that changes color when the
pH changes
• When the moles of H3O+ = moles of OH−, the
titration has reached its equivalence point
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Titration Curve:
Unknown Strong Base Added
to Strong Acid
Titration Curve
• A plot of pH vs. amount of added titrant
• The inflection point of the curve is the equivalence
point of the titration
• Prior to the equivalence point, the known solution in
the flask is in excess, so the pH is closest to its pH
• The pH of the equivalence point depends on the pH of
the salt solution
 equivalence point of neutral salt, pH = 7
 equivalence point of acidic salt, pH < 7
 equivalence point of basic salt, pH > 7
• Beyond the equivalence point, the unknown solution in
the burette is in excess, so the pH approaches its pH
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Tro: Chemistry: A Molecular Approach, 2/e
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
Because the
solutions are equal
concentration, and
1:1 stoichiometry, the
equivalence point is

at equal volumes
After Equivalence
(excess base)
Equivalence Point
equal moles of
HCl and NaOH
pH = 7.00
Before Equivalence
(excess acid)
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Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
•
•
•
•
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Initial pH = −log(0.100) = 1.00
Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
Before equivalence point
added 5.0 mL NaOH
5.0 x 10−4 mole NaOH added
mols before
HCl
NaCl
NaOH
2.50E-3
0
5.0E-4
mols change −5.0E-4 +5.0E-4 −5.0E-4
mols end
2.00E-3 5.0E-4
molarity, new 0.0667
77
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0.017
0
0
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Titration of 25 mL of 0.100 M HCl with
0.100 M NaOH
• HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(aq)
• To reach equivalence, the added moles NaOH =
•
•
initial moles of HCl = 2.50 x 10−3 moles
At equivalence, we have 0.00 mol HCl and 0.00
mol NaOH left over
Because the NaCl is a neutral salt, the pH at
equivalence = 7.00
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Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
•
•
•
•
2.5 x 10−3 mole NaOH added
HCl
0.100 mol NaOH
1L
 moles added NaOH
0
2.5E-3
0
molarity, new
0
0.050
0
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80
mols before
mols change
NaCl
NaOH
2.50E-3
0
3.0E-3
added
30.0
35.0
5.0
10.0
25.0
mL
mLNaOH
NaOH
25.0 mL
0.100
M
HCl
0.00050
0.00100
0.00200
0.00150
equivalence
mol point
NaOH
0.00250
HCl
pH = 12.22
11.96
1.18
1.37
7.00
1.00
added 15.0
40.0 mL NaOH
0.00150 mol HCl
0.00100
NaOH
pH = 1.60
12.36
added 20.0
50.0 mL NaOH
0.00250 mol HCl
0.00050
NaOH
pH = 1.95
12.52
−2.5E-3 +2.5E-3 −2.5E-3
mols end
0
2.5E-3
5.0E-4
molarity, new
0
0.045
0.0091
Copyright  2011 Pearson Education, Inc.
Practice – Calculate the pH of the solution that
results when 10.0 mL of 0.15 M NaOH is added to
50.0 mL of 0.25 M HNO3
Copyright  2011 Pearson Education, Inc.
Adding 0.100 M NaOH to 0.100 M HCl
3.0 x 10−3 mole NaOH added
HCl
NaOH
mols end
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Initial pH = −log(0.100) = 1.00
Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
added 30.0 mL NaOH
After equivalence point
L of added NaOH 
NaCl
mols before 2.50E-3
0
2.5E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
•
•
•
•
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Initial pH = −log(0.100) = 1.00
Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
At equivalence point
added 25.0 mL NaOH
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•
•
•
•
82
Copyright  2011 Pearson Education, Inc.
Practice – Calculate the pH of the solution that
results when 10.0 mL of 0.15 M NaOH is added to
50.0 mL of 0.25 M HNO3
HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l)
Initial pH = −log(0.250) = 0.60
Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
Before equivalence point added 10.0 mL NaOH
mols before
HNO3
NaNO3
NaOH
1.25E-2
0
1.5E-3
mols change −1.5E-3 +1.5E-3 −1.5E-3
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mols end
1.1E-3
1.5E-3
0
molarity, new
0.018
0.025
0
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Practice – Calculate the amount of 0.15 M NaOH
solution that must be added to 50.0 mL of 0.25 M
HNO3 to reach equivalence
Practice – Calculate the amount of 0.15 M NaOH
solution that must be added to 50.0 mL of 0.25 M
HNO3 to reach equivalence
•
•
•
•
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Practice – Calculate the pH of the solution that
results when 100.0 mL of 0.15 M NaOH is added to
50.0 mL of 0.25 M HNO3
Tro: Chemistry: A Molecular Approach, 2/e
•
•
•
•
87
HNO3
NaNO3
NaOH
mols before
1.25E-2
0
1.5E-2
mols change
−1.25E-2 +1.25E-2 −1.25E-2
mols end
0
1.25E-2
0.0025
molarity, new
0
0.0833
0.017
89
Initial pH = −log(0.250) = 0.60
Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
At equivalence point: moles of NaOH = 1.25 x 10−2
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•
•
•
86
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Copyright  2011 Pearson Education, Inc.
