Transcript zn soft free

Lecture Presentation
Chapter 4
Reactions in
Aqueous Solution
James F. Kirby
Quinnipiac University
Hamden, CT
Aqueous
Reactions
© 2015 Pearson Education, Inc.
LESSON 1
4-1 Aqueous Solution
Properties
Atoms,
Molecules,
and Ions
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Chemical Reactions
in
Aqueous Solution
Three Main Types of Chemical Reactions in Aqueous Solution
1. Precipitation reactions: Result in formation of an insoluble solid
(precipitate) that separates from the solution.
2. Acid-base or neutralization reactions: Result in formation of
water and an ionic compound (i.e., a salt) due to transfer of hydrogen
ions (i.e., protons).
3. Oxidation-reduction (redox) reactions: Result in formation of
new compounds due to transfer of electrons among reactants.
Solution: a homogeneous mixture
of 2 or more substances
Aqueous solutions
Solute: substance present in a
smaller amount
Solvent: substance present in a
(much) larger amount
A solution is a homogenous mixture of 2 or more
substances
The solute is(are) the substance(s) present in the
smaller amount(s)
The solvent is the substance present in the larger
amount
Solution
Solvent
Solute
Soft drink (l)
H2O
Sugar, CO2
Air (g)
N2
O2, Ar, CH4
Soft Solder (s)
Pb
Sn
Solutions
• Solutions are defined as
homogeneous mixtures of
two or more pure
substances.
• The solvent is present in
greatest abundance.
• All other substances are
solutes.
• When water is the solvent,
the solution is called an
aqueous solution.
Aqueous
Reactions
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Aqueous Solutions
• Substances can dissolve in water by different ways:
 Ionic Compounds dissolve by dissociation, where
water surrounds the separated ions.
 Molecular compounds interact with water, but most do
NOT dissociate.
 Some molecular substances react with water when
they dissolve.
Aqueous
Reactions
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NaCl (s)  Na+ (aq) + Cl- (aq)
H2O molecule
Polar molecule
Why is water such an effective solvent for ionic compounds?
H
d-
Negative region
(the O atom)
d+
O
Positive region
(the H atom)
H
d+
Hydration is the process in which an ion is surrounded
by water molecules arranged in a specific manner.
d-
d+
H2O
H
O
O H
H O
H
H
Na+
H
ClH
O
H
Polar compounds: hydrogen
bonding
• Negative “pole” of water pointed
toward H on -OH groups of polar
molecule; positive “pole” of water
pointed toward O on -OH group of
polar molecule.
• Clustering of H2O molecules
around entire molecules keeps the
molecules apart from each other.
Ionic compounds: hydration
• Negative “pole” of water pointed
toward cation; positive “pole” of
water pointed toward anion.
• Clustering of H2O molecules
around Na+ and Cl- ions keeps the
ions apart from each other.
H
HO
O
CH2
H
O H
OH
H
H
H
OH
O
H
O
OH
OH
O
H
H
Electrolytes and Nonelectrolytes
• An electrolyte is a substance that dissociates
into ions when dissolved in water.
• A nonelectrolyte may dissolve in water, but it
does not dissociate into ions when it does so.
Aqueous
Reactions
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Electrolytes
• A strong electrolyte dissociates completely when
dissolved in water.
• A weak electrolyte only dissociates partially when
dissolved in water.
• A nonelectrolyte does NOT dissociate in water. Aqueous
Reactions
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An electrolyte is a substance that, when dissolved in
water, results in a solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved,
results in a solution that does not conduct electricity.
nonelectrolyte
weak electrolyte
strong electrolyte
Substances that, when dissolved in H2O,
produce solutions that can conduct
electricity.
A substance that, when dissolved in H2O,
does not produces a solution that can
conduct electricity.
Conduct electricity in solution?
Cations (+) and Anions (-)
Strong Electrolyte – 100% dissociation
NaCl (s)
H 2O
Na+ (aq) + Cl- (aq)
Weak Electrolyte – not completely dissociated
CH3COOH
CH3COO- (aq) + H+ (aq)
Strengths of Electrolytes:
Complete, Partial, or No Ionization in Water
STRONG ELECTROLYTES: Complete Ionization
Hydrochloric acid
+
-
H (aq) + Cl (aq)
HCl (aq)
NaC2H3O2 (aq)
Sodium acetate
(salts in general)
+
Na (aq) +
C2H3O2
(aq)
Strengths of Electrolytes:
Complete, Partial, or No Ionization in Water
WEAK ELECTROLYTES: Partial Ionization
Acetic acid
HC2H3O2 (aq)
Reversible reaction: the reaction can
proceed in both directions
• Forward reaction occurs as fast as
back reaction --- equilibrium
NON-ELECTROLYTES: No Ionization
Glucose
C6H12O6 (aq)
H+ (aq) + C2H3O2- (aq)
Refer to Table 4.1,
p. 107,
Chang 7th ed.
For a partial list
of electrolytes and
nonelectrolytes.
.......
Nonelectrolyte does not conduct electricity?
No cations (+) and anions (-) in solution
C6H12O6 (s)
H 2O
C6H12O6 (aq)
Strong Electrolyte Weak Electrolyte
Nonelectrolyte
HCl
CH3COOH
(NH2)2CO
HNO3
HF
CH3OH
HClO4
HNO2
C2H5OH
NaOH
H2O
C12H22O11
Ionic Compounds
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Which solution, NaCl(aq) or CH3OH(aq), conducts
electricity?
a. NaCl(aq)
b. CH3OH(aq)
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Which solution, NaCl(aq) or CH3OH(aq), conducts
electricity?
a. NaCl(aq)
b. CH3OH(aq)
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Hwk: page 156-163:
15, 17, 19
Atoms,
Molecules,
and Ions
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LESSON 2
4-2 Precipitation Reactions
Atoms,
Molecules,
and Ions
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Precipitation Reactions
When two solutions containing soluble salts are
mixed, sometimes an insoluble salt will be
produced. A salt “falls” out of solution, like snow
out of the sky. This solid is called a precipitate.
Aqueous
Reactions
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Precipitation reactions: AgNO3 (aq) + NaCl (aq) 
AgCl (s) + NaNO3 (aq)
• Formation of an insoluble product, the precipitate
• Usually involve ionic compounds
+
NaCl (s)
NaCl (aq)
AgCl (s), NaNO3 (aq)
AgNO3 (aq)
Precipitate: insoluble solid that separates from the
solution
AgCl (s)
Precipitation reactions
and
1. Can we predict
if a precipitate will
form if we mix 2
solutions or add a
compound to a
solution?
2. If the answer to
No. 1 above is
“YES”, how do we
do it?
SOLUBILITY
The maximum amount of solute that will dissolve in a
given quantity of solvent at a specific temperature.
