Chapter
16
Applications of Aqueous
Equilibria
Chemistry 4th Edition
McMurry/Fay
Dr. Paul Charlesworth
Michigan Technological University
The Common-Ion Effect
01
•
Common Ion: Two dissolved solutes that contain
the same ion (cation or anion).
•
The presence of a common ion suppresses the
ionization of a weak acid or a weak base.
•
Common-Ion Effect: is the shift in equilibrium
caused by the addition of a compound having an
ion in common with the dissolved substance.
Prentice Hall ©2004
Chapter 16
Slide 2
The Common-Ion Effect
•
02
To determine the pH, we apply I.C.E. and then the
Henderson–Hasselbalch equation.

Conjugate Base
pH  pKa  log
Acid 
pKa   log Ka
•
When the concentration of HA and salt are high
(≥0.1 M) we can neglect the ionization of acid and
hydrolysis of salt.
Prentice Hall ©2004
Chapter 16
Slide 3
The Common-Ion Effect
03
•
Calculate the pH of a 0.20 M CH3COOH solution
with no salt added.
•
Calculate the pH of a solution containing 0.20 M
CH3COOH and 0.30 M CH3COONa.
•
What is the pH of a solution containing 0.30 M
HCOOH, before and after adding 0.52 M HCOOK?
Prentice Hall ©2004
Chapter 16
Slide 4
Buffer Solutions
01
•
A Buffer Solution: is a solution of (1) a weak acid
or a weak base and (2) its salt; both components
must be present.
•
A buffer solution has the ability to resist changes in
pH upon the addition of small amounts of either
acid or base.
•
Buffers are very important to biological systems.
Prentice Hall ©2004
Chapter 16
Slide 5
Buffer Solutions
Prentice Hall ©2004
Chapter 16
02
Slide 6
Buffer Solutions
03
•
Buffer solutions must contain relatively high acid
and base component concentrations, the buffer
capacity.
•
Acid and base component concentrations must not
react together.
•
The simplest buffer is prepared from equal
concentrations of acid and conjugate base.
Prentice Hall ©2004
Chapter 16
Slide 7
Buffer Solutions
•
04
Calculate the pH of a buffer system containing 1.0 M
CH3COOH and 1.0 M CH3COONa. What is the pH of the
system after the addition of 0.10 mole of gaseous HCl to
1.0 L of solution?
•
Calculate the pH of 0.30 M NH3/0.36 NH4Cl buffer system.
What is the pH after the addition of 20.0 mL of 0.050 M
NaOH to 80.0 mL of the buffer solution?
Prentice Hall ©2004
Chapter 16
Slide 8
Buffer Solutions
•
•
05
Buffer Preparation: Use the Henderson–
Hasselbalch equation in reverse.
1.
Choose weak acid with pKa close to required pH.
2.
Substitute into Henderson–Hasselbalch equation.
3.
Solve for the ratio of [conjugate base]/[acid].
This will give the mole ratio of conjugate base to
acid. The acid should always be 1.0.
Prentice Hall ©2004
Chapter 16
Slide 9
Buffer Solutions
•
06
Describe how you would prepare a “phosphate
buffer” with a pH of about 7.40.
•
How would you prepare a liter of “carbonate buffer”
at a pH of 10.10? You are provided with carbonic
acid (H2CO3), sodium hydrogen carbonate
(NaHCO3), and sodium carbonate (Na2CO3).
Prentice Hall ©2004
Chapter 16
Slide 10
Acid–Base Titrations
01
•
Titration: a procedure for determining the
concentration of a solution using another solution of
known concentration.
•
Titrations involving only strong acids or bases are
straightforward.
•
Titrations involving weak acids or bases are
complicated by hydrolysis of salt formed.
Prentice Hall ©2004
Chapter 16
Slide 11
Acid–Base Titrations
•
Strong Acid–Strong Base:
•
The equivalence point
02
is the point at which
equimolar amounts of
acid and base have
reacted.
Prentice Hall ©2004
Chapter 16
Slide 12
Acid–Base Titrations
•
03
The pH of a 25 mL 0.10 M HCl sample can be
determined after the addition of:
0. No addition of 0.10 M NaOH.
1. 10.0 mL (total) of 0.10 M NaOH.
