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4/19/2016
1
Copyright 2011 Pearson Education, Inc.
Chapter 16
Aqueous Ionic
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Copyright 2011 Pearson Education, Inc.
ethylene glycol
(aka 1,2–ethandiol)
The Danger of Antifreeze• Each year, thousands of pets and wildlife
species die from consuming antifreeze
• Most brands of antifreeze contain ethylene
glycolsweet taste
initial effect drunkenness
• Metabolized in the liver to glycolic acidHOCH2COOH
2
glycolic acid
(aka a-hydroxyethanoic acid)
Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
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 3Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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−
4Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
Making an Acid Buffer
5Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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)
you can also think of the H3O+ combining with the OH−
to make H2O; the H3O+ is then replaced by the shifting
equilibrium6Tro: Chemistry: A Molecular Approach, 2/e
4/19/2016
2
Copyright 2011 Pearson Education, Inc.
H2O
HA
How Buffers Work
HA + H3O+
A−
Added
HO−
new
A−
A−
7Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
How Acid Buffers Work:
Addition of Acid
HA(aq) + H2O(l) A−(aq) + H3O
+(aq)
• The buffer solution also contains significant 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
8Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
H2O
How Buffers Work
HA + H3O+A−
A−
Added
H3O+
new
HA
HA
9Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Common Ion Effect
HA(aq) + H2O(l) A−(aq) + H3O
+(aq)
• Adding a salt containing the anion NaA, which
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
10Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
Common Ion Effect
11Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
12
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
HC2H3O2 + H2O C2H3O2 + H3O
+
[HA] [A−] [H3O+]
initial 0.100 0.100 ≈ 0
change
equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
4/19/2016
3
Copyright 2011 Pearson Education, Inc.
[HA] [A−] [H3O+]
initial 0.100 0.100 0
change
equilibrium
Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
13
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
substitute into the
equilibrium
constant
expression
+x+xx
0.100 x 0.100 + x x
HC2H3O2 + H2O C2H3O2 + H3O
+
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
14
determine the
value of Ka
because Ka is very
small, approximate
the [HA]eq = [HA]init
and [A−]eq = [A−]init ,
then solve for x
[HA] [A−] [H3O+]
initial 0.100 0.100 ≈ 0
change −x +x +x
equilibrium 0.100 0.100 x0.100 x 0.100 +x
Ka for HC2H3O2 = 1.8 x 10−5
Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
15
check if the
approximation
is valid by
seeing if x <
5% of
[HC2H3O2]init
Ka for HC2H3O2 = 1.8 x 10−5
the approximation is valid
x = 1.8 x 10−5
[HA] [A−] [H3O+]
initial 0.100 0.100 ≈ 0
change −x +x +x
equilibrium 0.100 0.100 x
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
16
substitute x
into the
equilibrium
concentration
definitions
and solve x = 1.8 x 10−5
[HA] [A−] [H3O+]
initial 0.100 0.100 ≈ 0
change −x +x +x
equilibrium 0.100 0.100 1.8E-50.100 + x x 0.100 x
Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
17
substitute [H3O+]
into the formula
for pH and solve
[HA] [A−] [H3O+]
initial 0.100 0.100 ≈ 0
change −x +x +x
equilibrium 0.100 0.100 1.8E−5
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Example 16.1: What is the pH of a buffer that
is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
18
check by
substituting the
equilibrium
concentrations
back into the
equilibrium
constant
expression and
comparing the
calculated Ka to the
given Ka
the values
match
[HA] [A−] [H3O+]
initial 0.100 0.100 ≈ 0
change −x +x +x
equilibrium 0.100 0.100 1.8E−5
Ka for HC2H3O2 = 1.8 x 10−5
Tro: Chemistry: A Molecular Approach, 2/e
4/19/2016
4
Copyright 2011 Pearson Education, Inc.
Practice − What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
19Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice − What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
20
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
HF + H2O F + H3O+
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change
equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
[HA] [A−] [H3O+]
initial 0.14 0.071 0
change
equilibrium
Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
21
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
substitute into
the equilibrium
constant
expression
+x+xx
0.14 x 0.071 + x x
HF + H2O F + H3O+
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
pKa for HF = 3.15Ka for HF = 7.0 x 10−4
Practice – What is the pH of a buffer that is
0.14 M and 0.071 M KF?
22
determine the
value of Ka
because Ka is
very small,
approximate the
[HA]eq = [HA]init
and [A−]eq = [A−]init
solve for x
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change −x +x +x
equilibrium 0.012 0.100 x0.14 x 0.071 +x
Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
23
check if the
approximation
is valid by
seeing if
x < 5% of
[HC2H3O2]init
Ka for HF = 7.0 x 10−4
the approximation is valid
x = 1.4 x 10−3
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change −x +x +x
equilibrium 0.14 0.071 x
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
24
substitute x into
the equilibrium
concentration
definitions and
solve
x = 1.4 x 10−3
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change −x +x +x
equilibrium 0.14 0.072 1.4E-30.071 + x x 0.14 x
Tro: Chemistry: A Molecular Approach, 2/e
4/19/2016
5
Copyright 2011 Pearson Education, Inc.
Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
25
substitute
[H3O+] into the
formula for pH
and solve
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change −x +x +x
equilibrium 0.14 0.072 1.4E−3
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
26
check by
substituting the
equilibrium
concentrations
back into the
equilibrium
constant
expression and
comparing the
calculated Ka to
the given Ka
the values are
close enough
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change −x +x +x
equilibrium 0.14 0.072 1.4E−3
Ka for HF = 7.0 x 10−4
Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
Henderson-Hasselbalch Equation
• Calculating the pH of a buffer solution can be
simplified by using an equation derived from
the Ka expression called the Henderson-
Hasselbalch 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
27Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Deriving the Henderson-Hasselbalch
Equation
28Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
Example 16.2: What is the pH of a buffer that
is 0.050 M HC7H5O2 and 0.150 M NaC7H5O2?
29
assume the [HA] and
[A−] equilibrium
concentrations are
the same as the
initial
substitute into the
Henderson-
Hasselbalch
Equation
check the “x is small”
approximation
HC7H5O2 + H2O C7H5O2 + H3O
+
Ka for HC7H5O2 = 6.5 x 10−5
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
30Tro: Chemistry: A Molecular Approach, 2/e
4/19/2016
6
Copyright 2011 Pearson Education, Inc.
Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
31
find the pKa from the
given Ka
assume the [HA]
and [A−] equilibrium
concentrations are
the same as the
initial
substitute into the
Henderson-
Hasselbalch
equation
check the “x is
small”
approximation
HF + H2O F + H3O+
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Do I Use the Full Equilibrium Analysis or
the Henderson-Hasselbalch Equation?• The Henderson-Hasselbalch equation is
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:
a) the initial concentrations of acid and salt are not very dilute
b) the Ka is fairly small
• For most problems, this means that the initial acid and salt concentrations should be over 100 to 1000x larger than the value of Ka
32Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
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−]
33Tro: Chemistry: A Molecular Approach, 2/e 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?
34
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 A− OH−
mols before 0.100 0.100 0
mols added ─ ─ 0.010
mols after
Tro: Chemistry: A Molecular Approach, 2/e
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?
35
fill in the table
– tracking the
changes in the
number of
moles for each
component
HC2H3O2 + OH− C2H3O2 + H2O
HA A− OH−
mols before 0.100 0.100 ≈ 0
mols added ─ ─ 0.010
mols after 0.090 0.110 ≈ 0
Tro: Chemistry: A Molecular Approach, 2/e 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?
36
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 it with A−.
construct a
stoichiometry table
for the reaction
enter the initial
number of moles for
each
HC2H3O2 + OH− C2H3O2 + H2O
HA A– OH−
mols before 0.100 0.100 0.010
mols change
mols end
new molarity
Tro: Chemistry: A Molecular Approach, 2/e
4/19/2016
7
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?
37
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
divide by the liters
of solution to find
the new molarities
HC2H3O2 + OH− C2H3O2 + H2O
HA A− OH−
mols before 0.100 0.100 0.010
mols change
mols end
new molarity
−0.010−0.010 +0.010
00.1100.090
0.090 0.110 0
Tro: Chemistry: A Molecular Approach, 2/e 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?
38
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
+
[HA] [A−] [H3O+]
initial 0.090 0.110 ≈ 0
change
equilibrium
Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
[HA] [A−] [H3O+]
initial 0.090 0.110 ≈ 0
change
equilibrium
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?
39
represent the
change in the
concentrations in
terms of x
sum the columns
to find the
equilibrium
concentrations in
terms of x
substitute into
the equilibrium
constant
expression
+x+xx
0.090 x 0.110 + x x
HC2H3O2 + H2O C2H3O2 + H3O
+
Tro: Chemistry: A Molecular Approach, 2/e 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?
40
determine the
value of Ka
because Ka is very
small, approximate
the [HA]eq = [HA]init
and [A−]eq = [A−]init
solve for x
[HA] [A−] [H3O+]
initial 0.100 0.100 ≈ 0
change −x +x +x
equilibrium 0.090 0.110 x0.090 x 0.110 +x
Ka for HC2H3O2 = 1.8 x 10−5
Tro: Chemistry: A Molecular Approach, 2/e
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?