Practice – Calculate the pH of the solution that
results when 100.0 mL of 0.15 M NaOH is added to
50.0 mL of 0.25 M HNO3
HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l)
Initial pH = −log(0.250) = 0.60
Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
added 100.0 mL NaOH
After equivalence point
HNO3
NaNO3
NaOH
mols before
1.25E-2
0
1.5E-2
mols change
−1.25E-2 +1.25E-2 −1.25E-2
mols end
0
1.25E-2
0.0025
molarity, new
0
0.0833
0.017
Copyright  2011 Pearson Education, Inc.
Practice – Calculate the pH of the solution that
results when 100.0 mL of 0.15 M NaOH is added to
50.0 mL of 0.25 M HNO3
HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l)
Initial pH = −log(0.250) = 0.60
initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
added 100.0 mL NaOH
After equivalence point
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HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l)
Copyright  2011 Pearson Education, Inc.
Titration of a Strong Base with a Strong Acid
• If the titration is run
so that the acid is in
the burette and the
base is in the flask,
the titration curve
will be the reflection
of the one just
shown
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Titration of a Weak Acid with a Strong Base
• Titrating a weak acid with a strong base results in
•
•
•
•
differences in the titration curve at the equivalence
point and excess acid region
The initial pH is determined using the Ka of the
weak acid
The pH in the excess acid region is determined as
you would determine the pH of a buffer
The pH at the equivalence point is determined
using the Kb of the conjugate base of the weak acid
The pH after equivalence is dominated by the
excess strong base
Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH
• HCHO2(aq) + NaOH(aq)  NaCHO2(aq) + H2O(aq)
• Initial pH
[HCHO2] [CHO2−] [H3O+]
initial
change
0.100
0.000
≈0
−x
+x
+x
x
x
equilibrium 0.100 − x
Ka = 1.8 x 10−4
 the basicity from the conjugate base anion is negligible
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91
Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH
• HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• Before equivalence
added 5.0 mL NaOH
HA
A−
2.50E-3
0
0
mols added
–
–
5.0E-4
2.00E-3 5.0E-4
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• HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• At equivalence
CHO2−(aq) + H2O(l)  HCHO2(aq) + OH−(aq)
[HCHO2] [CHO2−] [OH−]
HA
A−
OH−
initial
0
0.0500
≈0
mols before
2.50E-3
0
0
change
+x –
−x –
+x
mols
added
2.50E-3
equilibrium
x 0 5.00E-2-x
mols
after
2.50E-3 x ≈ 0
≈0
Copyright  2011 Pearson Education, Inc.
93
25.0
added 30.0
5.0 mL
mLNaOH
NaOH
10.0
initial HCHO2 solution
equivalence
point
0.00200
0.00050
molmL
NaOH
0.00150
HCHO
2xs
added
35.0
NaOH
0.00250
mol HCHO
− 2
0.00250
mol CHO
pH
= 11.96
3.14mol
3.56
2 xs
0.00100
NaOH
pH = 2.37
−
[CHO
2 ]init = 0.0500 M
pH
= 12.22
[OH−]eq = 1.7 x 10−6
added
12.5 mL NaOH
pH = 8.23
added
40.0 mL NaOH
0.00125 mol HCHO2
0.00150 mol NaOH xs
pH = 3.74 = pKa
pH = 12.36
half-neutralization
added 30.0 mL NaOH
3.0 x 10−3 mole NaOH added
NaOH
mols before 2.50E-3
0
3.0E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
mols end
0
2.5E-3 5.0E-4
molarity, new
0
0.045
95
[OH−] = 1.7 x 10−6 M
Copyright  2011 Pearson Education, Inc.
• HCHO2(aq) + NaOH(aq)  NaCHO2(aq) + H2O(aq)
• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• After equivalence
A−
added 25.0 mL NaOH
Kb = 5.6 x 10−11
Adding NaOH to HCHO2
Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH
HA
Copyright  2011 Pearson Education, Inc.
Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH
OH−
mols before
mols after
Tro: Chemistry: A Molecular Approach, 2/e
added 50.0
15.0 mL NaOH
0.00100 mol HCHO
0.00250
NaOH 2xs
pH = 12.52
3.92
added 20.0 mL NaOH
0.00050 mol HCHO2
pH = 4.34
0.0091
Copyright  2011 Pearson Education, Inc.
Tro: Chemistry: A Molecular Approach, 2/e
96
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Titrating Weak Acid with a Strong Base
• The initial pH is that of the weak acid solution
calculate like a weak acid equilibrium problem
e.g., 15.5 and 15.6
calculate mol HAinit and mol A−init using reaction
stoichiometry
calculate pH with Henderson-Hasselbalch using
mol HAinit and mol A−init
• Half-neutralization pH = pKa
[A−]init = mol A−/total liters
calculate like a weak base equilibrium problem
e.g., 15.14
• Beyond equivalence point, the OH is in excess
[OH−] = mol MOH xs/total liters
[H3O+][OH−]=1 x 10−14
Copyright  2011 Pearson Education, Inc.
