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Reactions in Aqueous Solution
Chapter 4
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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
aqueous solutions
of KMnO4
2
An electrolyte is a substance that, when dissolved in
water, results in a solution that can conduct electricity.
weak electrolyte is a substance that, when dissolved in
water, results in a solution that can conduct small
electricity
A nonelectrolyte is a substance that, when dissolved,
results in a solution that does not conduct electricity.
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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)
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Ionization of acetic acid
CH3COOH
CH3COO- (aq) + H+ (aq)
A reversible reaction. The reaction can
occur in both directions.
Acetic acid is a weak electrolyte because its
ionization in water is incomplete.
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Hydration is the process in which an ion is surrounded
by water molecules arranged in a specific manner.
d-
d+
H2O
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Nonelectrolyte does not conduct electricity?
No cations (+) and anions (-) in solution
C6H12O6 (s)
H 2O
C6H12O6 (aq)
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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
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Solubility is the maximum amount of solute that will dissolve
in a given quantity of solvent at a specific temperature.
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Solubility Rules
1. Group IA and ammonium compounds are
soluble.
2. Acetates and nitrates are soluble.
3. Most chlorides, bromides, and iodides are
soluble.
Exceptions:
AgCl, Hg2Cl2, PbCl2;
AgBr, HgBr2, Hg2Br2, PbBr2;
AgI, HgI2, Hg2I2, PbI2
4. Most sulfates are soluble.
Exceptions:
CaSO4, SrSO4, BaSO4,
Ag2SO4, Hg2SO4, PbSO4
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5. Most carbonates are insoluble.
Exceptions:
Group IA carbonates and (NH4)2SO4
6. Most phosphates are insoluble.
Exceptions:
Group IA phosphates and (NH4)3PO4
7. Most sulfides are insoluble.
Exceptions:
Group IA sulfides and (NH4)2S
8. Most hydroxides are insoluble.
Exceptions:
Group IA hydroxides,
Ca(OH)2, Sr(OH)2, Ba(OH)2
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Decide whether the following reaction
occurs. If it does, write the molecular,
ionic, and net ionic equations.
KBr + MgSO4 
1. Determine the product formulas:
K+ and SO42− make K2SO4
Mg2+ and Br − make MgBr2
2. Determine whether the products are soluble:
K2SO4 is soluble
MgBr2 is soluble
KBr + MgSO4  no reaction
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Examples of Insoluble Compounds
CdS
PbS
Ni(OH)2
Al(OH)3
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Writing Net Ionic Equations
1. Write the balanced molecular equation.
2. Write the ionic equation showing the strong electrolytes
completely dissociated into cations and anions.
3. Cancel the spectator ions on both sides of the ionic equation
4. Check that charges and number of atoms are balanced in the
net ionic equation
Write the net ionic equation for the reaction of silver nitrate
with sodium chloride.
AgNO3 (aq) + NaCl (aq)
AgCl (s) + NaNO3 (aq)
Ag+ + NO3- + Na+ + Cl-
AgCl (s) + Na+ + NO3-
Ag+ + Cl-
AgCl (s)
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Properties of Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrus
fruits contain citric acid.
Cause color changes in plant dyes.
React with certain metals to produce hydrogen gas.
2HCl (aq) + Mg (s)
MgCl2 (aq) + H2 (g)
React with carbonates and bicarbonates
to produce carbon dioxide gas
2HCl (aq) + CaCO3 (s)
CaCl2 (aq) + CO2 (g) + H2O (l)
Aqueous acid solutions conduct electricity.
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Properties of Bases
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Cause color changes in plant dyes.
Aqueous base solutions conduct electricity.
Examples:
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Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
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Hydronium ion, hydrated proton, H3O+
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In Figure A, a solution
of HCl (a strong acid)
illustrated on a
molecular/ionic level,
shows the acid as all
ions.
In Figure B, a solution of
HF (a weak acid) also
illustrated on a
molecular/ionic level,
shows mostly molecules
with very few ions.
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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!
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Household Acids and
Bases
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Monoprotic acids
HCl
H+ + Cl-
HNO3
Strong electrolyte, strong acid
H+ + NO3H+ + CH3COO-
CH3COOH
Strong electrolyte, strong acid
Weak electrolyte, weak acid
Polyprotic Acid
An acid that results in two or more acidic hydrogens per molecule.
