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Chapter 4
Reactions in Aqueous Solutions
4.1 General Properties of Aqueous Solutions
• Solution - a homogeneous mixture
– Solute: the component that is dissolved
– Solvent: the component that does the dissolving
Generally, the component present in the greatest
quantity is considered to be the solvent. Aqueous
solutions are those in which water is the solvent.
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– Electrolyte: substance that dissolved in water
produces a solution that conducts electricity
• Contains ions:
–
Nonelectrolyte: substance that dissolved in water and
produces a solution that does not conduct electricity
•
Does not contain ions:
•
Dissociation - ionic compounds separate into
constituent ions when dissolved in solution
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• Ionization - formation of ions by molecular
compounds when dissolved
• Strong and weak electrolytes:
– Strong Electrolyte: 100% dissociation
• All water soluble ionic compounds, strong
acids and strong bases
– Weak electrolytes: Partially ionized in solution
• Exist mostly as the molecular form in
solution
• Weak acids and weak bases
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• Examples of weak electrolytes:
– Weak acids:
HC2H3O2(aq)

C2H3O2 (aq) + H+ (aq)
– Weak bases:
NH3 (aq) + H2O(l)

NH4+ (aq) + OH (aq)
(Note: double arrows indicate a reaction that occurs in
both directions - a state of dynamic equilibrium exists)
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nonelectrolyte
CH3OH
weak electrolyte strong electrolyte
H2CO3
NaOH
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4.2 Precipitation Reactions
• Precipitation (formation of a solid from two aqueous
solutions) occurs when product is insoluble
Solubility Rules of salts in water:

Most Nitrate (NO3-) salts are soluble

Most salts containing the alkali metal (group I) ions
(Li+, Na+, K+, Cs+, Rb+) and the ammonium ion
(NH4+) are soluble

Most chloride, bromide and iodide salts are soluble,
except salts containing Ag+, Pb2+ and Hg22+

Most sulfate (SO42-) salts are soluble,
except salts containing Ba2+, Pb2+, Hg22+ and Ca2+

Most hydroxide salts (OH-) salts are only slightly
soluble.
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The compounds Ba(OH)2, Sr(OH)2 and Ca(OH)2 are
marginally soluble. NaOH and KOH are soluble

Most sulfide (S2-), carbonate (CO32-), chromate
(CrO42-), and phosphate (PO43-) salts are only slightly
soluble
• Hydration: process by which polar water molecules
remove and surround individual ions from the solid.
Identify the precipitate:
Pb(NO3)2(aq) + 2NaCI(aq)  2NaNO3(?) + PbCI2 (?)
Which is soluble or insoluble in water:
Ba(NO3)2
soluble
AgCI
insoluble
Mg(OH)2
insoluble
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• Molecular equation: shows all compounds
represented by their chemical formulas
• Ionic equation: shows all strong electrolytes as ions
and all other substances (non- electrolytes, weak
electrolytes, gases) by their chemical formulas
Net Ionic equation: shows only the reacting species
in the chemical equation
– Eliminates spectator ions:
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Aqueous solutions of silver nitrate and sodium
sulfate are mixed. Write the net ionic reaction:
2AgNO3(aq)+Na2SO4(aq)
 2NaNO3(?)+Ag2SO4(?)
2Ag+(aq) + 2NO3(aq) + 2Na+(aq) + SO42(aq)
 2Na+(aq) + 2NO3(aq) + Ag2SO4(s)
2Ag+(aq) + SO42(aq)  Ag2SO4(s)
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4.3 Acid-Base Reactions:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
H+(aq) + OH-(aq)  H2O(l)
acid
base
water
All the other acids and bases are weak electrolytes.
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• Definitions of acids and bases:
– Arrhenius acid - produces H+ in solution
– Arrhenius base - produces OH in solution
_ Brønsted acid - proton donor
_ Brønsted base - proton acceptor
– Examples of a weak base and weak acid
• Ammonia with water:
• Hydrofluoric acid with water:
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• Types of acids:
– Monoprotic: one ionizable hydrogen
HCl + H2O  H3O+ + Cl
– Diprotic: two ionizable hydrogens
H2SO4 + H2O  H3O+ + HSO4
HSO4 + H2O  H3O+ + SO42
– Triprotic: three ionizable hydrogens:
H3PO4 + H2O  H3O+ + H2PO4
H2PO4 + H2O  H3O+ + HPO42
HPO42 + H2O  H3O+ + PO43
Polyprotic: generic term meaning more than one
ionizable hydrogen
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• Types of bases:
– Monobasic: One OH group
KOH  K+ + OH
– Dibasic: Two OH groups
Ba(OH)2  Ba2+ + 2OH
• Neutralization: Reaction between an acid and a base
Acid + Base  Salt + Water
Molecular equation:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Ionic equation:
H+(aq)+ Cl(aq) + Na+(aq) + OH(aq)
 Na+(aq) + Cl(aq) + H2O(l)
Net ionic equation: (formation of water)
H+(aq) + OH(aq)  H2O(l)
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4.4 Oxidation-Reduction Reactions
• Often called “redox” reactions
• Electrons are transferred between the reactants
– One substance is oxidized, loses electrons
•
Reducing agent
– Another substance is reduced, gains electrons
• Oxidizing agent
• Oxidation numbers change during
the reaction
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Example:
Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
– Zinc is losing 2 electrons and oxidized.
• Reducing agent
• Zn(s)  Zn2+(aq) + 2e
– Copper ions are gaining the 2 electrons.
• Oxidizing agent
• Cu2+(aq) + 2e  Cu(s)
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• Rules for assigning oxidation numbers

Oxidation state of an atom in an element is zero
e.g: Na(s), O2(g), O3(g) and Hg(l)

Oxidation state of an ion is its charge:
e.g: Na+ and Cl-

Fluorine in its compounds is always assigned
an oxidation state of -1

Oxygen is usually assigned an oxidation state of
-2, except in:
1. Peroxides is assigned an oxidation state of -1
(e.g. H2O2)
2. OF2 is assigned an oxidation state of +2
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 Hydrogen in its covalent compounds (with nonmetals) is assigned an oxidation state of +1 and
in its hydrides (with metals) is assigned an
oxidation state of -1.