Solubility Rules for Common Ionic Compounds
(applies to water at 25oC)
Li+, Na+, K+, Rb+, Cs+, NH4+,
NO3-, C2H3O2-, ClO4-, also
bicarbonate (HCO3-) and
chlorate (ClO3-)
Ag+, Hg22+, Pb2+
Ca2+, Sr2+, Ba2+,
Hg22+, Pb2+, also
silver (Ag+)
CO32-, PO43-, C2O42-,
also chromates (CrO42-)
and sulfites (SO32-)
S2-, OH-
Solubility of Ionic Compounds
• Not all ionic compounds dissolve in water.
• A list of solubility rules is used to decide
what combination of ions will dissolve.
Aqueous
Reactions
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Solubility Rules for Common Ionic Compounds in Water (at 25oC)
Soluble Compounds
Exceptions
All compounds containing alkali metal
ions (Li+, Na+, K+, Rb+, Cs+) and
ammonium ion (NH4+)
Nitrates (NO3-), acetates (C2H3O2-),
bicarbonates (HCO3-), chlorates (ClO3-),
perchlorates (ClO4-)
Halides (Cl-, Br-, I-)
Halides of silver (Ag+), mercury(I) (Hg22+), lead(II)
(Pb2+)
Sulfates (SO42-)
Sulfates of Group 2 metal ions (Be2+, Mg2+, Ca2+, Sr2+,
Ba2+), silver (Ag+), mercury(I) (Hg22+), lead(II) (Pb2+)
Insoluble Compounds
Exceptions
Sulfides (S2-)
Sulfides of Groups 1 and 2 metal ions (Li+, Na+, K+,
Rb+, Cs+, Be2+, Mg2+, Ca2+, Sr2+, Ba2+) and NH4+
Hydroxides (OH-)
Hydroxides of Group 1 metal ions (Li+, Na+, K+, Rb+,
Cs+), Ba2+, and NH4+
Carbonates (CO32-), oxalates (C2O42-),
phosphates (PO43-), chromates (CrO42-),
sulfites (SO32-)
Those of Group 1 metal ions (Li+, Na+, K+, Rb+, Cs+)
and NH4+
Solubility Rules for Common Ionic Compounds
In water at 250C
Soluble Compounds
Exceptions
Compounds containing alkali
metal ions and NH4+
NO3-, HCO3-, ClO3-
Cl-, Br-, ISO4
2-
Halides of Ag+, Hg22+, Pb2+
Sulfates of Ag+, Ca2+, Sr2+, Ba2+,
Hg2+, Pb2+
Insoluble Compounds
Exceptions
CO32-, PO43-, CrO42-, S2-
Compounds containing alkali
metal ions and NH4+
OH-
Compounds containing alkali
metal ions and Ba2+
Precipitation Reactions
Precipitate – insoluble solid that separates from solution
precipitate
Pb(NO3)2 (aq) + 2NaI (aq)
PbI2 (s) + 2NaNO3 (aq)
molecular equation
Pb2+ + 2NO3- + 2Na+ + 2I-
PbI2 (s) + 2Na+ + 2NO3-
ionic equation
Pb2+ + 2IPbI2
PbI2 (s)
net ionic equation
Na+ and NO3- are spectator ions
Ways to Write Metathesis Reactions
1) Molecular equation
2) Complete ionic equation
3) Net ionic equation
Aqueous
Reactions
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Molecular Equation
The molecular equation lists the reactants
and products without indicating the ionic
nature of the compounds.
AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Complete Ionic Equation
• In the complete ionic equation all strong
electrolytes (strong acids, strong bases, and soluble
ionic salts) are dissociated into their ions.
• This more accurately reflects the species that are
found in the reaction mixture.
Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq) 
AgCl(s) + K+(aq) + NO3−(aq)
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Net Ionic Equation
• To form the net ionic equation, cross out anything
that does not change from the left side of the
equation to the right.
• The ions crossed out are called spectator ions, K+
and NO3−, in this example.
• The remaining ions are the reactants that form the
product—an insoluble salt in a precipitation reaction,
as in this example.
Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq) 
AgCl(s) + K+(aq) + NO3−(aq)
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Writing Net Ionic Equations
1. Write a balanced molecular equation.
2. Dissociate all strong electrolytes.
3. Cross out anything that remains
unchanged from the left side to the
right side of the equation.
4. Write the net ionic equation with the
species that remain.
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Molecular and Ionic Equations
Spectator ion
Spectator ion
Spectator ions:
Ions that are not involved in the overall
reaction
Molecular and Ionic Equations
Recap:
1. Molecular equation
• Formulas of all compounds involved are written as
though all species existed as whole units.
• Doesn’t describe accurately what’s going on at the
microscopic level.
2. (Complete)Ionic equation
• Shows dissolved species as free ions.
• Spectator ions--not involved in overall reaction.
3. Net ionic equation
• Shows only the species that take part in the reaction.
Molecular and Ionic Equations
How to write ionic and net ionic equations:
1. Write a balanced molecular equation for the reaction.
2 AgNO3 (aq) + K2CrO4 (aq)  Ag2CrO4 (s) + 2 KNO3 (aq)
2. Rewrite the equation to show the dissociated ions that form in solution.
• Strong electrolytes ionize into cations and anions when dissolved in solution.
2 Ag+ (aq) + 2 NO3- (aq) + 2 K+ (aq) + CrO42- (aq) 
Ag2CrO4 (s) + 2 K+ (aq) + 2 NO3- (aq)
3. Identify and cancel spectator ions on both sides of the equation. Result: the net
ionic equation.
2 Ag+ (aq) + CrO42- (aq)  Ag2CrO4 (s)
(net ionic equation)
Molecular and Ionic Equations
How to write ionic and net ionic equations:
1. Write a balanced molecular equation for the reaction.
Pb(NO3)2 (aq) +2 NaI (aq)  PbI2 (s) + 2 NaNO3 (aq)
2. Rewrite the equation to show the dissociated ions that form in solution.
• Strong electrolytes ionize into cations and anions when dissolved in solution.
Pb2+ (aq) + 2 NO3- (aq) + 2 Na+ (aq) + 2 I- (aq) 
PbI2 (s) + 2 Na+ (aq) + 2 NO3- (aq)
3. Identify and cancel spectator ions on both sides of the equation. Result: the net
ionic equation.
Pb2+ (aq) + 2 I- (aq)  PbI2 (s)
(net ionic equation)
Molecular and Ionic Equations
How to write ionic and net ionic equations:
1. Write a balanced molecular equation for the reaction.
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
2. Rewrite the equation to show the dissociated ions that form in solution.
• Strong electrolytes ionize into cations and anions when dissolved in solution.
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq) 
AgCl (s) + K+ (aq) + NO3- (aq)
3. Identify and cancel spectator ions on both sides of the equation. Result: the net
ionic equation.
Ag+ (aq) + Cl- (aq)  AgCl (s)
(net ionic equation)
Writing Net Ionic Equations
1. Write the balanced molecular equation.
2. Write the ionic equation showing the strong electrolytes
3. Determine precipitate from solubility rules
4. Cancel the spectator ions on both sides of the ionic equation
© 2015 Pearson Education, Inc.