Example 16.7a: A 40.0 mL sample of 0.100 M HNO2
is titrated with 0.200 M KOH. Calculate the volume
of KOH at the equivalence point.
write an equation for
the reaction for B
with HA
so the resulting solution has only the conjugate
base anion in it before equilibrium is established
calculate the volume of added base as you did in
Example 4.8
becomes a buffer
97
• At the equivalence point, the mole HA = mol Base,
mol A− = original mole HA
• Before the equivalence point, the solution
Tro: Chemistry: A Molecular Approach, 2/e
Titrating Weak Acid with a Strong Base
HNO2 + KOH  NO2 + H2O
Tro: Chemistry: A Molecular Approach, 2/e
Example 16.7b: A 40.0 mL sample of 0.100 M HNO2
is titrated with 0.200 M KOH. Calculate the pH after
adding 5.00 mL KOH.
write an
equation for the
reaction for B
with HA
make a
stoichiometry
table and
determine the
moles of HA in
excess and
moles A made
Tro: Chemistry: A Molecular Approach, 2/e
99
Copyright  2011 Pearson Education, Inc.
Example 16.7b: A 40.0 mL sample of 0.100 M HNO2
is titrated with 0.200 M KOH. Calculate the pH after
adding 5.00 mL KOH.
HNO2 + H2O  NO2 + H3O+
write an
equation for the
reaction of HA
with H2O
Table 15.5 Ka = 4.6 x
10−4
NO2−
HNO2
0
mols before 0.00400
mols added
–
–
0.00300 0.00100
mols after
Tro: Chemistry: A Molecular Approach, 2/e
101
OH−
≈0
0.00100
≈0
Copyright  2011 Pearson Education, Inc.
HNO2
mols before
mols added
mols after
Tro: Chemistry: A Molecular Approach, 2/e
100
NO2−
OH−
0.00400
0
≈0
–
–
0.00100
0.00300 0.00100
≈0
Copyright  2011 Pearson Education, Inc.
Example 16.7b: A 40.0 mL sample of 0.100 M HNO2
is titrated with 0.200 M KOH. Calculate the pH at
the half-equivalence point.
write an
equation for the
reaction for B
with HA
determine the
moles of HAbefore
& moles of added
B
determine Ka
and pKa for HA
use the
HendersonHasselbalch
equation to
determine the
pH
HNO2 + KOH  NO2 + H2O
determine the
moles of HAbefore
& moles of
added B
use stoichiometry to
determine the volume
of added B
Copyright  2011 Pearson Education, Inc.
98
make a
stoichiometry
table and
determine the
moles of HA in
excess and
moles A made
HNO2 + KOH  NO2 + H2O
at half-equivalence, moles KOH = ½ mole HNO2
HNO2 NO2−
OH−
0
≈0
mols before 0.00400
0.00200
mols added
–
–
0.00200 0.00200
mols after
≈0
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102
Copyright  2011 Pearson Education, Inc.
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Example 16.7b: A 40.0 mL sample of 0.100 M HNO2
is titrated with 0.200 M KOH. Calculate the pH at
the half-equivalence point.
Titration Curve of a Weak Base with a
Strong Acid
HNO2 + H2O  NO2 + H3O+
write an equation
for the reaction of
HA with H2O
Table 15.5 Ka = 4.6 x 10-4
determine Ka and
pKa for HA
use the
HendersonHasselbalch
equation to
determine the pH
HNO2 NO2−
OH−
0
≈0
mols before 0.00400
0.00200
mols added
–
–
0.00200 0.00200
mols after
≈0
Tro: Chemistry: A Molecular Approach, 2/e
103
Copyright  2011 Pearson Education, Inc.
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the initial pH of the
NH3 solution.
Tro: Chemistry: A Molecular Approach, 2/e
105
Copyright  2011 Pearson Education, Inc.
Practice – Titration of 25.0 mL of 0.10 M NH3 with
0.10 M HCl. Calculate the initial pH of the NH3(aq)
pKb = 4.75
Kb = 10−4.75 = 1.8 x 10−5
• NH3(aq) + HCl(aq)  NH4Cl(aq)
+
• Initial: NH3(aq) + H2O(l)  NH4 (aq) + OH−(aq)
[HCl] [NH4+]
pKb = 4.75
Kb = 10−4.75 = 1.8 x 10−5
• NH3(aq) + HCl(aq)  NH4Cl(aq)
• Initial: NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)
[NH3]
initial
0
0
change
+x
+x
−x
equilibrium
x
x
0.10−x
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0.10
106
Copyright  2011 Pearson Education, Inc.
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the pH of the solution
after adding 5.0 mL of HCl.
[NH3]
initial
0
0
change
+x
+x
−x
equilibrium
x
x
0.10−x
Tro: Chemistry: A Molecular Approach, 2/e
Copyright  2011 Pearson Education, Inc.
Practice – Titration of 25.0 mL of 0.10 M NH3 with
0.10 M HCl. Calculate the initial pH of the NH3(aq)
[HCl] [NH4+]
Tro: Chemistry: A Molecular Approach, 2/e
104
0.10
107
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Tro: Chemistry: A Molecular Approach, 2/e
108
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Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the pH of the
solution after adding 5.0 mL of HCl.