41
check if the
approximation
is valid by
seeing if
x < 5% of
[HC2H3O2]init
Ka for HC2H3O2 = 1.8 x 10−5
the approximation is valid
x = 1.47 x 10−5
[HA] [A−] [H3O+]
initial 0.090 0.110 ≈ 0
change −x +x +x
equilibrium 0.090 0.110 x
Tro: Chemistry: A Molecular Approach, 2/e 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?
42
substitute x
into the
equilibrium
concentration
definitions
and solve x = 1.47 x 10−5
[HA] [A-] [H3O+]
initial 0.090 0.110 ≈ 0
change −x +x +x
equilibrium 0.090 0.110 1.5E-50.110 + x x 0.090 x
Tro: Chemistry: A Molecular Approach, 2/e
4/19/2016
8
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?
43
substitute
[H3O+] into
the formula
for pH and
solve
[HA] [A−] [H3O+]
initial 0.090 0.110 ≈ 0
change −x +x +x
equilibrium 0.090 0.110 1.5E−5
Tro: Chemistry: A Molecular Approach, 2/e 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?
44
check by
substituting the
equilibrium
concentrations
back into the
equilibrium
constant
expression and
comparing the
calculated Ka to
the given Ka
the values
match
[HA] [A−] [H3O+]
initial 0.090 0.110 ≈ 0
change −x +x +x
equilibrium 0.090 0.110 1.5E−5
Ka for HC2H3O2 = 1.8 x 10−5
Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
or, by using the
Henderson-Hasselbalch
Equation
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?
46
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−]
fll in the ICE table
HC2H3O2 + H2O C2H3O2 + H3O
+
[HA] [A−] [H3O+]
initial 0.090 0.110 ≈ 0
change
equilibrium
+x+xx
0.090 x 0.110 + x x
Tro: Chemistry: A Molecular Approach, 2/e
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?
47
find the pKa from
the given Ka
assume the [HA]
and [A−] equilibrium
concentrations are
the same as the
initial
HC2H3O2 + H2O C2H3O2 + H3O
+
Ka for HC2H3O2 = 1.8 x 10−5
[HA] [A−] [H3O+]
initial 0.090 0.110 ≈ 0
change −x +x +x
equilibrium 0.090 0.110 x
Tro: Chemistry: A Molecular Approach, 2/e 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?
48
substitute into the
Henderson-
Hasselbalch
equation
check the “x is
small”
approximation
HC2H3O2 + H2O C2H3O2 + H3O
+
pKa for HC2H3O2 = 4.745
Tro: Chemistry: A Molecular Approach, 2/e
4/19/2016
9
Copyright 2011 Pearson Education, Inc.
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
49
HC2H3O2 + H2O C2H3O2 + H3O
+
pKa for HC2H3O2 = 4.745
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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)
50Tro: Chemistry: A Molecular Approach, 2/e
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?
51
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
F− H3O+ HF
mols before 0.071 0 0.140
mols added – 0.020 –
mols after
Tro: Chemistry: A Molecular Approach, 2/e 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?
52
fill in the table
– tracking the
changes in the
number of
moles for each
component
F− + H3O+ HF + H2O
F− H3O+ HF
mols before 0.071 0 0.140
mols added – 0.020 –
mols after 0.051 0 0.160
Tro: Chemistry: A Molecular Approach, 2/e
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?
53
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
enter the initial
number of moles for
each
F− + H3O+ HF + H2O
F− H3O+ HF
mols before 0.071 0.020 0.140
mols change
mols after
new molarity
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F− H3O+ HF
mols before 0.071 0.020 0.140
mols change
mols after
new molarity
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?
54
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
divide by the liters
of solution to find
the new molarities
0.16000.051
F− + H3O+ HF + H2O
+0.020−0.020 −0.020
0.16000.051
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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?
55
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
substitute into the
Henderson-
Hasselbalch
equation
HF + H2O F + H3O+
[HF] [F−] [H3O+]
initial 0.160 0.051 ≈ 0
change −x +x +x
equilibrium 0.160 0.051 x
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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)
56Tro: Chemistry: A Molecular Approach, 2/e
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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−
• 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+
this does not affect the concentrations, just the way we are looking
at the reaction
Henderson-Hasselbalch Equation for
Basic Buffers
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Relationship between pKa and pKb
• 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?
59
find the pKa of the
conjugate acid (NH4+)
from the given Kb
assume the [B] and
[HB+] equilibrium
concentrations are the
same as the initial
substitute into the
Henderson-
Hasselbalch equation
check the “x is small”
approximation
NH3 + H2O NH4+ + OH−
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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
Henderson-Hasselbalch Equation for
Basic Buffers
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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?