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
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23
Identify each of the following species as a Brønsted acid, base,
or both. (a) HI, (b) CH3COO-, (c) H2PO4-
HI (aq)
H+ (aq) + I- (aq)
CH3COO- (aq) + H+ (aq)
H2PO4- (aq)
Brønsted acid
CH3COOH (aq)
H+ (aq) + HPO42- (aq)
H2PO4- (aq) + H+ (aq)
H3PO4 (aq)
Brønsted base
Brønsted acid
Brønsted base
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Neutralization Reaction
acid + base
HCl (aq) + NaOH (aq)
H+ + Cl- + Na+ + OH-
H+ + OH-
salt + water
NaCl (aq) + H2O
Na+ + Cl- + H2O
H2O
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Neutralization Reaction Involving a Weak
Electrolyte
weak acid + base
HCN (aq) + NaOH (aq)
HCN + Na+ + OH-
HCN + OH-
salt + water
NaCN (aq) + H2O
Na+ + CN- + H2O
CN- + H2O
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Neutralization Reaction Producing a Gas
acid + base
2HCl (aq) + Na2CO3 (aq)
2H+ + 2Cl- + 2Na+ + CO32-
2H+ + CO32-
salt + water + CO2
2NaCl (aq) + H2O +CO2
2Na+ + 2Cl- + H2O + CO2
H2O + CO2
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Acid−Base Reaction with Gas
Formation
Some salts, when treated with an acid,
produce a gas. Typically sulfides,
sulfites, and carbonates behave in this
way producing hydrogen sulfide, sulfur
trioxide, and carbon dioxide,
respectively.
The photo to the right shows
baking soda (sodium
hydrogen carbonate) reacting
with acetic acid in vinegar to
give bubbles of carbon
dioxide.
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Oxidation-Reduction Reactions
(electron transfer reactions)
2Mg
O2 + 4e-
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
2O2Reduction half-reaction (gain e-)
2Mg + O2 + 4e2Mg2+ + 2O2- + 4e29
2Mg + O2
2MgO
Zn (s) + CuSO4 (aq)
ZnSO4 (aq) + Cu (s)
Zn2+ + 2e- Zn is oxidized
Zn
Cu2+ + 2e-
Zn is the reducing agent
Cu Cu2+ is reduced Cu2+ is the oxidizing agent
Copper wire reacts with silver nitrate to form silver metal.
What is the oxidizing agent in the reaction?
Cu (s) + 2AgNO3 (aq)
Cu
Ag+ + 1e-
Cu(NO3)2 (aq) + 2Ag (s)
Cu2+ + 2eAg Ag+ is reduced
Ag+ is the oxidizing agent
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Oxidation number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an oxidation
number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal to
the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually –2. In H2O2
and O22- it is –1.
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4.4
4. The oxidation number of hydrogen is +1 except when
it is bonded to metals in binary compounds. In these
cases, its oxidation number is –1.
5. Group IA metals are +1, IIA metals are +2 and fluorine is
always –1.
6. The sum of the oxidation numbers of all the atoms in a
molecule or ion is equal to the charge on the
molecule or ion.
7. Oxidation numbers do not have to be integers.
Oxidation number of oxygen in the superoxide ion,
O2-, is –½.
-
HCO3
What are the oxidation numbers
of all the elements in HCO3- ?
O = –2
H = +1
3x(–2) + 1 + ? = –1
C = +4
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The Oxidation Numbers of Elements in their Compounds
33
What are the oxidation numbers of
all the elements in each of these
compounds?
NaIO3
IF7
K2Cr2O7
KIO3 K = +1 O = -2
3x(-2) + 1 + ? = 0
I = +5
Na2Cr2O7
IF7
F = -1
7x(-1) + ? = 0
I = +7
O = -2
Na = +1
7x(-2) + 2x(+1) + 2x(?) = 0
Cr = +6
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Potassium permanganate, KMnO4, is a
purple−colored compound; potassium
manganate, K2MnO4, is a green−colored
compound. Obtain the oxidation numbers
of the manganese in these compounds.
K
Mn
O
1(+1) + 1(oxidation number of Mn) + 4(−2) = 0
1 + 1(oxidation number of Mn) + (−8) = 0
(−7) + (oxidation number of Mn) = 0
Oxidation number of Mn = +7
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K
Mn
O
2(+1) + 1(oxidation number of Mn) + 4(−2) = 0
2 + 1(oxidation number of Mn) + (−8) = 0
(−6) + (oxidation number of Mn) = 0
Oxidation number of Mn = +6
In KMnO4, the oxidation number of Mn is +7.
In K2MnO4, the oxidation number of Mn is +6.
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What is the oxidation number of Cr in
dichromate, Cr2O72−?
