Chemistry and Chemical Reactivity
6th Edition
1
John C. Kotz
Paul M. Treichel
Gabriela C. Weaver
CHAPTER 5
Reactions in Aqueous Solution
Lectures written by John Kotz
©2006
2006
Brooks/Cole
Thomson
©
Brooks/Cole
- Thomson
Reactions
in Aqueous Solution
Many reactions involve ionic compounds,
especially reactions in water — aqueous
solutions.
KMnO4 in water
© 2006 Brooks/Cole - Thomson
K+(aq) + MnO4-(aq)
2
3
CCR, page 177
An Ionic Compound,
CuCl2, in Water
© 2006 Brooks/Cole - Thomson
Aqueous Solutions
How do we know ions are
present in aqueous
solutions?
The solutions conduct
electricity!
They are called
ELECTROLYTES
HCl, CuCl2, and NaCl are
strong electrolytes.
They dissociate
completely (or nearly so)
into ions.
© 2006 Brooks/Cole - Thomson
4
Aqueous Solutions
HCl, CuCl2, and NaCl are strong
electrolytes. They dissociate completely
(or nearly so) into ions.
© 2006 Brooks/Cole - Thomson
5
Aqueous
Solutions
Acetic acid ionizes only
to a small extent, so it
is a weak
electrolyte.
CH3CO2H(aq) --->
CH3CO2-(aq) + H+(aq)
© 2006 Brooks/Cole - Thomson
6
Aqueous
Solutions
Some compounds
dissolve in water but
do not conduct
electricity. They are
called nonelectrolytes.
Examples include:
sugar
ethanol
ethylene glycol
© 2006 Brooks/Cole - Thomson
7
Water Solubility of Ionic Compounds
If one ion from the “Soluble
Compd.” list is present in a
compound, the compound is
water soluble.
© 2006 Brooks/Cole - Thomson
8
9
Solubility: Which of the following is the only
INSOLUBLE salt in water?
1.
2.
3.
4.
© 2006 Brooks/Cole - Thomson
5.
0%
0%
1
2
0%
0%
3
4
0%
5
ACIDS
10
An acid -------> H+ in water
Some strong acids are
HCl
H2SO4
HClO4
HNO3
© 2006 Brooks/Cole - Thomson
hydrochloric
sulfuric
perchloric
nitric
HNO3
ACIDS
An acid -------> H+ in water
HCl(aq) ---> H+(aq) + Cl-(aq)
© 2006 Brooks/Cole - Thomson
11
12
The
Nature of
Acids
HCl
Cl-
H 2O
hydronium
ion
© 2006 Brooks/Cole - Thomson
H 3O+
Weak Acids
WEAK ACIDS = weak
electrolytes
CH3CO2H
acetic acid
H2CO3
carbonic acid
H3PO4
phosphoric acid
HF
hydrofluoric acid
© 2006 Brooks/Cole - Thomson
Acetic acid
13
ACIDS
Nonmetal oxides can be acids
CO2(aq) + H2O(liq)
---> H2CO3(aq)
SO3(aq) + H2O(liq)
---> H2SO4(aq)
and can come from burning
coal and oil.
© 2006 Brooks/Cole - Thomson
14
BASES
see Screen 5.9 and Table 5.2
Base ---> OH- in water
NaOH(aq)
NaOH is
a strong
base
© 2006 Brooks/Cole - Thomson
---> Na+(aq) + OH-(aq)
15
Ammonia, NH3
An Important Base
© 2006 Brooks/Cole - Thomson
16
BASES
Metal oxides are
bases
CaO(s) + H2O(liq)
--> Ca(OH)2(aq)
CaO in water. Indicator
shows solution is basic.
© 2006 Brooks/Cole - Thomson
17
18
Know the strong
acids & bases!
© 2006 Brooks/Cole - Thomson
19
Which compound below is an acid
in aqueous solution?
1.
2.
3.
4.
0%
1
© 2006 Brooks/Cole - Thomson
0%
0%
2
3
0%
4
Which compound below is a base
in aqueous solution?
1.
2.
3.
4.