61
find the pKb if given Kb
assume the [B] and
[HB+] equilibrium
concentrations are the
same as the initial
substitute into the
Henderson-Hasselbalch
equation base form,
find pOH
check the “x is small”
approximation
calculate pH from pOH
NH3 + H2O NH4+ + OH−
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Buffering Effectiveness
• A good buffer should be able to neutralize moderate amounts of added acid or base
• 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 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 factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base
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HA A− OH−
mols before 0.18 0.020 0
mols added ─ ─ 0.010
mols after 0.17 0.030 ≈ 0
Effect of Relative Amounts of Acid
and Conjugate Base
Buffer 1
0.100 mol HA & 0.100 mol A−
Initial pH = 5.00
Buffer 2
0.18 mol HA & 0.020 mol A−
Initial pH = 4.05pKa (HA) = 5.00
after adding 0.010 mol NaOH
pH = 5.09
HA + OH− A + H2O
HA A− OH−
mols before 0.100 0.100 0
mols added ─ ─ 0.010
mols after 0.090 0.110 ≈ 0
after adding 0.010 mol NaOH
pH = 4.25
A buffer is most effective with equal
concentrations of acid and base
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HA A− OH−
mols before 0.50 0.500 0
mols added ─ ─ 0.010
mols after 0.49 0.51 ≈ 0
HA A− OH−
mols before 0.050 0.050 0
mols added ─ ─ 0.010
mols after 0.040 0.060 ≈ 0
Effect of Absolute Concentrations of Acid
and Conjugate Base
Buffer 1
0.50 mol HA & 0.50 mol A−
Initial pH = 5.00
Buffer 2
0.050 mol HA & 0.050 mol A−
Initial pH = 5.00pKa (HA) = 5.00
after adding 0.010 mol NaOH
pH = 5.02
HA + OH− A + H2O
after adding 0.010 mol NaOH
pH = 5.18
A buffer is most effective when the
concentrations of acid and base are largest
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Buffering Capacity
a concentrated
buffer can
neutralize
more added
acid or base
than a dilute
buffer
65Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Effectiveness of Buffers
• A buffer will be most effective when the
[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
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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Formic acid, HCHO2 pKa = 3.74Formic acid, HCHO2 pKa = 3.74
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
Hypochlorous acid, HClOpKa = 7.54
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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Example 16.5b: What ratio of NaCHO2 : HCHO2
would be required to make a buffer with pH 4.25?
69
Formic acid, HCHO2, pKa = 3.74
To make a buffer with
pH 4.25, you would use
3.24 times as much
NaCHO2 as HCHO2
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – What ratio of NaC7H5O2 : HC7H5O2
would be required to make a buffer with pH 3.75?
Benzoic acid, HC7H5O2, pKa = 4.19
70Tro: Chemistry: A Molecular Approach, 2/e
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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
71Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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• In an acid-base titration, a solution of unknown
concentration (titrant) is slowly added to a solution of known concentration from a buretteuntil the reaction is completewhen the reaction is complete we have reached the
endpoint of the titration
• An indicator may be added to determine the endpointan 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
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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 solutionequivalence 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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Titration Curve:
Unknown Strong Base Added
to Strong Acid
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Before Equivalence
(excess acid)
After Equivalence
(excess base)
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
Equivalence Point
equal moles of
HCl and NaOH
pH = 7.00
Because the
solutions are equal
concentration, and
1:1 stoichiometry, the
equivalence point is
at equal volumes
77Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
HCl NaCl NaOH
mols before 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 0
molarity, new 0.0667 0.017 0
HCl NaCl NaOH
mols before 2.50E-3 0 5.0E-4
mols change
mols end
molarity, new
HCl NaCl NaOH
mols before 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 0
molarity, new
5.0 x 10−4 mole NaOH added
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
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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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HCl NaCl NaOH
mols before 2.50E-3 0 2.5E-3
mols change
mols end
molarity, new
HCl NaCl NaOH
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
HCl NaCl NaOH
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
2.5 x 10−3 mole NaOH added
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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HCl NaCl 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 0.0091
HCl NaCl NaOH
mols before 2.50E-3 0 3.0E-3
mols change
mols end
molarity, new
HCl NaCl 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
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
• After equivalence point added 30.0 mL NaOH3.0 x 10−3 mole NaOH added
L of added NaOH0.100 mol NaOH
1 L
moles added NaOH
Copyright 2011 Pearson Education, Inc.
