Transcript General Chemistry - Valdosta State University

Reactions in Aqueous
Solution
Chapter 5
Chapter 5
1
Compounds in Aqueous Solution
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
This is a reaction in which reactants are in solution
Solution – homogeneous mixture composed of two
parts:
solute – the medium which is dissolved
solvent – the medium which dissolves the solute.
Chapter 5
2
Compounds in Aqueous Solution
Compounds in Water
Some compounds conduct electricity when dissolved in
water – electrolytes
Those compounds which do not conduct electricity when
dissolved in water are called – nonelectrolytes
Chapter 5
3
Compounds in Aqueous Solution
Ionic Compounds in Water (Electrolytes)
The conductivity of the solution is due to the formation
of ions when the compound dissolves in water
H2O
NaCl( s) 
 Na  (aq)  Cl  (aq)
These ions are not the result of a chemical reaction, they
are the result of a dissociation of the molecule into ions
that compose the solid.
Chapter 5
4
Compounds in Aqueous Solution
Ionic Compounds in Water
Chapter 5
5
Compounds in Aqueous Solution
Molecular Compounds in Water(Nonelectrolytes)
In this case no ions are formed, the molecules just
disperse throughout the solvent.
sugar( s) 
 sugar(aq)
H 2O
Chapter 5
6
Compounds in Aqueous Solution
Molecular Compounds in Water(Nonelectrolytes)
There are exceptions to this, some molecules are strongly
attracted to water and will react with it.

4
NH 3  H 2O 
 NH  OH

HCl  H 2O 
 H 3O  Cl
Chapter 5


7
Compounds in Aqueous Solution
Strong and Weak Electrolytes
Strong electrolytes – A substance which completely
ionizes in water.
For example:

HCl  H 2O 
 H 3O  Cl
Chapter 5

8
Compounds in Aqueous Solution
Strong and Weak Electrolytes
Weak electrolyte: A substance which partially ionizes
when dissolved in water.
For example:

CH3CO2 H  H 2O  CH3CO2  H3O 
Chapter 5
9
Compounds in Aqueous Solution
Strong and Weak Electrolytes

CH3CO2 H  H 2O  CH3CO2  H3O 
Notice that the arrow in this reaction has two heads,
this indicates that two opposing reactions are
occurring simultaneously.
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10
Compounds in Aqueous Solution
Strong and Weak Electrolytes

CH 3CO2 H  H 2O  CH 3CO2  H 3O 
and

CH 3CO2  H 3O   CH 3CO2 H  H 2O
Since both reactions occur at the same time, this is
called a chemical equilibrium.
Chapter 5
11
Precipitation Reaction
Chapter 5
12
Precipitation Reaction
- A reaction which forms a solid (precipitate)
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
- AgCl is classified as an insoluble substance
Chapter 5
13
Precipitation Reaction
Net Ionic Equation
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
- AgNO3 and NaNO3 are electrolytes in solution so they
actually occur as free ions.
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) 
AgCl(s) + Na+(aq) + NO3-(aq)
Chapter 5
14
Precipitation Reaction
Net Ionic Equation
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) 
AgCl(s) + Na+(aq) + NO3-(aq)
- Notice that NO3-(aq) and Na+(aq) occur in both the left
and right side of the equation.
-These are called spectator ions.
Chapter 5
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Precipitation Reaction
Net Ionic Equation
Ag+(aq) + Cl-(aq) AgCl(s)
- With the spectator ions removed, the resulting
equation shows only the ions involved in the reaction
remain.
- This is a net ionic equation.
Chapter 5
16
Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
1. Most nitrates (NO3-) and acetates (CH3CO2-) are
soluble in water.
2. All chlorides are soluble except: Hg+, Ag+, Pb2+, Cu+
3. All sulfates are soluble except: Sr2+, Ba2+, Pb2+
4. Carbonates (CO32-), Phosphates (PO43-), Borates
(BO33-),Arsenates (AsO43-), and Arsenites (AsO33-)
are insoluble.
5. Hydroxides (OH-) of group Ia and Ba2+ and Sr2+ are
soluble.
6. Most sulfides (S2-) are insoluble.
Chapter 5
17
Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
Predict the solubility of the following compounds:
PbSO4
AgCH3CO2
(NH4)3PO4
KClO4
Chapter 5
18
Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
Predict the solubility of the following compounds:
PbSO4
Insoluble
AgCH3CO2
(NH4)3PO4
KClO4
Chapter 5
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Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
Predict the solubility of the following compounds:
PbSO4
Insoluble
AgCH3CO2
Soluble
(NH4)3PO4
KClO4
Chapter 5
20
Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
Predict the solubility of the following compounds:
PbSO4
Insoluble
AgCH3CO2
Soluble
(NH4)3PO4
Soluble
KClO4
Chapter 5
21
Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
Predict the solubility of the following compounds:
PbSO4
Insoluble
AgCH3CO2
Soluble
(NH4)3PO4
Soluble
KClO4
Soluble
Chapter 5
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Acids and Bases
Acid - substance which ionizes to form hydrogen cations
(H+) in solution
Examples:
Hydrochloric Acid
HCl
Nitric Acid
HNO3
Acetic Acid
CH3CO2H
Sulfuric Acid
H2SO4
Sulfuric acid can provide two H+’s - Diprotic acid,
The other acids can provide only one H+
- Monoprotic acid.
Chapter 5
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Acids and Bases
Diprotic acid
H2SO4  H+ + HSO4HSO4-