0%
1
© 2006 Brooks/Cole - Thomson
0%
0%
2
3
20
0%
4
21
Net Ionic
Equations
Mg(s) + 2 HCl(aq) --> H2(g) + MgCl2(aq)
We really should write
Mg(s) + 2 H+(aq) + 2 Cl-(aq) --->
H2(g) + Mg2+(aq) + 2 Cl-(aq)
The two Cl- ions are SPECTATOR IONS —
they do not participate. Could have used NO3-.
© 2006 Brooks/Cole - Thomson
Net Ionic Equations
Mg(s) + 2 HCl(aq)
--> H2(g) + MgCl2(aq)
Mg(s) + 2 H+(aq) + 2 Cl-(aq)
---> H2(g) + Mg2+(aq) + 2 Cl-(aq)
We leave the spectator ions out —
Mg(s) + 2 H+(aq) ---> H2(g) + Mg2+(aq)
to give the NET
© 2006 Brooks/Cole - Thomson
IONIC EQUATION
22
Chemical Reactions in Water
23
Sections 5.2 & 5.4-5.6—CD-ROM Ch. 5
We will look at
EXCHANGE
REACTIONS
AX + BY
AY + BX
The anions
exchange places
between cations.
© 2006 Brooks/Cole - Thomson
Pb(NO3) 2(aq) + 2 KI(aq)
----> PbI2(s) + 2 KNO3 (aq)
Precipitation Reactions
The “driving force” is the formation of an
insoluble compound — a precipitate.
Pb(NO3)2(aq) + 2 KI(aq) ----->
2 KNO3(aq) + PbI2(s)
Net ionic equation
Pb2+(aq) + 2 I-(aq) ---> PbI2(s)
© 2006 Brooks/Cole - Thomson
24
25
1.
2.
3.
4.
0%
1
© 2006 Brooks/Cole - Thomson
0%
0%
2
3
0%
4
Acid-Base Reactions
• The “driving force” is the formation of water.
NaOH(aq) + HCl(aq) --->
NaCl(aq) + H2O(liq)
• Net ionic equation
OH-(aq) + H+(aq)
---> H2O(liq)
• This applies to ALL reactions
of STRONG acids and bases.
© 2006 Brooks/Cole - Thomson
26
Acid-Base Reactions
CCR, page 191
© 2006 Brooks/Cole - Thomson
27
Acid-Base Reactions
• A-B reactions are sometimes called
NEUTRALIZATIONS because the solution is
neither acidic nor basic at the end.
• The other product of the A-B reaction is a SALT, MX.
HX + MOH ---> MX + H2O
Mn+ comes from base & Xn- comes from acid
This is one way to make compounds!
© 2006 Brooks/Cole - Thomson
28
Gas-Forming Reactions
This is primarily the chemistry of metal
carbonates.
CO2 and water ---> H2CO3
H2CO3(aq) + Ca2+ --->
2 H+(aq) + CaCO3(s) (limestone)
Adding acid reverses this reaction.
MCO3 + acid ---> CO2 + salt
© 2006 Brooks/Cole - Thomson
29
30
Gas-Forming
Reactions
CaCO3(s) + H2SO4(aq) --->
2 CaSO4(s) + H2CO3(aq)
Carbonic acid is unstable and forms CO2 & H2O
H2CO3(aq)
---> CO2 + water
(Antacid tablet has citric acid + NaHCO3)
© 2006 Brooks/Cole - Thomson
31
© 2006 Brooks/Cole - Thomson
32
Quantitative Aspects
of
Reactions in Solution
Sections 5.8-5.10
© 2006 Brooks/Cole - Thomson
Terminology
In solution we need to define the
• SOLVENT
the component whose
physical state is
preserved when
solution forms
• SOLUTE
the other solution component
© 2006 Brooks/Cole - Thomson
33
34
© 2006 Brooks/Cole - Thomson
35
I did ____ on the exam.
1.
2.
3.
4.
great
good
ok
Pretty bad
100%
0%
0%
0%
© 2006 Brooks/Cole - Thomson
1
Answer Now
0%
0%
2
3
0%
4
36
Concentration of Solute
The amount of solute in a solution
is given by its concentration.