added 5.0 mL NaOH
0.00200 mol HCl
pH = 1.18
added 10.0 mL NaOH
0.00150 mol HCl
pH = 1.37
added 25.0 mL NaOH
equivalence point
pH = 7.00
added 30.0 mL NaOH
0.00050 mol NaOH
pH = 11.96
added 40.0 mL NaOH
0.00150 mol NaOH
pH = 12.36
added 50.0 mL NaOH
0.00250 mol NaOH
pH = 12.52
added 35.0 mL NaOH
0.00100 mol NaOH
pH = 12.22
added 15.0 mL NaOH
0.00100 mol HCl
pH = 1.60
added 20.0 mL NaOH
0.00050 mol HCl
pH = 1.95
Adding 0.100 M NaOH to 0.100 M HCl
82
25.0 mL 0.100 M HCl
0.00250 mol HCl
pH = 1.00
Tro: Chemistry: A Molecular Approach, 2/e
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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
83Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
HNO3 NaNO3 NaOH
mols before 1.25E-2 0 1.5E-3
mols change −1.5E-3 +1.5E-3 −1.5E-3
mols end 1.1E-3 1.5E-3 0
molarity, new
HNO3 NaNO3 NaOH
mols before 1.25E-2 0 1.5E-3
mols change −1.5E-3 +1.5E-3 −1.5E-3
mols end 1.1E-3 1.5E-3 0
molarity, new 0.018 0.025 0
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
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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
85Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
• 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
• At equivalence point: moles of NaOH = 1.25 x 10−2
86Tro: Chemistry: A Molecular Approach, 2/e
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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
87Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
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
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
• After equivalence point added 100.0 mL NaOH
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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
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
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
• After equivalence point added 100.0 mL NaOH
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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
the basicity from the conjugate base anion is negligible91Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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 0.100 0.000 ≈ 0
change −x +x +x
equilibrium 0.100 − x x x
Ka = 1.8 x 10−4
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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 equivalenceadded 5.0 mL NaOH
HA A− OH−
mols before 2.50E-3 0 0
mols added – – 5.0E-4
mols after 2.00E-3 5.0E-4 ≈ 0
93Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
HA A− OH−
mols before 2.50E-3 0 0
mols added – – 2.50E-3
mols after 0 2.50E-3 ≈ 0
[HCHO2] [CHO2−] [OH−]
initial 0 0.0500 ≈ 0
change +x −x +x
equilibrium x 5.00E-2-x x
Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH
added 25.0 mL NaOHCHO2
−(aq) + H2O(l) HCHO2(aq) + OH−
(aq)
Kb = 5.6 x 10−11
[OH−] = 1.7 x 10−6 M
• 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
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HA A− 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
HA A− 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 0.0091
HA A− NaOH
mols before 2.50E-3 0 3.0E-3
mols change
mols end
molarity, new
Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH
95
added 30.0 mL NaOH
3.0 x 10−3 mole NaOH added
• 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
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added 35.0 mL NaOH
0.00100 mol NaOH xs
pH = 12.22
initial HCHO2 solution
0.00250 mol HCHO2
pH = 2.37
added 5.0 mL NaOH
0.00200 mol HCHO2
pH = 3.14
added 10.0 mL NaOH
0.00150 mol HCHO2
pH = 3.56
added 25.0 mL NaOH
equivalence point
0.00250 mol CHO2−
[CHO2−]init = 0.0500 M
[OH−]eq = 1.7 x 10−6
pH = 8.23
added 30.0 mL NaOH
0.00050 mol NaOH xs
pH = 11.96
added 20.0 mL NaOH
0.00050 mol HCHO2
pH = 4.34
added 15.0 mL NaOH
0.00100 mol HCHO2
pH = 3.92
added 12.5 mL NaOH
0.00125 mol HCHO2
pH = 3.74 = pKa
half-neutralization
Adding NaOH to HCHO2
added 40.0 mL NaOH
0.00150 mol NaOH xs
pH = 12.36
added 50.0 mL NaOH
0.00250 mol NaOH xs
pH = 12.52
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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
• Before the equivalence point, the solution becomes a buffer
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
97Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Titrating Weak Acid with a Strong Base
• At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established
mol A− = original mole HA
calculate the volume of added base as you did in Example 4.8
[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
98Tro: Chemistry: A Molecular Approach, 2/e
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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.
99
write an equation for
the reaction for B
with HA
use stoichiometry to
determine the volume
of added B
HNO2 + KOH NO2 + H2O
Tro: Chemistry: A Molecular Approach, 2/e 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.
100
write an
equation for the
reaction for B
with HA
determine the
moles of HAbefore
& moles of
added B
make a
stoichiometry
table and
determine the
moles of HA in
excess and
moles A made
HNO2 + KOH NO2 + H2O
HNO2 NO2− OH−
mols before 0.00400 0 ≈ 0
mols added – – 0.00100
mols after ≈ 00.00300 0.00100
Tro: Chemistry: A Molecular Approach, 2/e
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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 after
adding 5.00 mL KOH.