H+ + SO42-
Chapter 5
24
Acids and Bases
Base - substance which reacts with H+ ions.
Examples:
ammonia
sodium hydroxide
Chapter 5
NH3
NaOH
25
Acids and Bases
Acid-Base Reaction
H+ + OH-  H2O
- It is clear that the metal hydroxides (NaOH for
example) provide OH- by disassociation.
- Bases like ammonia make OH- by reacting with water
(ionization)
NH3 + H2O  NH4+ + OH-
Chapter 5
26
Acids and Bases
Strong and Weak Acids and Bases
Strong acids and bases are strong electrolytes.
Weak acids and bases are weak electrolytes.
Chapter 5
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Acids and Bases
Strong Acids
- The strength of acids and bases are concerned with
the ionization (or dissociation) of the substance, not its
chemical reactivity.
Example:
Hydrofluoric acid (HF) is a weak acid, but it is very
chemically reactive.
- this substance can’t be stored in glass bottles
because it reacts with glass (silicon dioxide).
Chapter 5
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Acids and Bases
Common Strong Acids and Bases
Common Strong Acids
Hydrochloric Acid
Hydrobromic Acid
Hydroiodic acid
Nitric Acid
Perchloric Acid
Sulfuric Acid
HCl
HBr
HI
HNO3
HClO4
H2SO4
Chapter 5
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Acids and Bases
Common Strong Acids and Bases
Common Strong Bases
Lithium Hydroxide
Sodium Hydroxide
Potassium Hydroxide
Chapter 5
LiOH
NaOH
KOH
30
Acids, Bases, and Salts
Neutralization Reaction
- Reaction between an acid and a base.
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
“The neutralization between acid and metal hydroxide
produces water and a salt”
Salt – an ionic compound whose cation comes from a
base and anion from an acid.
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Acids, Bases, and Salts
Neutralization Reaction
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
- despite the appearance of the equation, the reaction
actually takes place between the ions.
Chapter 5
32
Acids, Bases, and Salts
Neutralization Reaction
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Total Ionic Equation
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) 
H2O(l) + Na+(aq) + Cl-(aq)
Chapter 5
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Acids, Bases, and Salts
Neutralization Reaction
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Total Ionic Equation
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) 
H2O(l) + Na+(aq) + Cl-(aq)
Chapter 5
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Acids, Bases, and Salts
Neutralization Reaction
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Total Ionic Equation
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) 
H2O(l) + Na+(aq) + Cl-(aq)
Net Ionic Equation
H+(aq) + OH-(aq)  H2O(l)
Chapter 5
35
Gas-Forming Reactions
Metal Carbonates and Acid
2 HCl(aq) + Na2CO3(aq)  2 NaCl(aq) + H2O(l) +
CO2(g)
- Metal carbonates (or bicarbonates) always form a salt,
water and carbon dioxide
Chapter 5
36
Gas-Forming Reactions
Metal Sulfide and Acid
2 HCl(aq) + Na2S(s)  H2S(g) + 2 NaCl(aq)
- Metal sulfides form a salt and hydrogen sulfide.
Chapter 5
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Gas-Forming Reactions
Metal Sulfite and Acid
2 HCl(aq) + Na2SO3(s)  SO2(g) + 2 NaCl(aq) +
H2O(l)
- Metal sulfites form a salt, sulfur dioxide and water.
Chapter 5
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Gas-Forming Reactions
Ammonium Salt and Strong Base
NH4Cl(s) + NaOH(aq)  NH3(g) + NaCl(aq) + H2O(l)
- This reaction forms ammonia, salt and water
Chapter 5
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Oxidation-Reduction Reactions
- Reaction where electrons are exchanged.
2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)
Na(s)  Na+(aq) + 1 eoxidation – loss of electrons
2 H+(g) + 2 e-  H2(g)
reduction – gain of electrons
Chapter 5
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Oxidation-Reduction Reactions
- Reaction where electrons are exchanged.
2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)
- An alternate approach is to describe how one reagent
effects another.
- Reducing Agent, a substance that causes another substance
to be reduced.
Na(s)  Na+(aq) + 1 e- Oxidizing Agent, a substance that causes another substnace
to be oxidized.
2 H+(g) + 2 e-  H2(g)
Chapter 5
41
Oxidation-Reduction Reactions
Oxidation Numbers
1. Each atom of a pure element has an oxidation number of
zero(0).
2. For monatomic ions, the oxidation number equals the
charge on the ion.
3. Fluorine always has an oxidation state of -1 in compounds.
4. Cl, Br, and I always have oxidation numbers of -1, except
when combined with oxygen or fluorine.
5. The oxidation number of H is +1 and O is -2 in most
compounds.
6. The sum of the oxidation numbers must equal the charge
on the molecule or ion.
Chapter 5
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Oxidation-Reduction Reactions
Oxidation Numbers
Examples
PCl5
P 1( ) =
Cl 5( ) = ________
0
Chapter 5
43
Oxidation-Reduction Reactions
Oxidation Numbers
Example
PCl5
P 1( ? ) =
?
Cl 5(-1) = __-5____
0
Chapter 5
44
Oxidation-Reduction Reactions
Oxidation Numbers
Example
PCl5
P 1(+5) =
+5
Cl 5(-1) = __-5____
0
Chapter 5
45
Oxidation-Reduction Reactions
Oxidation Numbers
Example
CO32C 1( ) =
O 3( ) = _____
-2
Chapter 5
46
Oxidation-Reduction Reactions
Oxidation Numbers
Example
CO32C 1(?) =
?
O 3(-2) = __-6__
-2
Chapter 5
47
Oxidation-Reduction Reactions
Oxidation Numbers
Example
CO32C 1(+4) = +4
O 3(-2) = __-6__
-2
Chapter 5
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Oxidation-Reduction Reactions
Oxidation Numbers
Example
K2CrO4
K 2( ) =
O 4( ) =
Cr 1( ) = ________
0
Chapter 5
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Oxidation-Reduction Reactions
Oxidation Numbers
Example
K2CrO4
K 2(+1) =
+2
O 4(-2) =
-8
Cr 1( ? ) = ___?____
0
Chapter 5
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Oxidation-Reduction Reactions
Oxidation Numbers
Example
K2CrO4
K 2(+1) =
+2
O 4(-2) =
-8
Cr 1(+6) = ___+6__
0
Chapter 5
51
Solutions
Molarity(M)
Unit of concentration, moles of solute per liter of
solution.
Molesof solute
Molarity
Liters of solution
Chapter 5
52
Solutions
Example: What is the molarity(M) of a solution which
contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
17.51g
m olesNaCl 
58.45g / m ol
Chapter 5
53
Solutions
Example: What is the molarity(M) of a solution which
contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
17.51g
m olesNaCl 
 0.2996m ol
58.45g / m ol
Chapter 5
54
Solutions
Example: What is the molarity(M) of a solution which
contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
17.51g
m olesNaCl 
 0.2996m ol
58.45g / m ol
Solution volume
 1L 
Liters of solution 751m L