Molarity(M) =
moles solute
liters of solution
Concentration (M) = [ …]
© 2006 Brooks/Cole - Thomson
37
1.0 L of water
was used to
make 1.0 L of
solution. Notice
the water left
over.
CCR, page 206
© 2006 Brooks/Cole - Thomson
Preparing a Solution
Active Figure 5.18
© 2006 Brooks/Cole - Thomson
38
39
PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O
in enough water to make 250 mL of solution.
Calculate molarity.
Step 1: Calculate moles
of NiCl2•6H2O
1 mol
5.00 g •
= 0.0210 mol
237.7 g
Step 2: Calculate molarity
0.0210 mol
= 0.0841 M
0.250 L
[NiCl2•6 H2O ] = 0.0841 M
© 2006 Brooks/Cole - Thomson
The Nature of a CuCl2 Solution:
Ion Concentrations
CuCl2(aq) -->
Cu2+(aq) + 2 Cl-(aq)
If [CuCl2] = 0.30 M,
then
[Cu2+] = 0.30 M
[Cl-] = 2 x 0.30 M
© 2006 Brooks/Cole - Thomson
40
41
1.
2.
3.
4.
0%
1
© 2006 Brooks/Cole - Thomson
0%
0%
2
3
0%
4
42
USING MOLARITY
What mass of oxalic acid, H2C2O4, is
required to make 250. mL of a
0.0500 M solution?
Because
Conc (M) = moles/volume = mol/V
this means that
moles = M•V
© 2006 Brooks/Cole - Thomson
USING MOLARITY
What mass of oxalic acid, H2C2O4, is
required to make 250. mL of a 0.0500 M
solution?
moles = M•V
Step 1: Calculate moles of acid required.
(0.0500 mol/L)(0.250 L) = 0.0125 mol
Step 2: Calculate mass of acid required.
(0.0125 mol )(90.00 g/mol) =
© 2006 Brooks/Cole - Thomson
1.13 g
43
44
You have 60.0 mL of 0.25 M HCl. What
amount of HCl (moles) is present?
1.
2.
3.
4.
5.
© 2006 Brooks/Cole - Thomson
0.25 mol
0.60 mol
0.15 mol
0.025 mol
0.015 mol
0%
0%
1
2
0%
0%
3
4
0%
5
45
1.
2.
3.
4.
26.9 M
0.0635 M
0.254 M
0.762 M
0%
1
© 2006 Brooks/Cole - Thomson
0%
0%
2
3
0%
4
46
Preparing Solutions
• Weigh out a solid solute
and dissolve in a given
quantity of solvent.
• Dilute a concentrated
solution to give one
that is less
concentrated.
© 2006 Brooks/Cole - Thomson
47
Preparing a Solution by
Dilution
© 2006 Brooks/Cole - Thomson
48
PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH.
What do you do?
Add water to the 3.0 M solution to lower
its concentration to 0.50 M
Dilute the solution!
© 2006 Brooks/Cole - Thomson
49
PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
you do?
But how much water
do we add?
© 2006 Brooks/Cole - Thomson
50
PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
you do?
How much water is added?
The important point is that --->
moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solution
© 2006 Brooks/Cole - Thomson
PROBLEM: You have 50.0 mL of 3.0 M NaOH and
you want 0.50 M NaOH. What do you do?
Amount of NaOH in original solution =
M•V
=
(3.0 mol/L)(0.050 L) = 0.15 mol NaOH
Amount of NaOH in final solution must also =
0.15 mol NaOH
Volume of final solution =
(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L
or
© 2006 Brooks/Cole - Thomson
300 mL
51
52
You add 60.0 mL of 0.25 M HCl to a 500 mL volumetric flask and then
add water to the mark on the flask. What is the concentration of HCl
in the diluted solution?
1.
2.
3.
4.
5.