Which ions remain in solution after PbI2 precipitation
is complete?
a. K+ and I–
b. Pb2+ and I–
c. K+ and NO3–
d. Pb2+ and NO3–
© 2015 Pearson Education, Inc.
Which ions remain in solution after PbI2 precipitation
is complete?
a. K+ and I–
b. Pb2+ and I–
c. K+ and NO3–
d. Pb2+ and NO3–
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Hwk: page 156-163:
21, 23, 25, 27, 29, 93, 107
Quiz to Follow
Atoms,
Molecules,
and Ions
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LESSON 3
4-3 Acid Base Reactions
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrus
fruits contain citric acid.
React with certain metals to produce hydrogen gas.
React with carbonates and bicarbonates to produce carbon
dioxide gas
Bases
Have a bitter taste.
Feel slippery. Many soaps contain bases.
4.3
General Properties of Acids and Bases
Acids
• Sour taste
• Cause color changes in plant
dyes (change litmus from blue to
red)
• React with certain metals (e.g.,
Zn, Mg, Fe) to produce hydrogen
gas:
Mg (s) + 2HCl (aq)  MgCl2 (aq)
+ H2 (g)
• React with carbonates (CO32-) &
bicarbonates (HCO3-) to produce carbon
dioxide (CO2) gas:
2HCl(aq) + CaCO3(s) 
CaCl2(aq) + H2O(l) + CO2(g)
HCl(aq) + NaHCO3(s) 
NaCl(aq) + H2O(l) + CO2(g)
• Aqueous acid solutions conduct electricity
Bases
• Bitter taste
• Feel slippery
• Cause color changes in plant
dyes (change litmus from red to
blue)
• Aqueous base solutions conduct
electricity
Acids and Bases Around the House…
Acids
• Aspirin
(acetylsalicylic acid)
• Orange juice
(citric acid)
• Vinegar
(dilute solution of acetic acid)
• Muriatic acid
(concentrated hydrochloric acid)
Bases
• Ammonia
(concentrated ammonium hydroxide)
• Washing soda
(sodium carbonate)
• Baking soda
(sodium bicarbonate)
• Drain cleaner
(sodium & potassium hydroxides)
Basic Compounds in Antacids
54
Acids and Metals
Acids react with metals
• such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn
• to produce hydrogen gas and the salt of the metal
Molecular equations:
2K(s) + 2HCl(aq)  2KCl(aq) + H2(g)
metal
acid
salt
hydrogen gas
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
metal
acid
salt
hydrogen gas
55
Learning Check
Write a balanced equation for the reaction of magnesium
metal with HCl(aq). Label the metal, acid, and salt.
56
Learning Check
Write a balanced equation for the reaction of magnesium
metal with HCl(aq).
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
metal
acid
salt
57
Gas-Forming Reactions
 Some metathesis reactions do not give the
product expected.
 When a carbonate or bicarbonate reacts with
an acid, the products are a salt, carbon
dioxide, and water.
CaCO3(s) + 2 HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)
NaHCO3(aq) + HBr(aq) NaBr(aq) + CO2(g) + H2O(l)
Aqueous
Reactions
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Acids and Carbonates
Acids react
• with carbonates and hydrogen carbonates
• to produce carbon dioxide gas, a salt, and water
2HCl(aq) + CaCO3(s)  CO2(g) + CaCl2(aq) + H2O(l)
acid
carbonate
carbon
salt
water
dioxide
HCl(aq) + NaHCO3(s)  CO2(g) + NaCl (aq) + H2O(l)
acid
bicarbonate
carbon
salt
water
dioxide
59
Learning Check
Write a balanced equation for the following reactions:
A. MgCO3(s) + HBr(aq) 
B. HCl(aq) + NaHCO3(aq) 
60
Solution
Write a balanced equation for the following reactions:
A. MgCO3(s) + 2HBr(aq)  MgBr2(aq) + CO2(g) + H2O(l)
carbonate
acid
salt
carbon water
dioxide
B. HCl(aq) + NaHCO3(aq)  NaCl(aq) + CO2(g) + H2O(l)
acid
bicarbonate
salt
carbon water
dioxide
61
Neutralization Reactions
In a neutralization reaction,
• an acid reacts with a base to produce salt and water
• the acid HCl reacts with NaOH to produce salt and water
• the salt formed is the anion from the acid and cation of
the base
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
acid
base
salt
water
62
Neutralization Reactions
In neutralization reactions,
• if we write the strong acid and strong base as ions, we
see that H+ reacts with OH− to form water, leaving the ions
Na+ and Cl- in solution:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) 
Na+(aq) + Cl-(aq) + H2O(l)
• the overall reaction is H3O+ from the acid and OH- from
the base form water:
H+(aq) + OH-(aq)  H2O(l)
63
Acid-base reactions: HCl (aq) + NaOH (aq)  H2O (aq) + NaCl (aq)
0.2 M
NaOH (aq)
0.2 M HCl (aq)
H2O (l), NaCl (aq)
Acid-base reactions: Neutralization reactions
• Acid and base essentially cancel each other out (neutralize
each other) to produce water and a salt:
HCl (aq) + NaOH (aq)  H2O (l) + NaCl (aq)
(acid)
(base)
(water)
(salt)
NH4OH (aq) + HCl (aq)  H2O (l) + NH4Cl (aq)
(base)
(acid)
(water)
(salt)
Acid-Base Neutralization Reactions
H2O
acid + base  salt + water
An ionic compound
made up of a cation
other than H+ and an
anion other than OHor O2-
H+ donor
H+ acceptor
Acid-Base Reactions
In an acid–base reaction, the acid (H2O above)
donates a proton (H+) to the base (NH3 above).
Reactions between an acid and a base are called
neutralization reactions.
When the base is a metal hydroxide, water and a
salt (an ionic compound) are produced.
Aqueous
Reactions
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Neutralization Reactions
When a strong acid (like HCl) reacts with a strong
base (like NaOH), the net ionic equation is circled
below:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) 
Na+(aq) + Cl−(aq) + H2O(l)
H+(aq) + OH−(aq)  H2O(l)
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Guide for Balancing Neutralization
Reactions
69
Balancing Neutralization Reactions
Write the balanced equation for the neutralization of
magnesium hydroxide and nitric acid.
Step 1 Write the reactants and products.
Mg(OH)2 + HNO3
Step 2 Balance the H+ in the acid with the OH- in the base.
Mg(OH)2 + 2HNO3
Step 3 Balance the H2O with H+ and the OH-.
Mg(OH)2 + 2HNO3  salt + 2H2O
Step 4 Write the salt from the remaining ions.
Mg(OH)2 + 2HNO3  Mg(NO3)2 + 2H2O
70
Learning Check
Select the correct group of coefficients for each of the
following neutralization equations.