• NH3(aq) + HCl(aq)  NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• Before equivalence: after adding 5.0 mL of HCl
NH4+(aq) + H2O(l)  NH4+(aq) + H2O(l)
NH3
NH4Cl
HCl
mols before
2.50E-3
0
5.0E-4
mols change
mols end
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the pH of the solution
after adding 5.0 mL of HCl.
• NH3(aq) + HCl(aq)  NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• Before equivalence: after adding 5.0 mL of HCl
pKb = 4.75
pKa = 14.00 − 4.75 = 9.25
NH3
NH4Cl
HCl
mols before
2.50E-3
0
5.0E-4
−5.0E-4 −5.0E-4 −5.0E-4
mols change
−5.0E-4 −5.0E-4 −5.0E-4
2.00E-3
0
mols end
2.00E-3
5.0E-4
0
0
molarity, new
0.0667
0.017
0
molarity, new
0.0667
5.0E-4
0.017
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Copyright  2011 Pearson Education, Inc.
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the pH of the
solution at equivalence.
Tro: Chemistry: A Molecular Approach, 2/e
111
Copyright  2011 Pearson Education, Inc.
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the pH of the
solution at equivalence.
NH3(aq) + HCl(aq)  NH4Cl(aq)
at equivalence [NH4Cl] = 0.050 M
Tro: Chemistry: A Molecular Approach, 2/e
Copyright  2011 Pearson Education, Inc.
110
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the pH of the
solution at equivalence.
• NH3(aq) + HCl(aq)  NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• At equivalence mol NH3 = mol HCl = 2.50 x 10−3
NH3
NH4Cl
HCl
mols before
2.50E-3
0
2.5E-3
mols change
−2.5E-3 +2.5E-3 −2.5E-3
mols end
0
2.5E-3
0
molarity, new
0
0.050
0
Tro: Chemistry: A Molecular Approach, 2/e
112
added 25.0 mL HCl
Copyright  2011 Pearson Education, Inc.
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the pH of the
solution after adding 30.0 mL of HCl.
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
[NH3]
[NH4+]
[H3O+]
initial
0
0.050
≈0
change
+x
−x
+x
equilibrium
x
0.050−x
x
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113
Copyright  2011 Pearson Education, Inc.
Tro: Chemistry: A Molecular Approach, 2/e
114
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Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the pH of the
solution after adding 30.0 mL of HCl.
• NH3(aq) + HCl(aq)  NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• After equivalence: after adding 30.0 mL HCl
NH3
NH4Cl
HCl
mols before
2.50E-3
0
3.0E-3
mols change
−2.5E-3 +2.5E-3 −2.5E-3
mols end
0
2.5E-3
5.0E-4
molarity, new
0
0.045
0.0091
Titration of a Polyprotic Acid
• If Ka1 >> Ka2, there will be two equivalence
points in the titration
the closer the Ka’s are to each other, the less
distinguishable the equivalence points are
titration of 25.0 mL of 0.100 M
H2SO3 with 0.100 M NaOH
when you mix a strong acid, HCl,
with a weak acid, NH4+, you only
need to consider the strong acid
Tro: Chemistry: A Molecular Approach, 2/e
115
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Monitoring pH During a Titration
Tro: Chemistry: A Molecular Approach, 2/e
116
Copyright  2011 Pearson Education, Inc.
Monitoring pH During a Titration
• The general method for monitoring the pH during
the course of a titration is to measure the
conductivity of the solution due to the [H3O+]
 using a probe that specifically measures just H3O+
• The endpoint of the titration is reached at the
equivalence point in the titration – at the inflection
point of the titration curve
• If you just need to know the amount of titrant
added to reach the endpoint, we often monitor the
titration with an indicator
Tro: Chemistry: A Molecular Approach, 2/e
117
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Tro: Chemistry: A Molecular Approach, 2/e
118
Copyright  2011 Pearson Education, Inc.