0.015 M
0.025 M
0.030 M
0.060 M
0.050 M
© 2006 Brooks/Cole - Thomson
0%
0%
1
2
0%
0%
3
4
0%
5
Preparing Solutions by
Dilution
A shortcut
Cinitial • Vinitial = Cfinal • Vfinal
© 2006 Brooks/Cole - Thomson
53
54
© 2006 Brooks/Cole - Thomson
SOLUTION STOICHIOMETRY
Section 5.10
• Zinc reacts with
acids to produce H2
gas.
• Have 10.0 g of Zn
• What volume of
2.50 M HCl is
needed to convert
the Zn completely?
© 2006 Brooks/Cole - Thomson
55
GENERAL PLAN FOR STOICHIOMETRY
CALCULATIONS
Mass
HCl
Mass
zinc
Moles
zinc
Stoichiometric
factor
Moles
HCl
Volume
HCl
© 2006 Brooks/Cole - Thomson
56
57
Zinc reacts with acids to produce H2 gas. If you
have 10.0 g of Zn, what volume of 2.50 M HCl is
needed to convert the Zn completely?
Step 1: Write the balanced equation
Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g)
Step 2: Calculate amount of Zn
1.00 mol Zn
10.0 g Zn •
= 0.153 mol Zn
65.39 g Zn
Step 3: Use the stoichiometric factor
© 2006 Brooks/Cole - Thomson
58
Zinc reacts with acids to produce H2 gas. If you
have 10.0 g of Zn, what volume of 2.50 M HCl is
needed to convert the Zn completely?
Step 3: Use the stoichiometric factor
2 mol HCl
0.153 mol Zn •
= 0.306 mol HCl
1 mol Zn
Step 4: Calculate volume of HCl req’d
1.00 L
0.306 mol HCl •
= 0.122 L HCl
2.50 mol
© 2006 Brooks/Cole - Thomson
ACID-BASE REACTIONS
Titrations
H2C2O4(aq) + 2 NaOH(aq) --->
acid
base
Na2C2O4(aq) + 2 H2O(liq)
Carry out this reaction using a TITRATION.
Oxalic acid,
H2C2O4
© 2006 Brooks/Cole - Thomson
59
Setup for titrating an acid with a base
Active Figure 5.23
© 2006 Brooks/Cole - Thomson
60
Titration
© 2006 Brooks/Cole - Thomson
1. Add solution from the
buret.
2. Reagent (base) reacts
with compound (acid) in
solution in the flask.
3. Indicator shows when
exact stoichiometric
reaction has occurred.
4. Net ionic equation
H+ + OH- --> H2O
5. At equivalence point
moles H+ = moles OH-
61
LAB PROBLEM #1: Standardize a
solution of NaOH — i.e., accurately
determine its concentration.
1.065 g of H2C2O4 (oxalic acid) requires
35.62 mL of NaOH for titration to an
equivalence point. What is the
concentra-tion of the NaOH?
© 2006 Brooks/Cole - Thomson
62
63
1.065 g of H2C2O4 (oxalic acid) requires 35.62
mL of NaOH for titration to an equivalence
point. What is the concentration of the NaOH?
Step 1: Calculate amount of H2C2O4
1 mol
1.065 g •
= 0.0118 mol
90.04 g
Step 2: Calculate amount of NaOH req’d
2 mol NaOH
0.0118 mol acid •
= 0.0236 mol NaOH
1 mol acid
© 2006 Brooks/Cole - Thomson
64
1.065 g of H2C2O4 (oxalic acid) requires 35.62
mL of NaOH for titration to an equivalence
point. What is the concentration of the NaOH?
Step 1: Calculate amount of H2C2O4
= 0.0118 mol acid
Step 2: Calculate amount of NaOH req’d
= 0.0236 mol NaOH
Step 3: Calculate concentration of NaOH
0.0236 mol NaOH
 0.663 M
0.03562 L
[NaOH] = 0.663 M
© 2006 Brooks/Cole - Thomson
65
LAB PROBLEM #2:
Use standardized NaOH to determine
the amount of an acid in an unknown.
Apples contain malic acid, C4H6O5.
C4H6O5(aq) + 2 NaOH(aq) --->
Na2C4H4O5(aq) + 2 H2O(liq)
76.80 g of apple requires 34.56 mL of 0.663 M
NaOH for titration. What is weight % of malic
acid?