1. HCl(aq) + Al(OH)3(aq)  AlCl3(aq) + H2O(l)
A. 1, 3, 3, 1
B. 3, 1, 1, 1
C. 3, 1, 1, 3
2. Ba(OH)2(aq) + H3PO4(aq)  Ba3(PO4)2(s) + H2O(l)
A. 3, 2, 2, 2
B. 3, 2, 1, 6
C. 2, 3, 1, 6
71
Solution
1. HCl(aq) + Al(OH)3(aq)  AlCl3(aq) + H2O(l)
Step 1 Write the reactants and products.
HCl + Al(OH)3
Step 2
Balance the H+ in the acid with the OH- in the
base.
3HCl + Al(OH)3
Step 3 Balance the H2O with H+ and the OH-.
3HCl + Al(OH)3  salt + 3H2O
Step 4 Write the salt from the remaining ions.
3HCl(aq) + Al(OH)3(aq)  AlCl3(aq) + 3H2O(l)
The answer is C. 3, 1, 1, 3.
72
Solution
2. Ba(OH)2(aq) + H3PO4(aq)  Ba3(PO4)2(s) + H2O(l)
Step 1 Write the reactants and products.
Ba(OH)2 + H3PO4
Step 2 Balance the H+ in the acid with the OH- in the base.
3Ba(OH)2 + 2H3PO4
Step 3 Balance the H2O with H+ and the OH-.
3Ba(OH)2 + 2H3PO4  salt + 6H2O
Step 4 Write the salt from the remaining ions.
3Ba(OH)2(aq) + 2H3PO4(aq)
 Ba3(PO4)2(s) + 6H2O(l)
The answer is B. 3, 2, 1, 6.
73
Learning Check
Write the neutralization reactions for stomach acid, HCl,
and the ingredients in Mylanta.
Mylanta: Al(OH)3 and Mg(OH)2
74
Solution
Write the neutralization reactions for stomach acid, HCl, and
the ingredients in Mylanta.
Mylanta: For Al(OH)3:
Step 1 Write the reactants and products.
Al(OH)3 + HCl
Step 2 Balance the H+ in the acid with the OH- in the base.
Al(OH)3 + 3HCl
Step 3 Balance the H2O with H+ and the OH-.
Al(OH)3 + 3HCl  salt + 3H2O
Step 4 Write the salt from the remaining ions.
Al(OH)3(aq) + 3HCl(aq)  AlCl3(aq) + 3H2O(l)
75
Solution
Write the neutralization reactions for stomach acid, HCl, and
the ingredients in Mylanta.
Mylanta: For Mg(OH)2:
Step 1 Write the reactants and products.
Mg(OH)2 + HCl
Step 2 Balance the H+ in the acid with the OH- in the base.
Mg(OH)2 + 2HCl
Step 3 Balance the H2O with H+ and the OH-.
Mg(OH)2 + 2HCl  salt + 2H2O
Step 4 Write the salt from the remaining ions.
2HCl(aq) + Mg(OH)2(aq)  MgCl2(aq) + 2H2O(l)
76
Sample Problem 8.8 Balancing Equations of Acids
Write the balanced equation for the neutralization of HCl(aq) and Ba(OH)2(s).
Solution
Step 1 Write the reactants and products.
HCl(aq) + Ba(OH)2(s) → salt + H2O(l)
Step 2 Balance the H+ in the acid with the OH– in the base. Placing a coefficient of 2 in front of HCl
provides 2H+ for the 2OH– in Ba(OH)2.
2HCl(aq) + Ba(OH)2(s) → salt + H2O(l)
Step 3 Balance the H2O with the H+ and the OH–. Use a coefficient of 2 in front of H2O to balance 2H+
and 2OH–.
2HCl(aq) + Ba(OH)2(s) → salt + 2H2O(l)
Step 4 Write the salt from the remaining ions. Use the ions Ba2+ and 2Cl– to write the formula of the
BaCl2.
salt,
2HCl(aq) + Ba(OH)2(s) → BaCl2(aq) + 2H2O(l)
Study Check 8.8
Write the balanced equation for the reaction between H2SO4(aq) and NaHCO3(aq).
Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
4.3
Acid-Base Definitions:
1. Arrhenius acids & bases:
Formulated by the Swedish chemist Svante Arrhenius (1859-1927) in
the late 19th century, to classify substances whose properties in
aqueous solution were well known.
• Acids: Substances that ionize in water to produce H+ ions.
HCl (aq)  H+ (aq) + Cl- (aq)
• Bases: Substances that ionize in water to produce OH- ions.
NaOH (aq)  Na+ (aq) + OH- (aq)
A Brønsted acid is a proton donor
A Brønsted base is a proton acceptor
base
acid
acid
base
A Brønsted acid must contain at least one
ionizable proton!
4.3
Acid-Base Definitions:
2. Bronsted-Lowry acids & bases:
Formulated simultaneously by the Danish chemist Johannes Bronsted
(1879-1947), and the British chemist Thomas Lowry (1874-1936), in
1923 to cover substances dissolved in water and other solvents (e.g.,
alcohols).
• Acids: Substances that donate H+ ions (a.k.a. protons).
HC2H3O2 (aq) = H+ (aq) + C2H3O2- (aq)
(acetic acid)
(acetate anion)
• Bases: Substances that accept H+ ions.
C2H3O2- (aq) + H2O (l) =
HC2H3O2 (aq) + OH- (aq)
(acetate anion)
(acetic acid)
Acids
• The Swedish physicist
and chemist S. A.
Arrhenius defined acids
as substances that
increase the
concentration of H+
when dissolved in water.
• Both the Danish chemist
J. N. Brønsted and the
British chemist T. M.
Lowry defined them as
proton donors.
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Bases
• Arrhenius defined bases
as substances that
increase the
concentration of OH−
when dissolved in water.
• Brønsted and Lowry
defined them as proton
acceptors.
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Bronsted-Lowry Acids & Bases
HC2H3O2 (aq) = H+ (aq) + C2H3O2- (aq)
(acetic acid)
(proton)
should be
written as…
(acetate anion)
Can’t exist alone in solution due to
its strong attraction to the negative
pole (the O atom) in H2O.
Consequently, the proton is
coordinated to H2O molecules (i.e.,
hydrated) in aqueous solution.
HC2H3O2(aq) + H2O(l) = H3O+(aq) + C2H3O2-(aq)
(acetic acid)
(hydronium ion)
Chemists write the hydrogen ion as “H+” in
acid-base reactions as a shorthand, in which
the presence of water as solvent is understood.
(acetate anion)
Bronsted-Lowry Acids & Bases
Monoprotic Acids: Those acids which, upon ionization in H2O, donate
only one proton.
Examples:
HCl (aq)  H+ (aq) + Cl- (aq)
HCl is a strong electrolyte--ionizes completely in H2O.
HC2H3O2 (aq) = H+ (aq) + C2H3O2- (aq)
HC2H3O2 is a weak electrolyte--ionizes partially in H2O.
Bronsted-Lowry Acids & Bases
Polyprotic Acids: Those acids which, upon ionization in H2O, donate
more than one proton.