Phenolphthalein
Indicators
• Many dyes change color depending on the pH of the
solution
• These dyes are weak acids, establishing an
equilibrium with the H2O and H3O+ in the solution
HInd(aq) + H2O(l)  Ind(aq) + H3O+(aq)
• The color of the solution depends on the relative
concentrations of Ind:HInd
 when Ind:HInd ≈ 1, the color will be mix of the colors of
Ind and HInd
 when Ind:HInd > 10, the color will be mix of the colors
of Ind
 when Ind:HInd < 0.1, the color will be mix of the colors
of HInd
Tro: Chemistry: A Molecular Approach, 2/e
119
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Tro: Chemistry: A Molecular Approach, 2/e
120
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Monitoring a Titration with an
Indicator
Methyl Red
H
C
(CH3)2N
H
C
C
H
C
C
C
H
N
H
C
OH-
C
N
N
• For most titrations, the titration curve shows a
H
C
C
CH
C
C
H
H
C
H
C
NaOOC
H
C
C
C
H
N
H
C
H
H3O+
(CH3)2N
N
N
C
C
H
•
CH
C
C
H
NaOOC
very large change in pH for very small
additions of titrant near the equivalence point
An indicator can therefore be used to
determine the endpoint of the titration if it
changes color within the same range as the
rapid change in pH
pKa of HInd ≈ pH at equivalence point
Tro: Chemistry: A Molecular Approach, 2/e
121
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Tro: Chemistry: A Molecular Approach, 2/e
Acid-Base Indicators
122
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Solubility Equilibria
• All ionic compounds dissolve in water to some
degree
however, many compounds have such low solubility
in water that we classify them as insoluble
• We can apply the concepts of equilibrium to
salts dissolving, and use the equilibrium
constant for the process to measure relative
solubilities in water
Tro: Chemistry: A Molecular Approach, 2/e
123
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Tro: Chemistry: A Molecular Approach, 2/e
124
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Tro: Chemistry: A Molecular Approach, 2/e
126
Copyright  2011 Pearson Education, Inc.
Solubility Product
• The equilibrium constant for the dissociation of a
•
•
•
•
solid salt into its aqueous ions is called the
solubility product, Ksp
For an ionic solid MnXm, the dissociation reaction is:
MnXm(s)  nMm+(aq) + mXn−(aq)
The solubility product would be
Ksp = [Mm+]n[Xn−]m
For example, the dissociation reaction for PbCl2 is
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
And its equilibrium constant is
Ksp = [Pb2+][Cl−]2
Tro: Chemistry: A Molecular Approach, 2/e
125
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Example 16.8: Calculate the molar solubility of
PbCl2 in pure water at 25 C
Molar Solubility
• Solubility is the amount of solute that will
dissolve in a given amount of solution
at a particular temperature
• The molar solubility is the number of moles of
solute that will dissolve in a liter of solution
the molarity of the dissolved solute in a saturated
solution
for the general reaction MnXm(s)  nMm+(aq) + mXn−(aq)
Tro: Chemistry: A Molecular Approach, 2/e
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127
Example 16.8: Calculate the molar solubility of
PbCl2 in pure water at 25 C
substitute into
the Ksp
expression
find the value of
Ksp from Table
16.2, plug into
the equation,
and solve for S
[Pb2+]
Ksp = [Pb2+][Cl−]2
[Pb2+]
Initial
[Cl−]
0
0
Change
+S
+2S
Equilibrium
S
2S
Tro: Chemistry: A Molecular Approach, 2/e
Copyright  2011 Pearson Education, Inc.
128
Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25 C is 1.05 x 10−2 M
Initial
[Cl−]
0
0
Change
+S
+2S
Equilibrium
S
2S
Copyright  2011 Pearson Education, Inc.
129
Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25 C is 1.05 x 10−2 M
create an ICE
table defining
the change in
terms of the
solubility of the
solid
create an ICE
table defining
the change in
terms of the
solubility of the
solid
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
Ksp = (S)(2S)2
Tro: Chemistry: A Molecular Approach, 2/e
write the
dissociation
reaction and Ksp
expression
write the
dissociation
reaction and Ksp
expression
PbBr2(s)  Pb2+(aq) + 2 Br−(aq)
Ksp = [Pb2+][Br−]2
initial
[Pb2+]
[Br−]
0
0
Tro: Chemistry: A Molecular Approach, 2/e
Copyright  2011 Pearson Education, Inc.
130
Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25 C is 1.05 x 10−2 M
substitute into
the Ksp
expression
plug into the
equation and
solve
Ksp = [Pb2+][Br−]2
Ksp = (1.05 x 10−2)(2.10 x 10−2)2
initial
[Pb2+]
[Br−]
0
0
change
+(1.05 x 10−2) +2(1.05 x 10−2)
change
+(1.05 x 10−2) +2(1.05 x 10−2)
equilibrium
(1.05 x 10−2)
equilibrium
(1.05 x 10−2)
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131
(2.10 x 10−2)
Copyright  2011 Pearson Education, Inc.
Tro: Chemistry: A Molecular Approach, 2/e
132
(2.10 x 10−2)
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The Effect of Common Ion on
Solubility
Ksp and Relative Solubility
• Molar solubility is related to Ksp
• But you cannot always compare solubilities of
compounds by comparing their Ksps
• To compare Ksps, the compounds must have
the same dissociation stoichiometry
• Addition of a soluble salt that contains one of
•
the ions of the “insoluble” salt, decreases the
solubility of the “insoluble” salt
For example, addition of NaCl to the solubility
equilibrium of solid PbCl2 decreases the
solubility of PbCl2
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
addition of Cl− shifts the
equilibrium to the left
Tro: Chemistry: A Molecular Approach, 2/e
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133
Example 16.10: Calculate the molar solubility
of CaF2 in 0.100 M NaF at 25 C
write the
dissociation
reaction and Ksp
expression
create an ICE
table defining
the change in
terms of the
solubility of the
solid
CaF2(s)  Ca2+(aq) + 2 F−(aq)
Ksp = [Ca2+][F−]2
[Ca2+]
initial
0
[F−]
0.100
change
+S
+2S
equilibrium
S
0.100 + 2S
Tro: Chemistry: A Molecular Approach, 2/e
Copyright  2011 Pearson Education, Inc.