© 2006 Brooks/Cole - Thomson
66
76.80 g of apple requires 34.56 mL of
0.663 M NaOH for titration. What is
weight % of malic acid?
Step 1:
C•V =
=
Step 2:
Calculate amount of NaOH used.
(0.663 M)(0.03456 L)
0.0229 mol NaOH
Calculate amount of acid titrated.
1 mol acid
0.0229 mol NaOH •
2 mol NaOH
= 0.0115 mol acid
© 2006 Brooks/Cole - Thomson
67
76.80 g of apple requires 34.56 mL of
0.663 M NaOH for titration. What is
weight % of malic acid?
Step 1:
=
Step 2:
=
Calculate amount of NaOH used.
0.0229 mol NaOH
Calculate amount of acid titrated
0.0115 mol acid
Step 3: Calculate mass of acid titrated.
134 g
0.0115 mol acid •
= 1.54 g
mol
© 2006 Brooks/Cole - Thomson
68
76.80 g of apple requires 34.56 mL of
0.663 M NaOH for titration. What is
weight % of malic acid?
Step 1:
=
Step 2:
=
Step 3:
=
Calculate amount of NaOH used.
0.0229 mol NaOH
Calculate amount of acid titrated
0.0115 mol acid
Calculate mass of acid titrated.
1.54 g acid
Step 4: Calculate % malic acid.
1.54 g
• 100% = 2.01%
76.80 g
© 2006 Brooks/Cole - Thomson
69
Oxidation-Reduction Reactions
Section 5.7
Thermite reaction
Fe2O3(s) + 2 Al(s)
---->
2 Fe(s) + Al2O3(s)
© 2006 Brooks/Cole - Thomson
70
EXCHANGE: Precipitation Reactions
EXCHANGE
Gas-Forming
Reactions
REACTIONS
REDOX
REACTIONS
© 2006 Brooks/Cole - Thomson
EXCHANGE
Acid-Base
Reactions
71
REDOX REACTIONS
REDOX = reduction & oxidation
2 H2(g) + O2(g) ---> 2 H2O(liq)
© 2006 Brooks/Cole - Thomson
72
REDOX REACTIONS
REDOX = reduction & oxidation
Corrosion of aluminum
2 Al(s) + 3 Cu2+(aq) ---> 2 Al3+(aq) + 3 Cu(s)
© 2006 Brooks/Cole - Thomson
73
REDOX REACTIONS
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
In all reactions if
something has been
oxidized then
something has also
been reduced
© 2006 Brooks/Cole - Thomson
74
REDOX REACTIONS
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
© 2006 Brooks/Cole - Thomson
75
Why Study Redox Reactions
Batteries
Corrosion
Manufacturing metals
© 2006 Brooks/Cole - Thomson
Fuels
76
REDOX REACTIONS
Redox reactions are characterized by
ELECTRON TRANSFER between an
electron donor and electron acceptor.
Transfer leads to—
1. increase in oxidation number
of some element = OXIDATION
2. decrease in oxidation number
of some element = REDUCTION
© 2006 Brooks/Cole - Thomson
OXIDATION NUMBERS
The electric charge an element
APPEARS to have when electrons are
counted by some arbitrary rules:
1. Each atom in free element has ox. no. = 0.
Zn
O2
I2
S8
2. In simple ions, ox. no. = charge on ion.
-1 for Cl+2 for Mg2+
© 2006 Brooks/Cole - Thomson
77
78
Group 1 = +1, group 2 = +2
F is –1
H is +1
O is –2
The remaining atoms are
determined by algebra. (Neutral
species have oxidation numbers
that add to zero; charged species
add to the expressed charge.)
© 2006 Brooks/Cole - Thomson
OXIDATION NUMBERS
NH3
N =
ClO-
Cl =
H3PO4
P =
MnO4-
Mn =
Cr2O72C3H8
© 2006 Brooks/Cole - Thomson
Cr =
C =
Oxidation
number of F
in HF?