 1000m L
Chapter 5
55
Solutions
Example: What is the molarity(M) of a solution which
contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): 58.45g/mol
17.51g
m olesNaCl 
 0.2996m ol
58.45g / m ol
Solution volume
 1L 
Liters of solution 751m L
  0.751L
 1000m L
Chapter 5
56
Solutions
Example: What is the molarity(M) of a solution which
contains 17.51g of NaCl in 751mL of solution?
Molesof solute
Molarity
Liters of solution
Chapter 5
57
Solutions
Example: What is the molarity(M) of a solution which
contains 17.51g of NaCl in 751mL of solution?
0.2996m ol
m olarity
0.751L
Chapter 5
58
Solutions
Example: What is the molarity(M) of a solution which
contains 17.51g of NaCl in 751mL of solution?
0.2996m ol
m olarity
0.751L
 0.399M
Chapter 5
59
Solutions
Molarity
Chapter 5
60
Solutions
Dilution
MdiluteVdilute = MconcentratedVconcentrated
Chapter 5
61
Solutions
Dilution
Example: What volume of 6.00M NaOH is required to
make 500mL of 0.100M NaOH?
Mconcentrated = 6.00M
Vconcentrated = ?
Mdilute = 0.100M
Vdilute = 500mL
0.100M(500mL) = 6.00M(Vconcentrated)
Vconcentrated = 8.33mL
Chapter 5
62
pH Scale
Concentration scale for acids and bases.