134
Example 16.10: Calculate the molar solubility
of CaF2 in 0.100 M NaF at 25 C
substitute into
the Ksp
expression,
assume S is
small
find the value of
Ksp from Table
16.2, plug into
the equation,
and solve for S
Ksp = [Ca2+][F−]2
Ksp = (S)(0.100 + 2S)2
Ksp = (S)(0.100)2
initial
change
equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
135
Copyright  2011 Pearson Education, Inc.
Practice – Determine the concentration of Ag+
ions in seawater that has a [Cl−] of 0.55 M
Ksp of AgCl = 1.77 x 10−10
Tro: Chemistry: A Molecular Approach, 2/e
137
Copyright  2011 Pearson Education, Inc.
[F−]
0
0.100
+S
+2S
S
0.100 + 2S
Copyright  2011 Pearson Education, Inc.
Practice – Determine the concentration of Ag+
ions in seawater that has a [Cl−] of 0.55 M
write the
dissociation
reaction and Ksp
expression
create an ICE
table defining
the change in
terms of the
solubility of the
solid
Tro: Chemistry: A Molecular Approach, 2/e
136
[Ca2+]
AgCl(s)  Ag+(aq) + Cl−(aq)
Ksp = [Ag+][Cl−]
initial
[Ag+]
[Cl−]
0
0.55
change
+S
+S
equilibrium
S
0.55 + S
Tro: Chemistry: A Molecular Approach, 2/e
138
Copyright  2011 Pearson Education, Inc.
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Practice – Determine the concentration of Ag+
ions in seawater that has a [Cl−] of 0.55 M
The Effect of pH on Solubility
• For insoluble ionic hydroxides, the higher the pH,
substitute into
the Ksp
expression,
assume S is
small
the lower the solubility of the ionic hydroxide
Ksp = [Ag+][Cl−]
 and the lower the pH, the higher the solubility
 higher pH = increased [OH−]
Ksp = (S)(0.55 + S)
Ksp = (S)(0.55)
find the value of
Ksp from Table
16.2, plug into
the equation,
and solve for S
Initial
[Ag+]
[Cl−]
0
0.55
Change
+S
+S
Equilibrium
S
0.55 + S
Tro: Chemistry: A Molecular Approach, 2/e
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M(OH)n(s)  Mn+(aq) + nOH−(aq)
• For insoluble ionic compounds that contain anions
of weak acids, the lower the pH, the higher the
solubility
M2(CO3)n(s)  2 Mn+(aq) + nCO32−(aq)
H3O+(aq) + CO32− (aq)  HCO3− (aq) + H2O(l)
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Precipitation
• Precipitation will occur when the concentrations of
the ions exceed the solubility of the ionic
compound
• If we compare the reaction quotient, Q, for the
current solution concentrations to the value of Ksp,
we can determine if precipitation will occur
 Q = Ksp, the solution is saturated, no precipitation
 Q < Ksp, the solution is unsaturated, no precipitation
 Q > Ksp, the solution would be above saturation, the salt
above saturation will precipitate
• Some solutions with Q > Ksp will not precipitate
unless disturbed – these are called
supersaturated solutions
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Selective Precipitation
• A solution containing several different cations
•
can often be separated by addition of a reagent
that will form an insoluble salt with one of the
ions, but not the others
A successful reagent can precipitate with more
than one of the cations, as long as their Ksp
values are significantly different
a supersaturated solution will
precipitate if a seed crystal is
added
precipitation
occurs if Q > Ksp
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Example 16.12: Will a precipitate form when we mix
Pb(NO3)2(aq) with NaBr(aq) if the concentrations after
mixing are 0.0150 M and 0.0350 M respectively?
write the equation for
the reaction
determine the ion
concentrations of the
original salts
determine the Ksp for
any “insoluble”
product
Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq)
Pb(NO3)2 = 0.0150 M
Pb2+ = 0.0150 M,
NO3− = 2(0.0150 M)
write the dissociation
reaction for the
insoluble product
NaBr = 0.0350 M
Na+ = 0.0350 M,
Br− = 0.0350 M
Ksp of PbBr2 = 4.67 x 10–6
PbBr2(s)  Pb2+(aq) + 2 Br−(aq)
calculate Q, using the
ion concentrations
compare Q to Ksp. If
Q > Ksp, precipitation
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Q < Ksp, so no precipitation
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Practice – Will a precipitate form when we mix
Ca(NO3)2(aq) with NaOH(aq) if the concentrations after
mixing are both 0.0175 M?