101
write an
equation for the
reaction of HA
with H2O
determine Ka
and pKa for HA
use the
Henderson-
Hasselbalch
equation to
determine the
pH
HNO2 + H2O NO2 + H3O
+
HNO2 NO2− OH−
mols before 0.00400 0 ≈ 0
mols added – – 0.00100
mols after 0.00300 0.00100 ≈ 0
Table 15.5 Ka = 4.6 x 10−4
Tro: Chemistry: A Molecular Approach, 2/e 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.
102
write an
equation for the
reaction for B
with HA
determine the
moles of HAbefore
& moles of added
B
make a
stoichiometry
table and
determine the
moles of HA in
excess and
moles A made
HNO2 + KOH NO2 + H2O
HNO2 NO2− OH−
mols before 0.00400 0 ≈ 0
mols added – – 0.00200
mols after ≈ 00.00200 0.00200
at half-equivalence, moles KOH = ½ mole HNO2
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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.
103
write an equation
for the reaction of
HA with H2O
determine Ka and
pKa for HA
use the
Henderson-
Hasselbalch
equation to
determine the pH
HNO2 + H2O NO2 + H3O
+
HNO2 NO2− OH−
mols before 0.00400 0 ≈ 0
mols added – – 0.00200
mols after 0.00200 0.00200 ≈ 0
Table 15.5 Ka = 4.6 x 10-4
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Titration Curve of a Weak Base with a
Strong Acid
104Tro: Chemistry: A Molecular Approach, 2/e
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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 initial pH of the
NH3 solution.
105Tro: 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)
106
• NH3(aq) + HCl(aq) NH4Cl(aq)
• Initial: NH3(aq) + H2O(l) NH4+
(aq) + OH−(aq)
[HCl] [NH4+] [NH3]
initial 0 0 0.10
change +x +x −x
equilibrium x x 0.10−x
pKb = 4.75
Kb = 10−4.75 = 1.8 x 10−5
Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Titration of 25.0 mL of 0.10 M NH3 with
0.10 M HCl. Calculate the initial pH of the NH3(aq)
107
• NH3(aq) + HCl(aq) NH4Cl(aq)
• Initial: NH3(aq) + H2O(l) NH4+
(aq) + OH−(aq)
[HCl] [NH4+] [NH3]
initial 0 0 0.10
change +x +x −x
equilibrium x x 0.10−x
pKb = 4.75
Kb = 10−4.75 = 1.8 x 10−5
Tro: Chemistry: A Molecular Approach, 2/e 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.
108Tro: Chemistry: A Molecular Approach, 2/e
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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.
109
• 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
NH3 NH4Cl HCl
mols before 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 0
molarity, new 0.0667 0.017 0
NH4+
(aq) + H2O(l) NH4+
(aq) + H2O(l) pKb = 4.75
pKa = 14.00 − 4.75 = 9.25
Tro: Chemistry: A Molecular Approach, 2/e 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(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
NH3 NH4Cl HCl
mols before 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 0
molarity, new 0.0667 0.017 0
110Tro: Chemistry: A Molecular Approach, 2/e
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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 at equivalence.
111Tro: Chemistry: A Molecular Approach, 2/e 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.
112
• 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
added 25.0 mL HClNH3 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
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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 at equivalence.
113
NH3(aq) + HCl(aq) NH4Cl(aq)
at equivalence [NH4Cl] = 0.050 M
[NH3] [NH4+] [H3O
+]
initial 0 0.050 ≈ 0
change +x −x +x
equilibrium x 0.050−x x
NH4+
(aq) + H2O(l) NH3(aq) + H3O+
(aq)
Tro: Chemistry: A Molecular Approach, 2/e 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.
114Tro: Chemistry: A Molecular Approach, 2/e
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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.