pH   log[H ]
• The square brackets around the H+ indicate that
the concentration of H+ is in molarity.
• So, a change of 1 pH unit indicates a 10X change in
H+ concentration.
[ H  ] 10 pH
Chapter 5
63
Solution Stoichiometry
- We can now use molarity to determine stoichiometric
quantities.
Example
How many grams of hydrogen gas are produced when
20.0 mL of 1.75M HCl is allowed to react with 15.0g of
sodium metal?
2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
Chapter 5
64
Solution Stoichiometry
2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
- Convert quantities to moles
m olesHCl  1.75M 0.0200L  
m olesNa 
15.0 g

23.0 g / m ol
Chapter 5
65
Solution Stoichiometry
2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
- Convert quantities to moles
m olesHCl  1.75M 0.0200L  0.0350m ol
m olesNa 
15.0 g
 0.652m ol
23.0 g / m ol
Chapter 5
66
Solution Stoichiometry
2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
- Convert quantities to moles
m olesHCl  1.75M 0.0200L  0.0350m ol
15.0 g
m olesNa 
 0.652m ol
23.0 g / m ol
- Determine limiting reagent
0.0350

2
0.652
Na

2
HCl
Chapter 5
67
Solution Stoichiometry
2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
- Convert quantities to moles
m olesHCl  1.75M 0.0200L  0.0350m ol
15.0 g
m olesNa 
 0.652m ol
23.0 g / m ol
- Determine limiting reagent
HCl
Na
0.0350
 0.0175
2
0.652
 0.326
2
Chapter 5
68
Solution Stoichiometry
2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
- Calculate moles of H2
1H2
x

2 HCl 0.0350m ol
Chapter 5
69
Solution Stoichiometry
2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
- Calculate moles of H2
1H2
x

2 HCl 0.0350m ol
x  0.0175m ol H 2
Chapter 5
70
Solution Stoichiometry
2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
- Calculate moles of H2
1H2
x

2 HCl 0.0350m ol
x  0.0175m ol H 2
- Calculate grams of H2
g H 2  0.0175mol2.02g / mol 
Chapter 5
71
Solution Stoichiometry
2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
- Calculate moles of H2
1H2
x

2 HCl 0.0350m ol
x  0.0175m ol H 2
- Calculate grams of H2
g H 2  0.0175mol2.02g / mol  0.0353g
Chapter 5
72
Solution Stoichiometry
Titrations
Chapter 5
73
Practice Problems
4,10, 16, 20, 24, 30, 36, 38, 44, 60, 62, 68
Chapter 5
74