Ksp of Ca(OH)2 = 4.68 x 10−6
Practice – Will a precipitate form when we mix
Ca(NO3)2(aq) with NaOH(aq) if the concentrations after
mixing are both 0.0175 M?
write the equation for
the reaction
determine the ion
concentrations of the
original salts
determine the Ksp for
any “insoluble”
product
Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)
Ca(NO3)2 = 0.0175 M
Ca2+ = 0.0175 M,
NO3− = 2(0.0175 M)
write the dissociation
reaction for the
insoluble product
NaOH = 0.0175 M
Na+ = 0.0175 M,
OH− = 0.0175 M
Ksp of Ca(OH)2 = 4.68 x 10–6
Ca(OH)2(s)  Ca2+(aq) + 2 OH−(aq)
calculate Q, using the
ion concentrations
compare Q to Ksp. If
Q > Ksp, precipitation
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Example 16.13: What is the minimum [OH−]
necessary to just begin to precipitate Mg2+ (with
[0.059]) from seawater?
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Q > Ksp, so precipitation
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Practice – What is the minimum concentration of
Ca(NO3)2(aq) that will precipitate Ca(OH)2 from
0.0175 M NaOH(aq)?
Ksp of Ca(OH)2 = 4.68 x 10−6
Mg(OH)2(s)  Mg2+(aq) + 2 OH−(aq)
precipitating may just occur when Q = Ksp
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Practice – What is the minimum concentration of
Ca(NO3)2(aq) that will precipitate Ca(OH)2 from
0.0175 M NaOH(aq)?
Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)
Ca(OH)2(s)  Ca2+(aq) + 2 OH−(aq)
precipitating may just occur when Q = Ksp
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Example 16.14: What is the [Mg2+] when Ca2+
(with [0.011]) just begins to precipitate from
seawater?
Ca(OH)2(s)  Ca2+(aq) + 2 OH−(aq)
precipitating may just occur when Q = Ksp
[Ca(NO3)2] = [Ca2+] = 0.0153 M
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Example 16.14: What is the [Mg2+] when Ca2+
(with [0.011]) just begins to precipitate from
seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10−6 M
Practice – A solution is made by mixing Pb(NO3)2(aq) with
AgNO3(aq) so both compounds have a concentration of
0.0010 M. NaCl(s) is added to precipitate out both AgCl(s)
and PbCl2(aq). What is the [Ag+] concentration when the
Pb2+ just begins to precipitate?
precipitating Ca2+ begins when [OH−] = 2.06 x 10−2 M
Mg(OH)2(s)  Mg2+(aq) + 2 OH−(aq)
when Ca2+ just
begins to
precipitate out, the
[Mg2+] has dropped
from 0.059 M to
4.8 x 10−10 M
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Practice – What is the [Ag+] concentration when
the Pb2+(0.0010 M) just begins to precipitate?
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
AgCl(s)  Ag+(aq) + Cl−(aq)
precipitating may just occur when Q = Ksp
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Practice – What is the [Ag+] concentration when
the Pb2+(0.0010 M) just begins to precipitate
precipitating Ag+ begins when [Cl−] = 1.77 x 10−7 M
precipitating Pb2+ begins when [Cl−] = 1.08 x 10−1 M
AgCl(s)  Ag+(aq) + Cl−(aq)
when Pb2+ just
begins to
precipitate out, the
[Ag+] has dropped
from 0.0010 M to
1.6 x 10−9 M
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Practice – What is the [Ag+] concentration when
the Pb2+(0.0010 M) just begins to precipitate?
Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
precipitating may just occur when Q = Ksp
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Qualitative Analysis
• An analytical scheme that utilizes selective
precipitation to identify the ions present in a
solution is called a qualitative analysis
scheme
wet chemistry
• A sample containing several ions is subjected
to the addition of several precipitating agents
• Addition of each reagent causes one of the
ions present to precipitate out
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Qualitative Analysis
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Group 2
Group 1
• Group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+,
• Group one cations are Ag+, Pb2+ and Hg22+
• All these cations form compounds with Cl− that
are insoluble in water
as long as the concentration is large enough
PbCl2 may be borderline
•
•
Pb2+, Sb3+, and Hg2+
All these cations form compounds with HS− and
S2− that are insoluble in water at low pH
Precipitated by the addition of H2S in HCl
molar solubility of PbCl2 = 1.43 x 10−2 M
• Precipitated by the addition of HCl
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Group 3
•
•
Cr3+,
precipitated as sulfides; as well as
Fe3+, and Al3+ precipitated as hydroxides
All these cations form compounds with S2− that
are insoluble in water at high pH
Precipitated by the addition of H2S in NaOH
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Group 4
• Group three cations are Fe2+, Co2+, Zn2+, Mn2+,
Ni2+
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• Group four cations are Mg2+, Ca2+, Ba2+
• All these cations form compounds with PO43−
that are insoluble in water at high pH
• Precipitated by the addition of (NH4)2HPO4
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Group 5
• Group five cations are Na+, K+, NH4+
• All these cations form compounds that are
•
soluble in water – they do not precipitate
Identified by the color of their flame
Complex Ion Formation
• Transition metals tend to be good Lewis acids
• They often bond to one or more H2O molecules to
form a hydrated ion
 H2O is the Lewis base, donating electron pairs to form
coordinate covalent bonds
Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq)
• Ions that form by combining a cation with several
anions or neutral molecules are called complex
ions
 e.g., Ag(H2O)2+
• The attached ions or molecules are called
ligands
 e.g., H2O
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Complex Ion Equilibria
• If a ligand is added to a solution that forms a
stronger bond than the current ligand, it will
replace the current ligand
Ag(H2O)2+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) + 2 H2O(l)
generally H2O is not included, because its complex
ion is always present in aqueous solution
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
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Formation Constants
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Formation Constant
• The reaction between an ion and ligands to form
•
a complex ion is called a complex ion
formation reaction
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
The equilibrium constant for the formation
reaction is called the formation constant, Kf
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Example 16.15: 200.0 mL of 1.5 x 10−3 M
Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3.