Example: Sulfuric acid (H2SO4) -- a diprotic acid
(donates 2 H+ upon ionization in H2O)
H2SO4 (aq)  H+ (aq) + HSO4- (aq)
HSO4- (aq) = H+ (aq) + SO42- (aq)
H2SO4 is a strong electrolyte, ionizes completely into H+ and
HSO4-, as represented by the single arrow.
HSO4- is a weak electrolyte, ionizes partially into H+ and SO42-,
as represented by the double arrow.
Example: Phosphoric acid (H3PO4) -- a triprotic acid
(donates 3 H+ upon ionization in H2O)
H3PO4 (aq) = H+ (aq) + H2PO4- (aq)
H2PO4- (aq) = H+ (aq) + HPO42- (aq)
HPO42- (aq) = H+ (aq) + PO43- (aq)
Monoprotic acids
HCl
H+ + Cl-
HNO3
H+ + NO3H+ + CH3COO-
CH3COOH
Strong electrolyte, strong acid
Strong electrolyte, strong acid
Weak electrolyte, weak acid
Diprotic acids
H2SO4
H+ + HSO4-
Strong electrolyte, strong acid
HSO4-
H+ + SO42-
Weak electrolyte, weak acid
Triprotic acids
H3PO4
H2PO4HPO42-
H+ + H2PO4H+ + HPO42H+ + PO43-
Weak electrolyte, weak acid
Weak electrolyte, weak acid
Weak electrolyte, weak acid
4.3
STRONG ACIDS
Ionize completely…see list
HCl  H+ + Cl-
Strong or Weak?
• Strong acids completely dissociate in water;
weak acids only partially dissociate.
• Strong bases dissociate to metal cations and
hydroxide anions in water; weak bases only
partially react to produce hydroxide anions.
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Weak Acid
=> an acid that only partially dissociates to H+ ions in water.
ex)
HF(aq) ↔ H+ (aq) + F- (aq)
note the double arrow indicates that the reaction does
not go to completion
Bronsted-Lowry Acids & Bases
Monobases: Those bases which, in H2O, can accept only one proton.
Example: Hydroxide ion (OH-)
H+ (aq) + OH- (aq)  H2O (l)
Example: Ammonia (NH3)
NH3 (aq) + H+ (aq) = NH4+ (aq)
Polybases: Those bases which, in H2O, can accept more than one proton.
Example: Phosphate anion (PO43- ) -- a tribase (can accept 3 H+)
PO43- (aq) + H+ (aq) = HPO42- (aq)
HPO42- (aq) + H+ (aq) = H2PO4- (aq)
H2PO4- (aq) + H+ (aq) = H3PO4 (aq)
In H2O, weak Bronsted bases undergo a hydrolysis reaction. This
term means that the weak Bronsted base reacts with H2O to form an
acid and hydroxide ion (OH-).
NH3 (aq) + H2O (l) = NH4+ (aq) + OH- (aq)
PO43- (aq) + H2O (l) = HPO42- (aq) + OH- (aq)
HPO42- (aq) + H2O (l) = H2PO4- (aq) + OH- (aq)
H2PO4- (aq) + H2O (l) = H3PO4 (aq) + OH- (aq)
Strong Bronsted bases ionize completely in H2O.
NaOH (aq)  Na+ (aq) + OH- (aq)
Ba(OH)2 (aq)  Ba2+ (aq) + 2OH- (aq)
Weak Bases:
a base that is only partially dissociated to form OH- ions in
water.
ex)
NH3(aq) + H2O(l)
↔ NH4 + (aq)
+ OH-
Neutralization Reaction
1) Strong Acid – Strong Base Reaction
acid + base
salt + water
All salts are strong electrolytes.
HCl (aq) + NaOH (aq)
H+ + Cl- + Na+ + OH-
H+ + OH-
NaCl (aq) + H2O
Na+ + Cl- + H2O
H2O
2) Weak Acid – Strong Base
Two Step Reaction
(1) HB(aq) ↔ H+ (aq)
+
B- (aq)
(2) H+ (aq) + OH- (aq) → HOH
(3) Net Ionic: HB(aq) + OH- (aq) → B- (aq) + HOH
ex) HF(aq) + OH- (aq) → HOH + F- (aq)
3) Strong Acid – Weak Base
Two Step Reaction
(1) B + HOH → BH+ (aq) + OH- (aq)
(2) H+ (aq) + OH- (aq) → HOH
(3) Net Ionic: H+ (aq) + B(aq) →
ex) H+ (aq) + NH3(aq) →
NH4 + (aq)
BH + (aq)
Hwk: page 156-163:
31, 33, 35, 37, 39, 41
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
LESSON 4
4-4 Oxidation-Reduction
Reactions
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
Oxidation-reduction (redox) reactions:
H2O2 (aq) + 2KI (aq) + 2HCl (aq)  I2 (aq) + 2KCl (aq)
I2 (aq) + 2Na2S2O3 (aq)  2NaI (aq) + Na2S4O6 (aq)
+
+
Na2S2O3 (aq)
+
KI (aq)
1M
HCl
(aq)
0.1 M
H2 O2
(aq)
I2 (aq),
KCl (aq),
NaI (aq),
Na2S4O6 (aq)
Oxidation-Reduction Reactions
2 Na0  2 Na+ + 2 eLoss of electrons from Na metal to form Na+ cations
OXIDATION
- 2 e-
Cl2 (g) + 2 Na (s)  2 NaCl (s)
+ 2 e-
Cl20 + 2 e-  2 ClGain of electrons from Cl2 gas to form Cl- anions
REDUCTION
Oxidation-Reduction Reactions
2 Na0  2 Na+ + 2 e-
OXIDATION HALF-REACTION
Cl20 + 2 e-  2 Cl-
REDUCTION HALF-REACTION
0
Cl2 + 2
Na0
 2
Na+ and Clcombine
to form
NaCl
Na+
+ 2
Cl-
Add
the half-reactions…
Electrons cancel
each other
Cl2 (g) + 2 Na (s)  2 NaCl (s)
OVERALL EQUATION: SUM OF THE HALF-REACTIONS
O
I
L
xidation
s
oss (of electrons)
R
I
G
eduction
s
ain (of electrons)
Oxidation-Reduction Reactions
•
•
•
•
Loss of electrons is oxidation.
Gain of electrons is reduction.
One cannot occur without the other.
The reactions are often called redox reactions.
How many electrons does each oxygen atom gain
during the course of this reaction?
a.
None
b.
One
c.
Two
d.
Four
How many electrons does each oxygen atom gain
during the course of this reaction?
a.
None
b.
One
c.
Two
d.