115
• 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
when you mix a strong acid, HCl,
with a weak acid, NH4+, you only
need to consider the strong acidTro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
116Tro: Chemistry: A Molecular Approach, 2/e
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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
117Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Monitoring pH During a Titration
118Tro: Chemistry: A Molecular Approach, 2/e
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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 HInd119Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Phenolphthalein
120Tro: Chemistry: A Molecular Approach, 2/e
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Methyl Red
C
C CH
CH
CH
CH
C
CH
CH
C
CH
CH
(CH3)2N N N N
H
NaOOC
C
C CH
CH
CH
CH
C
CH
CH
C
CH
CH
(CH3)2N N N N
NaOOC
H3O+ OH-
121Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Monitoring a Titration with an
Indicator
• For most titrations, the titration curve shows a
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
122Tro: Chemistry: A Molecular Approach, 2/e
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Acid-Base Indicators
123Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
124Tro: Chemistry: A Molecular Approach, 2/e
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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
125Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.126Tro: Chemistry: A Molecular Approach, 2/e
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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)
127Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Example 16.8: Calculate the molar solubility of
PbCl2 in pure water at 25 C
128
write the
dissociation
reaction and Ksp
expression
create an ICE
table defining
the change in
terms of the
solubility of the
solid
[Pb2+] [Cl−]
Initial 0 0
Change +S +2S
Equilibrium S 2S
PbCl2(s) Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
Tro: Chemistry: A Molecular Approach, 2/e
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Example 16.8: Calculate the molar solubility of
PbCl2 in pure water at 25 C
129
substitute into
the Ksp
expression
find the value of
Ksp from Table
16.2, plug into
the equation,
and solve for S
[Pb2+] [Cl−]
Initial 0 0
Change +S +2S
Equilibrium S 2S
Ksp = [Pb2+][Cl−]2
Ksp = (S)(2S)2
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25 C is 1.05 x 10−2 M
130Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25 C is 1.05 x 10−2 M
131
write the
dissociation
reaction and Ksp
expression
create an ICE
table defining
the change in
terms of the
solubility of the
solid
[Pb2+] [Br−]
initial 0 0
change +(1.05 x 10−2) +2(1.05 x 10−2)
equilibrium (1.05 x 10−2) (2.10 x 10−2)
PbBr2(s) Pb2+(aq) + 2 Br−(aq)
Ksp = [Pb2+][Br−]2
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – Determine the Ksp of PbBr2 if its molar
solubility in water at 25 C is 1.05 x 10−2 M
132
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
[Pb2+] [Br−]
initial 0 0
change +(1.05 x 10−2) +2(1.05 x 10−2)
equilibrium (1.05 x 10−2) (2.10 x 10−2)
Tro: Chemistry: A Molecular Approach, 2/e
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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
133Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
The Effect of Common Ion on
Solubility• 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
134Tro: Chemistry: A Molecular Approach, 2/e
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Example 16.10: Calculate the molar solubility
of CaF2 in 0.100 M NaF at 25 C
135
write the
dissociation
reaction and Ksp
expression
create an ICE
table defining
the change in
terms of the
solubility of the
solid
[Ca2+] [F−]
initial 0 0.100
change +S +2S
equilibrium S 0.100 + 2S
CaF2(s) Ca2+(aq) + 2 F−(aq)
Ksp = [Ca2+][F−]2
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Example 16.10: Calculate the molar solubility
of CaF2 in 0.100 M NaF at 25 C
136
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
[Ca2+] [F−]
initial 0 0.100
change +S +2S
equilibrium S 0.100 + 2S
Ksp = [Ca2+][F−]2
Ksp = (S)(0.100 + 2S)2
Ksp = (S)(0.100)2
Tro: Chemistry: A Molecular Approach, 2/e
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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
137Tro: Chemistry: A Molecular Approach, 2/e 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
[Ag+] [Cl−]
initial 0 0.55
change +S +S
equilibrium S 0.55 + S
AgCl(s) Ag+(aq) + Cl−(aq)
Ksp = [Ag+][Cl−]
138Tro: Chemistry: A Molecular Approach, 2/e
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Practice – Determine the concentration of Ag+
ions in seawater that has a [Cl−] of 0.55 M
139
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
[Ag+] [Cl−]
Initial 0 0.55
Change +S +S
Equilibrium S 0.55 + S
Ksp = [Ag+][Cl−]
Ksp = (S)(0.55 + S)
Ksp = (S)(0.55)
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
The Effect of pH on Solubility
• For insoluble ionic hydroxides, the higher the pH,
the lower the solubility of the ionic hydroxide
and the lower the pH, the higher the solubility
higher pH = increased [OH−]
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) + CO3
2− (aq) HCO3− (aq) + H2O(l)
140Tro: Chemistry: A Molecular Approach, 2/e
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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 occurQ = 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
141Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
precipitation
occurs if Q > Ksp
a supersaturated solution will
precipitate if a seed crystal is
added
142Tro: Chemistry: A Molecular Approach, 2/e
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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
143Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
write the dissociation
reaction for the
insoluble product
calculate Q, using the
ion concentrations
compare Q to Ksp. If
Q > Ksp, precipitation
Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq)
Ksp of PbBr2 = 4.67 x 10–6
PbBr2(s) Pb2+(aq) + 2 Br−
(aq)
Pb(NO3)2 = 0.0150 M
Pb2+ = 0.0150 M,
NO3− = 2(0.0150 M)
NaBr = 0.0350 M
Na+ = 0.0350 M,
Br− = 0.0350 M
Q < Ksp, so no precipitation144Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
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
145Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
write the dissociation
reaction for the
insoluble product
calculate Q, using the
ion concentrations
compare Q to Ksp. If
Q > Ksp, precipitation
Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)
Ksp of Ca(OH)2 = 4.68 x 10–6
Ca(OH)2(s) Ca2+(aq) + 2 OH−
(aq)
Ca(NO3)2 = 0.0175 M
Ca2+ = 0.0175 M,
NO3− = 2(0.0175 M)
NaOH = 0.0175 M
Na+ = 0.0175 M,
OH− = 0.0175 M
Q > Ksp, so precipitation
146Tro: Chemistry: A Molecular Approach, 2/e
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Example 16.13: What is the minimum [OH−]
necessary to just begin to precipitate Mg2+ (with
[0.059]) from seawater?