What is the [Cu2+] at equilibrium?
Write the
formation
reaction and Kf
expression.
Look up Kf value
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
determine the
concentration of
ions in the
diluted solutions
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Example 16.15: 200.0 mL of 1.5 x 10−3 M
Cu(NO3)2 is mixed with 250.0 mL of 0.20 M
NH3. What is the [Cu2+] at equilibrium?
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Create an ICE
table. Because
Kf is large,
assume all the
Cu2+ is
converted into
complex ion,
then the system
returns to
equilibrium.
initial
change
[Cu2+]
[NH3]
[Cu(NH3)22+]
6.7E-4
0.11
0
≈−6.7E-4 ≈−4(6.7E-4)
equilibrium
x
0.11
Example 16.15: 200.0 mL of 1.5 x 10-3 M
Cu(NO3)2 is mixed with 250.0 mL of 0.20 M
NH3. What is the [Cu2+] at equilibrium?
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
substitute in
and solve
for x
confirm the
“x is small”
approximation
≈+6.7E-4
initial
[NH3]
0.11
≈−6.7E-4 ≈−4(6.7E-4)
change
6.7E-4
[Cu2+]
6.7E-4
equilibrium
x
0.11
[Cu(NH3)22+]
0
≈+6.7E-4
6.7E-4
2.7 x 10−13 << 6.7 x 10−4, so the approximation is valid
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Practice – What is [HgI42−] when 125 mL of 0.0010 M
KI is reacted with 75 mL of 0.0010 M HgCl2?
4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq)
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Practice – What is [HgI42−] when 125 mL of 0.0010 M
KI is reacted with 75 mL of 0.0010 M HgCl2?
4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq)
Write the
formation
reaction and Kf
expression.
Look up Kf value
Hg2+(aq) + 4 I−(aq)  HgI42−(aq)
determine the
concentration of
ions in the
diluted solutions
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Practice – What is [HgI42−] when 125 mL of 0.0010 M
KI is reacted with 75 mL of 0.0010 M HgCl2?
4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq)
Hg2+(aq) + 4 I−(aq)  HgI42−(aq)
Create an ICE
table. Because
Kf is large,
assume all the
lim. rgt. is
converted into
complex ion,
then the system
returns to
equilibrium.
change
equilibrium
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[Hg2+]
[I−]
[HgI42−]
3.75E-4
6.25E-4
0
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Practice – What is [HgI42−] when 125 mL of 0.0010 M
KI is reacted with 75 mL of 0.0010 M HgCl2?
4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq)
Hg2+(aq) + 4 I−(aq)  HgI42−(aq)
substitute in
and solve for
x
confirm the
“x is small”
approximation
I− is the limiting reagent
initial
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initial
[Hg2+]
[I−]
[HgI42−]
3.75E-4
6.25E-4
0
≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)
change
equilibrium
2.19E-4
x
1.56E-4
≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)
2.19E-4
x
1.56E-4
2 x 10−8 << 1.6 x 10−4, so the approximation is valid
[HgI42−] = 1.6 x 10−4
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The Effect of Complex Ion Formation
on Solubility
• The solubility of an ionic compound that
contains a metal cation that forms a complex
ion increases in the presence of aqueous
ligands
AgCl(s)  Ag+(aq) + Cl−(aq)
Ksp = 1.77 x 10−10
+
+
Ag (aq) + 2 NH3(aq)  Ag(NH3)2 (aq) Kf = 1.7 x 107
• Adding NH3 to a solution in equilibrium with
AgCl(s) increases the solubility of Ag+
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Solubility of Amphoteric
Metal Hydroxides
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Al3+
• Many metal hydroxides are insoluble
• All metal hydroxides become more soluble in acidic
solution
 shifting the equilibrium to the right by removing
176
OH−
• Some metal hydroxides also become more soluble
in basic solution
 acting as a Lewis base forming a complex ion
• Substances that behave as both an acid and base
are said to be amphoteric
• Some cations that form amphoteric hydroxides
• Al3+ is hydrated in water to form an acidic
solution
Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+(aq) + H3O+(aq)
• Addition of OH− drives the equilibrium to the right
and continues to remove H from the molecules
Al(H2O)5(OH)2+(aq) + OH−(aq)  Al(H2O)4(OH)2+(aq) + H2O (l)
Al(H2O)4(OH)2+(aq) + OH−(aq)  Al(H2O)3(OH)3(s) + H2O (l)
include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+
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