Four
Identifying Redox Reactions
Reducing Agent
•
•
•
•
Causes reduction
Loses one or more electrons
Undergoes oxidation
Oxidation number of atom increases
Oxidizing Agent
•
•
•
•
Causes oxidation
Gains one or more electrons
Undergoes reduction
Oxidation number of atom decreases
e-
eCl
Cl
Na
OXIDIZING AGENT
Accepts e- from Na, causes Na
to be oxidized to Na+
+ e-
+ e-
Na
REDUCING AGENT
Donates e- to Cl2, causes Cl
to be reduced to Cl-
-e-
- e-
Na+
Cl-
Cl-
Na+
Oxidation Number
• a.k.a. oxidation state --- the number of charges an atom would have in
a molecule or an ionic compound if electrons were transferred
completely.
oxidation numbers
0
Formation of an
ionic compound:
0
+1
-1
Cl2 (g) + 2 Na (s)  2 NaCl (s)
oxidation numbers
Formation of a
molecular
compound:
0
0
+5
-2
P4 (s) + 5 O2 (g)  P4O10 (s)
Oxidation Number
Seven rules for assigning oxidation numbers:
1.
In FREE elements (uncombined e.g. H2, P4, S8, Ag, Au, Fe), each atom has
an oxidation number of ZERO.
2. For ions containing only one atom (monatomic ions), the oxidation
number is equal to the charge on the ion.
Examples familiar to you: Alkali metals (Group 1) – ox. # is +1.
Alkaline earth metals (Group 2) – ox. # is +2.
Aluminum ion – ox. # is +3 (all its compounds)
3.
The oxidation number of oxygen in most compounds (e.g. MgO and H2O)
is –2. However, in H2O2 (hydrogen peroxide) and O22- (peroxide anion),
the ox. # is –1.
4.
The oxidation number of hydrogen is +1, EXCEPT when it is bonded to
metals in binary compounds (e.g. LiH and CaH2).
Oxidation Number
Seven rules for assigning oxidation numbers:
5.
Fluorine has an oxidation number of –1 in ALL its compounds. Other
halogens (Cl, Br, I) have NEGATIVE oxidation numbers when they occur
as HALIDE ions in their compounds. When combined with oxygen (e.g.
in ClO4-, HClO4), the halogens have POSITIVE oxidation numbers.
6.
In a NEUTRAL molecule or ionic compound, the SUM of the oxidation
numbers of ALL the atoms must equal ZERO. In a POLYATOMIC ION,
the sum of oxidation numbers of all the elements in the ion must equal the
NET CHARGE ON THE ION.
7.
Oxidation numbers do not have to be integers. Example: ox. # of O in the
superoxide ion, O2-,is –1/2.
IT’S A WISE IDEA TO REFER TO THE OXIDATION TABLE I
GAVE YOU AND BECOME FAMILIAR WITH THE
OXIDATION NUMBERS OF THE VARIOUS ELEMENTS,
ESPECIALLY THE COMMON OXIDATION NUMBERS.
Oxidation Number
Additional information:
1.
Metallic elements have only positive oxidation numbers. Nonmetallic
elements, however, may have either positive or negative oxidation
numbers.
2.
The HIGHEST oxidation number an element in Groups 1A through 7A
can have is its group number. The halogens, for example, can possess an
oxidation number of +7.
3.
The transition metals (Groups 1B through 8B) can possess several
oxidation numbers.
Oxidation Numbers
To determine if an oxidation–reduction
reaction has occurred, we assign an oxidation
number to each element in a neutral
compound or charged entity.
Rules to Assign Oxidation Numbers
• Elements in their elemental form have an
oxidation number of zero.
• The oxidation number of a monatomic ion
is the same as its charge.
Rules to Assign Oxidation Numbers
• Nonmetals tend to have negative oxidation
numbers, although some are positive in
certain compounds or ions.
– Oxygen has an oxidation number of −2, except
in the peroxide ion, in which it has an oxidation
number of −1.
– Hydrogen is −1 when bonded to a metal, +1
when bonded to a nonmetal.
Rules to Assign Oxidation Numbers
• Nonmetals tend to have negative oxidation
numbers, although some are positive in
certain compounds or ions.
– Fluorine always has an oxidation number of −1.
– The other halogens have an oxidation number
of −1 when they are negative; they can have
positive oxidation numbers, most notably in
oxyanions.
Rules to Assign Oxidation Numbers
• The sum of the oxidation numbers in a
neutral compound is zero.
• The sum of the oxidation numbers in a
polyatomic ion is the charge on the ion.
Oxide vs Peroxide
Oxidation Number
Assign oxidation numbers to all the elements in the following
compounds and ions:
S4O62-
KMnO4
Fe (IO4)2
Fe(CN)64-
PtCl62-
NF3
Oxidation Number
Consider the reaction of aqueous copper (II) sulfate with zinc metal:
CuSO4 (aq) + Zn (s)  Cu (s) + ZnSO4 (aq)
The products are Cu metal and zinc sulfate.
Which reactant is oxidized?
Zn0  Zn2+ + 2e-
Which reactant is reduced?
Cu2+ + 2e-  Cu0
Oxidation number
of Cu2+ ions is +2
Overall
Reaction:
Oxidation number
of Cu metal is 0
Cu2+ (aq) + Zn0 (s)  Cu0 (s) + Zn2+ (aq)
Identifying Redox Reactions
Oxidizing Agent
reduction
0
4Fe(s) + 3O2(g)
0
Reducing Agent
Copyright © 2008
Pearson Prentice Hall,
2Fe2 O3 (s)
+3
oxidation
Chapter 4/142
-2
Identifying Redox Reactions
Oxidizing Agent
reduction
+3
0
2Fe2O3 (s) + 3 C (s)
4Fe(s) + 3 C O2 (g)
0
Reducing Agent
Copyright © 2008
Pearson Prentice Hall,
+4
oxidation
Chapter 4/143
Ex)
Zn +
2 HCl
 ZnCl2 + H2
Identify what species is oxidized and reduced?
Identify oxidizing and reducing agents.
Ex)
2Al
+
3Cl2  2AlCl3
Identify what species is oxidized and reduced?
Identify oxidizing and reducing agents.
Ex)
CH4
+
2O2
 CO2 +
2H2O
Identify what species is oxidized and reduced?
Identify oxidizing and reducing agents.
Ex)
PbO
+ CO
 Pb + CO2
Identify what species is oxidized and reduced?
Identify oxidizing and reducing agents.
Displacement Reactions
In displacement
reactions, ions
oxidize an element.
In this reaction,
silver ions oxidize
copper metal:
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
The reverse reaction does NOT occur. Why not?
Why does the solution in Fig. 4.14 turn blue?
a.
Cu2+ (aq) ions are blue.
b.
NO3– (aq) ions are blue.
c.
Ag(s) is blue.
d.
H2O is blue.
Why does the solution in Fig. 4.14 turn blue?
a.
Cu2+ (aq) ions are blue.
b.
NO3– (aq) ions are blue.
c.
Ag(s) is blue.
d.
H2O is blue.
Activity Series
• Elements
higher on the
activity series
are more
reactive.
• They are more
likely to exist
as ions.