2. 25.0 mL (total) of 0.10 M NaOH.
3. 35.0 mL (total) of 0.10 M NaOH.
Prentice Hall ©2004
Chapter 16
Slide 13
Acid–Base Titrations
04
Weak Acid–Strong Base:
• The conjugate base
hydrolyzes to form
weak acid and OH–.
• At equivalence point
only the conjugate base
is present.
• pH at equivalence point
will always be >7.
•
Prentice Hall ©2004
Chapter 16
Slide 14
Acid–Base Titrations
Prentice Hall ©2004
Chapter 16
05
Slide 15
Acid–Base Titrations
•
06
The pH of a 25 mL 0.10 M CH3COOH sample can
be determined after the addition of:
0. No addition of 0.10 M NaOH.
1. 10.0 mL (total) of 0.10 M NaOH.
2. 25.0 mL (total) of 0.10 M NaOH.
3. 35.0 mL (total) of 0.10 M NaOH.
Prentice Hall ©2004
Chapter 16
Slide 16
Acid–Base Titrations
•
07
Exactly 100 mL of 0.10 M nitrous acid are titrated with
a 0.10 M NaOH solution. Calculate the pH for:
1. The initial solution.
2. The point at which 80 mL of base have been added.
3. The equivalence point.
4. The point at which 105 mL of base have been added.
Prentice Hall ©2004
Chapter 16
Slide 17
Acid–Base Titrations
09
Strong Acid–Weak Base:
• The (conjugate) acid
hydrolyzes to form
weak base and H3O+.
• At equivalence point
only the (conjugate)
acid is present.
• pH at equivalence point
will always be <7.
•
Prentice Hall ©2004
Chapter 16
Slide 18
Acid–Base Titrations
•
10
Calculate the pH at the equivalence point when
25 mL of 0.10 M NH3 is titrated with a 0.10 M HCl
solution.
•
Calculate the pH at the equivalence point in the
titration of 50 mL of 0.10 M methylamine with a
0.20 M HCl solution.
Prentice Hall ©2004
Chapter 16
Slide 19
Acid–Base Titrations
•
11
Polyprotic Acids:
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Chapter 16
Slide 20
Solubility Equilibria
•
01
Solubility Product: is the product of
the molar concentrations of
constituent ions and provides a
measure of a compound’s solubility.
MX2(s) æ M2+(aq) + 2 X–(aq)
Ksp = [M2+][X–]2
Prentice Hall ©2004
Chapter 16
Slide 21
Solubility Equilibria
Al(OH)3
BaCO3
BaF2
BaSO4
Bi2S3
CdS
CaCO3
CaF2
Ca(OH)2
Ca3(PO4)2
Cr(OH)3
CoS
CuBr
Prentice Hall ©2004
1.8 x 10–33
8.1 x 10–9
1.7 x 10–6
1.1 x 10–10
1.6 x 10–72
8.0 x 10–28
8.7 x 10–9
4.0 x 10–11
8.0 x 10–6
1.2 x 10–26
3.0 x 10–29
4.0 x 10–21
4.2 x 10–8
CuI
Cu(OH)2
CuS
Fe(OH)2
Fe(OH)3
FeS
PbCO3
PbCl2
PbCrO4
PbF2
PbI2
PbS
MgCO3
Mg(OH)2
5.1 x 10–12
2.2 x 10–20
6.0 x 10–37
1.6 x 10–14
1.1 x 10–36
6.0 x 10–19
3.3 x 10–14
2.4 x 10–4
2.0 x 10–14
4.1 x 10–8
1.4 x 10–8
3.4 x 10–28
4.0 x 10–5
1.2 x 10–11
Chapter 16
02
MnS
3.0 x 10–14
Hg2Cl2 3.5 x 10–18
HgS
4.0 x 10–54
NiS
1.4 x 10–24
AgBr
7.7 x 10–13
Ag2CO3 8.1 x 10–12
AgCl
1.6 x 10–10
Ag2SO4 1.4 x 10–5
Ag2S
6.0 x 10–51
SrCO3 1.6 x 10–9
SrSO4 3.8 x 10–7
SnS
1.0 x 10–26
Zn(OH)2 1.8 x 10–14
ZnS
3.0 x 10–23
Slide 22
Solubility Equilibria
03
•
The solubility of calcium sulfate (CaSO4) is found
experimentally to be 0.67 g/L. Calculate the value
of Ksp for calcium sulfate.