Cr
O
2(oxidation number of Cr) + 7(−2) = −2
2(oxidation number of Cr) + (−14) = −2
2(oxidation number of Cr) = +12
Oxidation number of Cr = +6
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Oxidation
The half−reaction in which there is a loss of
electrons by a species (or an increase in
oxidation number).
Reduction
The half−reaction in which there is a gain of
electrons by a species (or a decrease in oxidation
number).
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Oxidizing Agent
A species that oxidizes another species; it is
itself reduced.
Reducing Agent
A species that reduces another species; it is itself
oxidized.
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Types of Oxidation-Reduction Reactions
Combination Reaction
A reaction in which two substances combine to form a
third substance.
A+B
0
0
2Al + 3Br2
C
+3 -1
2AlBr3
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For example:
2Na(s) + Cl2(g)  2NaCl(s)
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Decomposition Reaction
C
+1 +5 -2
2KClO3
A+B
+1 -1
0
2KCl + 3O2
A reaction in which a single
compound reacts to give two or
more substances.
For example:
2HgO(s)  2Hg(l) + O2(g)
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Types of Oxidation-Reduction Reactions
Combustion Reaction
A reaction in which a substance reacts with oxygen,
usually with the rapid release of heat to produce a flame.
A + O2
B
0
0
S + O2
0
0
2Mg + O2
+4 -2
SO2
+2 -2
2MgO
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Combustion Reaction
For example:
4Fe(s) + 3O2(g)  2Fe2O3(s)
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Types of Oxidation-Reduction Reactions
Displacement Reaction
A reaction in which an element reacts with a compound,
displacing another element from it.
A + BC
0
+1
+2
Sr + 2H2O
+4
0
TiCl4 + 2Mg
0
AC + B
-1
Cl2 + 2KBr
0
Sr(OH)2 + H2
0
Hydrogen Displacement
+2
Ti + 2MgCl2
-1
Metal Displacement
0
2KCl + Br2
Halogen Displacement
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Displacement Reaction
For example:
Zn(s) + 2HCl(aq) 
H2(g) + ZnCl2(aq)
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Types of Oxidation-Reduction Reactions
Disproportionation Reaction
The same element is simultaneously oxidized
and reduced.
Example:
reduced
+1
0
Cl2 + 2OH-
-1
ClO- + Cl- + H2O
oxidized
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Classify each of the following reactions.
Ca2+ + CO32NH3 + H+
Zn + 2HCl
Ca + F2
CaCO3
NH4+
ZnCl2 + H2
CaF2
Precipitation
Acid-Base
Redox (H2 Displacement)
Redox (Combination)
Mg(s)  Mg2+(aq) + 2e−
(oxidation)
Fe3+(aq) + 3e−  Fe(s)
(reduction)
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Solution Stoichiometry
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution
Molality= amount of substance in mol of solute/mass in Kg of the solvent
What mass of KI is required to make 500. mL of a
2.80 M KI solution?
M KI
M KI
volume of KI solution
500. mL x
1L
1000 mL
moles KI
x
2.80 mol KI
1 L soln
x
166 g KI
1 mol KI
grams KI
= 232 g KI
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To prepare a solution, add the measured amount
of solute to a volumetric flask, then add water to
bring the solution to the mark on the flask.
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Preparing a Solution of Known Concentration
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Dilution is the procedure for preparing a less concentrated
solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
=
Moles of solute
after dilution (f)
MiVi
=
MfVf
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Diluting a
solution
quantitatively
requires specific
glassware.
The photo at the
right shows a
volumetric flask
used in dilution.
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You place a 1.52−g of potassium
dichromate, K2Cr2O7, into a 50.0−mL
volumetric flask. You then add water to
bring the solution up to the mark on the
neck of the flask. What is the molarity of
K2Cr2O7 in the solution?
Molar mass of K2Cr2O7 is 294 g.
1 mol
1.52 g
294 g
 0.103 M
-3
50.0  10 L
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A saturated stock solution of NaCl is 6.00
M. How much of this stock solution is
needed to prepare 1.00−L of physiological
saline soluiton (0.154 M)?
M iVi  M fV f
M fVf
Vi 
Mi
(0.154 M )(1.00 L)
Vi 
6.00 M
Vi  0.0257L or 25.7 mL
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How would you prepare 30.0 mL of 0.100 M HNO3
from a stock solution of 2.00 M HNO3?
MiVi = MfVf
Mi = 2.00 M Mf = 0.100 M Vf = 0.0300 L
Vi =
MfVf
Mi
Vi = ? L
= 0.100 M x 0.0300 L = 0.00150 L = 1.50 mL
2.00 M
Dilute 1.50 mL of acid with water to a total volume
of 30.0 mL.
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