Metal/Acid Displacement Reactions
• The elements above hydrogen will react with
acids to produce hydrogen gas.
• The metal is oxidized to a cation.
How many moles of hydrogen gas would be
produced for every mole of magnesium added
into the HCl solution?
a.
None
b.
One
c.
Two
d.
Four
How many moles of hydrogen gas would be
produced for every mole of magnesium added
into the HCl solution?
a.
None
b.
One
c.
Two
d.
Four
Hwk: page 156-163:
45, 49, 51, 53
Quiz to Follow
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
LESSON 5
4-5 Concentration of
Solutions
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
grams of solute = M . V . (molar mass)
Molarity
• The quantity of solute in a solution can matter to
a chemist.
• We call the amount dissolved its concentration.
• Molarity is one way to measure the
concentration of a solution:
Molarity (M) =
moles of solute
volume of solution in liters
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Mixing a Solution
• To create a solution of a known molarity, weigh out
a known mass (and, therefore, number of moles) of
the solute.
• Then add solute to a volumetric flask, and add
solvent to the line on the neck of the flask.
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Dilution Formula
M1. V1 = M2 . V2
moles of solute
before dilution
moles of solute
after dilution
Dilution
• One can also dilute a more concentrated
solution by
– using a pipet to deliver a volume of the solution to a
new volumetric flask, and
– adding solvent to the line on the neck of the new flask.
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Dilution
The molarity of the new solution can be
determined from the equation
Mc  Vc = Md  Vd,
where Mc and Md are the molarity of the
concentrated and dilute solutions,
respectively, and Vc and Vd are the
volumes of the two solutions.
Aqueous
Reactions
© 2015 Pearson Education, Inc.
What is the concentration of a solution made by
dissolving 0.2500 g of K2CrO4 (potassium
chromate) in 500.0 mL of distilled H2O?
What is the concentration of a new solution made
by dilution of 10.00 mL of the above K2CrO4
solution to 1000.0 mL with distilled H2O?
How many grams of K2CrO4 are contained in the
new solution?
Hwk: page 156-163:
60, 62, 63, 65, 66, 67, 69, 73, 75, 77
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
LESSON 6
4-6 Solution Stoichiometry
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
Using Molarities in
Stoichiometric Calculations
Aqueous
Reactions
© 2015 Pearson Education, Inc.
Titration
A titration is an analytical technique in which one
can calculate the concentration of a solute in a
solution.
Aqueous
Reactions
© 2015 Pearson Education, Inc.
© 2015 Pearson Education, Inc.
How would the volume of standard solution added
change if that solution were Ba(OH)2(aq) instead
of NaOH(aq)?
a. Increase by one-half the volume used for titration
with NaOH.
b. Increase by two the volume used for titration with NaOH.
c. Decrease by two the volume used for titration with NaOH.
d. Decrease by one-half the volume used for titration
with NaOH.
© 2015 Pearson Education, Inc.
How would the volume of standard solution added
change if that solution were Ba(OH)2(aq) instead
of NaOH(aq)?
a. Increase by one-half the volume used for titration
with NaOH.
b. Increase by two the volume used for titration with NaOH.
c. Decrease by two the volume used for titration with NaOH.
d. Decrease by one-half the volume used for titration
with NaOH.
© 2015 Pearson Education, Inc.
Titration
• A solution of known concentration, called a
standard solution, is used to determine the
unknown concentration of another solution.
• The reaction is complete at the equivalence
Aqueous
point.
Reactions
© 2015 Pearson Education, Inc.
Titration of
0.200 M NaOH
with 0.200 M HCL
Start: 5.00 mL of 0.200 M NaOH
diluted to 50.0 mL with distilled
H2 O
Indicator: Bromothymol blue
G
B
Y
pH 6.0
pH 7.6
pH 7.0
pH 12.30
0.0 mL HCl
added
pH 7.00
5.0 mL HCl
added
pH 3.14
5.2 mL HCl
added
See handout
Hwk: page 156-163:
79, 81, 83, 85, 87,
Atoms,
Molecules,
and Ions
© 2015 Pearson Education, Inc.
Review Questions
Chapter 4
Stoichiometry
© 2015 Pearson Education, Inc.
A homogeneous mixture of two
or more components is
referred to as
a.
b.
c.
d.
a solute.
a solution.
an electrolyte.
a mess.
© 2015 Pearson Education, Inc.
A homogeneous mixture of two
or more components is
referred to as
a.
b.
c.
d.
a solute.
a solution.
an electrolyte.
a mess.
© 2015 Pearson Education, Inc.
The solvent in a sample of
soda pop is
a.
b.
c.
d.
sugar.
carbon dioxide.
water.
air.
© 2015 Pearson Education, Inc.
The solvent in a sample of
soda pop is
a.
b.
c.
d.
sugar.
carbon dioxide.
water.
air.
© 2015 Pearson Education, Inc.
The gaseous solute in a
sample of soda pop is
a.
b.
c.
d.
sugar.
carbon dioxide.
water.
air.
© 2015 Pearson Education, Inc.
The gaseous solute in a
sample of soda pop is
a.
b.
c.
d.
sugar.
carbon dioxide.
water.
air.
© 2015 Pearson Education, Inc.
Gatorade® and other sports
drinks conduct electricity
because they contain
a.
b.
c.
d.
water.
sugar.
air.
electrolytes.
© 2015 Pearson Education, Inc.
Gatorade® and other sports
drinks conduct electricity
because they contain
a.
b.
c.
d.
water.
sugar.
air.
electrolytes.
© 2015 Pearson Education, Inc.
When Fe(NO3)2 dissolves
in water, the particles in
solution are
a.
b.
c.
d.
Fe+ and (NO3)2–.
Fe2+ and 2 NO3–.
Fe and 2 NO3.
Fe and N2 and 3 O2.
© 2015 Pearson Education, Inc.
When Fe(NO3)2 dissolves
in water, the particles in
solution are
a.
b.
c.
d.
Fe+ and (NO3)2–.
Fe2+ and 2 NO3–.
Fe and 2 NO3.
Fe and N2 and 3 O2.
© 2015 Pearson Education, Inc.
Which set includes only
substances that produce
electrolytes in water?
a.
b.
c.
d.
NaBr, KCl, MgSO4
C6H12O6, CH3OH, C6H6
HCl, NH3, Cl2, N2
SiO2, CaCO3, H2SO4
© 2015 Pearson Education, Inc.
Which set includes only
substances that produce
electrolytes in water?
a.
b.
c.
d.
NaBr, KCl, MgSO4
C6H12O6, CH3OH, C6H6
HCl, NH3, Cl2, N2
SiO2, CaCO3, H2SO4
© 2015 Pearson Education, Inc.
Which compound below is not
soluble in water?
a.
b.
c.
d.
NaBr
KNO3
MgSO4
ZnS
© 2015 Pearson Education, Inc.
Which compound below is not
soluble in water?
a.
b.
c.
d.
NaBr
KNO3
MgSO4
ZnS
© 2015 Pearson Education, Inc.