precipitating may just occur when Q = Ksp
Mg(OH)2(s) Mg2+(aq) + 2 OH−
(aq)
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
148Tro: Chemistry: A Molecular Approach, 2/e
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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)?
precipitating may just occur when Q = Ksp
[Ca(NO3)2] = [Ca2+] = 0.0153 M
Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)
Ca(OH)2(s) Ca2+(aq) + 2 OH−
(aq)
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.150
Example 16.14: What is the [Mg2+] when Ca2+
(with [0.011]) just begins to precipitate from
seawater?
precipitating may just occur when Q = Ksp
Ca(OH)2(s) Ca2+(aq) + 2 OH−
(aq)
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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
precipitating Ca2+ begins when [OH−] = 2.06 x 10−2 M
when Ca2+ just
begins to
precipitate out, the
[Mg2+] has dropped
from 0.059 M to
4.8 x 10−10 M
Mg(OH)2(s) Mg2+(aq) + 2 OH−
(aq)
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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?
152Tro: Chemistry: A Molecular Approach, 2/e
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Practice – What is the [Ag+] concentration when
the Pb2+(0.0010 M) just begins to precipitate?
precipitating may just occur when Q = Ksp
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
AgCl(s) Ag+(aq) + Cl−(aq)
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.154
Practice – What is the [Ag+] concentration when
the Pb2+(0.0010 M) just begins to precipitate?
precipitating may just occur when Q = Ksp
Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)
PbCl2(s) Pb2+(aq) + 2 Cl−(aq)
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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
when Pb2+ just
begins to
precipitate out, the
[Ag+] has dropped
from 0.0010 M to
1.6 x 10−9 M
AgCl(s) Ag+(aq) + Cl−(aq)
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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 1
• 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
molar solubility of PbCl2 = 1.43 x 10−2 M
• Precipitated by the addition of HCl
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Group 2
• Group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+,
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
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Group 3
• Group three cations are Fe2+, Co2+, Zn2+, Mn2+,
Ni2+ precipitated as sulfides; as well as Cr3+,
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 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
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Complex Ion Formation
• Transition metals tend to be good Lewis acids
• They often bond to one or more H2O molecules to form a hydrated ionH2O 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 ionse.g., Ag(H2O)2
+
• The attached ions or molecules are called ligandse.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 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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Formation Constants
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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
determine the
concentration of
ions in the
diluted solutions
Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)
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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?
Create an ICE
table. Because
Kf is large,
assume all the
Cu2+ is
converted into
complex ion,
then the system
returns to
equilibrium.
[Cu2+] [NH3] [Cu(NH3)22+]
initial 6.7E-4 0.11 0
change ≈−6.7E-4 ≈−4(6.7E-4) ≈+6.7E-4
equilibrium x 0.11 6.7E-4
Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)
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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)substitute in
and solve
for x
confirm the
“x is small”
approxima-
tion
[Cu2+] [NH3] [Cu(NH3)22+]
initial 6.7E-4 0.11 0
change ≈−6.7E-4 ≈−4(6.7E-4) ≈+6.7E-4
equilibrium x 0.11 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)
171Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
determine the
concentration of
ions in the
diluted solutions
Hg2+(aq) + 4 I−(aq) HgI42−(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)
Create an ICE
table. Because
Kf is large,
assume all the
lim. rgt. is
converted into
complex ion,
then the system
returns to
equilibrium.
[Hg2+] [I−] [HgI42−]
initial 3.75E-4 6.25E-4 0
change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)
equilibrium 2.19E-4 x 1.56E-4
Hg2+(aq) + 4 I−(aq) HgI42−(aq)
I− is the limiting reagent
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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)
substitute in
and solve for
x
confirm the
“x is small”
approximation
2 x 10−8 << 1.6 x 10−4, so the approximation is valid
Hg2+(aq) + 4 I−(aq) HgI42−(aq)
[HgI42−] = 1.6 x 10−4
[Hg2+] [I−] [HgI42−]
initial 3.75E-4 6.25E-4 0
change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)
equilibrium 2.19E-4 x 1.56E-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• Many metal hydroxides are insoluble
• All metal hydroxides become more soluble in acidic solution shifting the equilibrium to the right by removing OH−
• Some metal hydroxides also become more soluble in basic solutionacting 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 include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+
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Al3+
• 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)
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