79
80
In the demonstration of the reaction of Zn with VO2+, the
oxidation number of V in VO2+ is:
1.
2.
3.
4.
5.
© 2006 Brooks/Cole - Thomson
-5
-3
0
+3
+5
0%
0%
1
2
0%
0%
3
4
0%
5
Recognizing a Redox Reaction
Corrosion of aluminum
2 Al(s) + 3 Cu2+(aq) --> 2 Al3+(aq) + 3 Cu(s)
Al(s) --> Al3+(aq) + 3 e• Ox. no. of Al increases as e- are donated by the metal.
• Therefore, Al is OXIDIZED
• Al is the REDUCING AGENT in this balanced halfreaction.
© 2006 Brooks/Cole - Thomson
81
Recognizing a Redox Reaction
82
Corrosion of aluminum
2 Al(s) + 3 Cu2+(aq) --> 2 Al3+(aq) + 3 Cu(s)
Cu2+(aq) + 2 e- --> Cu(s)
• Ox. no. of Cu decreases as e- are accepted by the ion.
• Therefore, Cu is REDUCED
• Cu is the OXIDIZING AGENT in this balanced halfreaction.
© 2006 Brooks/Cole - Thomson
Recognizing a Redox Reaction
Notice that the 2 half-reactions add up to
give the overall reaction
—if we use 2 mol of Al and 3 mol of Cu2+.
2 Al(s) --> 2 Al3+(aq) + 6 e3 Cu2+(aq) + 6 e- --> 3 Cu(s)
----------------------------------------------------------2 Al(s) + 3 Cu2+(aq) ---> 2 Al3+(aq) + 3 Cu(s)
Final eqn. is balanced for mass and charge.
© 2006 Brooks/Cole - Thomson
83
Examples of Redox Reactions
Metal + halogen
2 Al + 3 Br2 ---> Al2Br6
© 2006 Brooks/Cole - Thomson
84
Examples of Redox Reactions
Nonmetal (P) + Oxygen
---> P4O10
Metal (Mg) + Oxygen
---> MgO
© 2006 Brooks/Cole - Thomson
85
86
Recognizing a Redox
Reaction
See Table 5.4
Reaction Type
Oxidation
Reduction
In terms of oxygen
gain
loss
In terms of halogen
gain
loss
In terms of electrons
loss
gain
© 2006 Brooks/Cole - Thomson
Common Oxidizing and
Reducing Agents
87
See Table 5.4
Metals
(Cu) are
reducing
agents
Metals
(Na, K,
Mg, Fe)
are
reducing
agents
HNO3 is an
oxidizing
agent
Cu + HNO3 -->
2 K + 2 H2O -->
Cu2+ + NO2
2 KOH + H2
© 2006 Brooks/Cole - Thomson
Examples of Redox Reactions
88
Metal + acid
Mg + HCl
Mg = reducing agent
H+ = oxidizing agent
Metal + acid
Cu + HNO3
Cu = reducing agent
HNO3 = oxidizing agent
© 2006 Brooks/Cole - Thomson
89
Copper metal reacts with nitric acid (HNO3) to give
Cu2+ ions and NO2 gas.
1.
2.
3.
4.
0%
1
© 2006 Brooks/Cole - Thomson
0%
0%
2
3
0%
4
90
1.
2.
3.
an oxidation-reduction
reaction.
a precipitation reaction.
an acid-base reaction.
0%
1
© 2006 Brooks/Cole - Thomson
0%
2
0%
3
91
1.
2.
3.
an oxidation-reduction
reaction.
a precipitation reaction.
an acid-base reaction.
0%
1
© 2006 Brooks/Cole - Thomson
0%
2
0%
3
92
1.
2.
3.
an oxidation-reduction
reaction.
a precipitation reaction.
an acid-base reaction.
0%
1
© 2006 Brooks/Cole - Thomson
0%
2
0%
3
93
Please make your selection...
1.
2.
3.
4.
Choice One
Choice Two
Choice Three
Choice Four
100%
0%
0%
0%
© 2006 Brooks/Cole - Thomson
1
Answer Now
0%
0%
2
3
0%
4