•
The solubility of lead chromate (PbCrO4) is
4.5 x 10–5 g/L. Calculate the solubility product of
this compound.
•
Calculate the solubility of copper(II) hydroxide,
Cu(OH)2, in g/L.
Prentice Hall ©2004
Chapter 16
Slide 23
Solubility Equilibria
•
04
Ion Product (Q): solubility equivalent of the
reaction quotient. It is used to determine whether a
precipitate will form.
Q < Ksp
Unsaturated
Q = Ksp
Saturated
Q > Ksp
Supersaturated; precipitate forms.
Prentice Hall ©2004
Chapter 16
Slide 24
Solubility Equilibria
•
05
Exactly 200 mL of 0.0040 M BaCl2 are added to
exactly 600 mL of 0.0080 M K2SO4. Will a
precipitate form?
•
If 2.00 mL of 0.200 M NaOH are added to 1.00 L of
0.100 M CaCl2, will precipitation occur?
Prentice Hall ©2004
Chapter 16
Slide 25
The Common-Ion Effect and
Solubility
•
01
The solubility product (Ksp) is an equilibrium
constant; precipitation will occur when the ion
product exceeds the Ksp for a compound.
•
If AgNO3 is added to saturated AgCl, the increase
in [Ag+] will cause AgCl to precipitate.
Q = [Ag+]0 [Cl–]0 > Ksp
Prentice Hall ©2004
Chapter 16
Slide 26
The Common-Ion Effect and
Solubility
Prentice Hall ©2004
Chapter 16
02
Slide 27
The Common-Ion Effect and
Solubility
Prentice Hall ©2004
Chapter 16
03
Slide 28
The Common-Ion Effect and
Solubility
•
04
Calculate the solubility of silver chloride (in g/L) in a
6.5 x 10–3 M silver chloride solution.
•
Calculate the solubility of AgBr (in g/L) in:
(a) pure water
(b) 0.0010 M NaBr
Prentice Hall ©2004
Chapter 16
Slide 29
Complex Ion Equilibria
and Solubility
•
01
A complex ion is an ion containing a central metal
cation bonded to one or more molecules or ions.
•
Most metal cations are transition metals because
they have more than one oxidation state.
•
The formation constant (Kf) is the equilibrium
constant for the complex ion formation.
Prentice Hall ©2004
Chapter 16
Slide 30
Complex Ion Equilibria
and Solubility
Prentice Hall ©2004
Chapter 16
02
Slide 31
Complex Ion Equilibria
and Solubility
Prentice Hall ©2004
Chapter 16
03
Slide 32
Complex Ion Equilibria
and Solubility
ION
Kf
Ag(NH3)2+
1.5 x 107
Ag(CN)2–
1.0 x 1021
Cu(CN)42–
1.0 x 1025
Cu(NH3)42+
5.0 x 1013
Cd(CN)42–
7.1 x 1016
CdI42–
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ION
HgCl42–
HgI42–
Hg(CN)42–
Co(NH3)63+
Zn(NH3)42+
04
Kf
1.7 x 1016
3.0 x 1030
2.5 x 1041
5.0 x 1031
2.9 x 109
2.0 x 106
Chapter 16
Slide 33
Complex Ion Equilibria
and Solubility
•
05
A 0.20 mole quantity of CuSO4 is added to a liter of
1.20 M NH3 solution. What is the concentration of
Cu2+ ions at equilibrium?
•
If 2.50 g of CuSO4 are dissolved in 9.0 x 102 mL of
0.30 M NH3, what are the concentrations of Cu2+,
Cu(NH3)42+, and NH3 at equilibrium?
Prentice Hall ©2004
Chapter 16
Slide 34
Complex Ion Equilibria
and Solubility
•
06
Calculate the molar solubility of AgCl in a 1.0 M
NH3 solution.
•
Calculate the molar solubility of AgBr in a 1.0 M
NH3 solution.
Prentice Hall ©2004
Chapter 16
Slide 35
Complex Ion Equilibria
and Solubility
Prentice Hall ©2004
Chapter 16
07
Slide 36