Pb(NO3)2 + 2 KI  PbI2 + 2 KNO3
The physical evidence that the
above reaction occurs is
a.
b.
c.
d.
an explosion.
the formation of a gas.
that the solution boils.
the formation of a precipitate.
© 2015 Pearson Education, Inc.
Pb(NO3)2 + 2 KI  PbI2 + 2 KNO3
The physical evidence that the
above reaction occurs is
a.
b.
c.
d.
an explosion.
the formation of a gas.
that the solution boils.
the formation of a precipitate.
© 2015 Pearson Education, Inc.
Which pair of compounds will
produce a precipitate if solutions
of appropriate concentrations
are mixed together?
a.
b.
c.
d.
H2SO4 and NaOH
HNO3 and CaCl2
Ba(NO3)2 and Na3PO4
LiCl and SrI2
© 2015 Pearson Education, Inc.
Which pair of compounds will
produce a precipitate if solutions
of appropriate concentrations
are mixed together?
a.
b.
c.
d.
H2SO4 and NaOH
HNO3 and CaCl2
Ba(NO3)2 and Na3PO4
LiCl and SrI2
© 2015 Pearson Education, Inc.
When an acid reacts with a
base, the result is
a.
b.
c.
d.
cancellation.
elimination.
neutralization.
adduct formation.
© 2015 Pearson Education, Inc.
When an acid reacts with a
base, the result is
a.
b.
c.
d.
cancellation.
elimination.
neutralization.
adduct formation.
© 2015 Pearson Education, Inc.
When nitric acid is neutralized
by potassium hydroxide, the
spectator ions are
a.
b.
c.
d.
K+ and NO3–.
H+ and OH–.
H+ and NO3–.
K+ and OH–.
© 2015 Pearson Education, Inc.
When nitric acid is neutralized
by potassium hydroxide, the
spectator ions are
a.
b.
c.
d.
K+ and NO3–.
H+ and OH–.
H+ and NO3–.
K+ and OH–.
© 2015 Pearson Education, Inc.
Which compound below is not
a strong acid?
a.
b.
c.
d.
HC2H3O2
H2SO4
HNO3
HBr
© 2015 Pearson Education, Inc.
Which compound below is not
a strong acid?
a.
b.
c.
d.
HC2H3O2
H2SO4
HNO3
HBr
© 2015 Pearson Education, Inc.
When an atom undergoes
oxidation, it ______ electrons.
a.
b.
c.
d.
gains
loses
retains
balances
© 2015 Pearson Education, Inc.
When an atom undergoes
oxidation, it ______ electrons.
a.
b.
c.
d.
gains
loses
retains
balances
© 2015 Pearson Education, Inc.
When an atom undergoes
reduction, it ______ electrons.
a.
b.
c.
d.
gains
loses
retains
balances
© 2015 Pearson Education, Inc.
When an atom undergoes
reduction, it ______ electrons.
a.
b.
c.
d.
gains
loses
retains
balances
© 2015 Pearson Education, Inc.
When Zn(s) reacts with
HCl(aq) to produce H2(g) and
ZnCl2(aq), the zinc is _____
because it _____ electrons.
a.
b.
c.
d.
reduced; gains
reduced; loses
oxidized; gains
oxidized; loses
© 2015 Pearson Education, Inc.
When Zn(s) reacts with
HCl(aq) to produce H2(g) and
ZnCl2(aq), the zinc is _____
because it _____ electrons.
a.
b.
c.
d.
reduced; gains
reduced; loses
oxidized; gains
oxidized; loses
© 2015 Pearson Education, Inc.
Al + H+  Al3+ + H2
When the oxidation–reduction
reaction above is correctly
balanced, the coefficients are
a.
b.
c.
d.
1, 2  1, 1.
1, 3  1, 2.
2, 3  2, 3.
2, 6  2, 3.
© 2015 Pearson Education, Inc.
Al + H+  Al3+ + H2
When the oxidation–reduction
reaction above is correctly
balanced, the coefficients are
a.
b.
c.
d.
1, 2  1, 1.
1, 3  1, 2.
2, 3  2, 3.
2, 6  2, 3.
© 2015 Pearson Education, Inc.
In the list shown, the metal
that is most easily oxidized
is ___.
a.
b.
c.
d.
Li
Ca
Fe
Cu
© 2015 Pearson Education, Inc.
In the list shown, the metal
that is most easily oxidized
is ___.
a.
b.
c.
d.
Li
Ca
Fe
Cu
© 2015 Pearson Education, Inc.
A solution is prepared by
dissolving 35.0 g of NaCl in
water to make 500 mL of
solution. What is the molarity?
a.
b.
c.
d.
7.00 M
3.04 M
1.97 M
1.20 M
© 2015 Pearson Education, Inc.
A solution is prepared by
dissolving 35.0 g of NaCl in
water to make 500 mL of
solution. What is the molarity?
a.
b.
c.
d.
7.00 M
3.04 M
1.97 M
1.20 M
© 2015 Pearson Education, Inc.
250.0 mL of 0.100 M AgNO3
solution contains _____ g of
silver nitrate.
a.
b.
c.
d.
4.25
8.50
17.0
34.0
© 2015 Pearson Education, Inc.
250.0 mL of 0.100 M AgNO3
solution contains _____ g of
silver nitrate.
a.
b.
c.
d.
4.25
8.50
17.0
34.0
© 2015 Pearson Education, Inc.
To make 250.0 mL of 0.500 M
KI solution, _____ mL of
6.00 M KI must be used.
a.
b.
c.
d.
20.8
41.7
500.0
3000.0
© 2015 Pearson Education, Inc.
To make 250.0 mL of 0.500 M
KI solution, _____ mL of
6.00 M KI must be used.
a.
b.
c.
d.
20.8
41.7
500.0
3000.0
© 2015 Pearson Education, Inc.
HCl + KOH  KCl + H2O
25.00 mL of HCl was titrated
using 18.50 mL of 0.1554 M
NaOH. The concentration of
HCl is
a.
b.
c.
d.
0.0777 M.
0.1150 M.
0.2100 M.
0.2875 M.
© 2015 Pearson Education, Inc.
HCl + KOH  KCl + H2O
25.00 mL of HCl was titrated
using 18.50 mL of 0.1554 M
NaOH. The concentration of
HCl is
a.
b.
c.
d.
0.0777 M.
0.1150 M.
0.2100 M.
0.2875 M.
© 2015 Pearson Education, Inc.
HX + NaOH  NaX + H2O
229 mg of HX was titrated
using 29.33 mL of 0.0965 M
NaOH. What is element X?
a.
b.
c.
d.
F
Cl
Br
I
© 2015 Pearson Education, Inc.
HX + NaOH  NaX + H2O
229 mg of HX was titrated
using 29.33 mL of 0.0965 M
NaOH. What is element X?
a.
b.
c.
d.
F
Cl
Br
I
© 2015 Pearson Education, Inc.