The sum of the oxidation states for a neutral
compound must be zero.
Oxidation numbers must sum to the overall
charge of the species:
SO42 = 2
(O is usually 2)
X + 4(2) = 2
X  8 = 2

X = +6
(S is +6)
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Assign oxidation numbers for all elements in each
species:
MgBr2
(Mg +2, Br 1)
ClO2
(Cl +1 , O 2)
Al2 (SO4) 3
(Al +3, SO4 2]
KMnO4
(K +1, MnO4 1]
Displacement reactions:
– A common reaction: active metal replaces
(displaces) a metal ion from a solution:
Mg(s) + CuCl2(aq)  Cu(s) + MgCl2(aq)
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Activity series:
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• Balancing redox reactions:
– Electrons (charge) must be balanced as well as
number and types of atoms
– Consider this net ionic reaction:
Al(s) + Ni2+(aq)  Al3+(aq) + Ni(s)
– Divide reaction into two half-reactions:
Al(s)  Al3+(aq) + 3e
Ni2+(aq) + 2e  Ni(s)
(oxidation)
(reduction)
– Make electrons gained equal electrons lost:
2 [Al(s)  Al3+(aq) + 3e ]
3 [Ni2+(aq) + 2e  Ni(s) ]
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– Cancel electrons and write balanced net ionic
reaction:
2Al(s)  2Al3+(aq) + 6e
3Ni2+(aq) + 6e  3Ni(s)
- Sum the two half-reactions:
2Al(s) + 3Ni2+(aq)  2Al3+(aq) + 3Ni(s)
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• Combination Reactions:
– Many combination reactions may also be
classified as redox reactions
– Consider the reaction between Hydrogen gas
reacts and oxygen gas:
2H2(g) + O2(g)  2H2O(l)
Identify the substance oxidized and the Substance
reduced.
• Decomposition reactions:
– Many decomposition reactions may also be
classified as redox reactions
– Consider strong heating of Potassium chlorate:
2KClO3(s)  2KCl(s) + 3O2(g)
Identify substances oxidized and reduced.
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• Disproportionation reactions:
– One element undergoes both oxidation and
reduction
– Consider:
• Combustion reactions:
– Common example, hydrocarbon fuel reacts with
oxygen to produce carbon dioxide and water
– Consider:
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4.5 Concentration of Solutions
• Concentration is the amount of solute dissolved in a
given amount of solvent.
• Qualitative expressions of concentration
– Concentrated – higher ratio of solute to solvent
– Dilute - smaller ratio of solute to solvent
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• Quantitative concentration term:
– Molarity is the ratio of moles solute per liter of
solution:
– Symbols: M or [ ]
Calculate the molarity of a solution prepared by
dissolving 45.00 grams of KI into a total volume of
500.0 mL.
45.00 g KI 1 mol KI 1000 mL


 0.5422 M
500.0 mL 166.0 g KI
1L
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How many milliliters of 3.50 M NaOH can be prepared
from 75.00 grams of the solid?
75.00 g NaOH 
1 mol NaOH
1L
1000 mL


 536 mL
40.00 g NaOH 3.50 mol NaOH
1L
• Dilution:
– Process of preparing a less concentrated solution
from a more concentrated one.
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For the next experiment the class will need 250. mL of 0.10
M CuCl2. There is a bottle of 2.0 M CuCl2. Describe how to
prepare this solution. How much of the 2.0 M solution do we
need?
McLc = MdLd
(2.0 M) (Lc) = (0.10 M) (250.mL)
Lc = 12.5 mL
• Solution Stoichiometry:
– Soluble ionic compounds dissociate completely
in solution.
– Using mole ratios we can calculate the
concentration of all species in solution.
NaCl
dissociates into
Na+ and Cl
Na2SO4
dissociates into
2Na+ and SO42
AlCl3
dissociates into
Al3+ and 3Cl
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Find the concentration of all species in a 0.25 M
solution of MgCl2
MgCl2  Mg2+ + 2Cl
Given:
MgCl2 = 0.25 M
[Mg2+ ] = 0.25 M (1:1 ratio)
[Cl ] = 0.50 M (1:2 ratio)
Using the square bracket notation, express the molar
concentration for all species in the following solutions:
0.42 M Ba(OH)2
[Ba2+ ] = 0.42 M
(1:1 ratio)
[OH ] = 0.84 M
(2:1 ratio)
1.2 M NH4Cl
[NH4+ ] = 1.2 M
(1:1 ratio)
[Cl ]
(1:1 ratio)
= 1.2 M
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4.6 Aqueous Reactions and Chemical Analysis
• Types of quantitative analysis:
– Gravimetric analysis:
Example:
(mass analysis)
precipitation reaction
– Volumetric analysis:
Example:
(volume analysis)
titration
• Gravimetric Analysis:
One form: isolation of a precipitate
A 0.825 g sample of an ionic compound containing
chloride ions and an unknown metal is dissolved in water
and treated with excess silver nitrate. If 1.725 g of AgCl
precipitate forms, what is the percent by mass of Cl in the
original sample?
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• Volumetric analysis:
– Commonly accomplished by titration
• Addition of a solution of known concentration
(standard solution) to another solution of
unknown concentration.
– Standardization is the determination of the exact
concentration of a solution.
– Equivalence point represents completion of the
reaction.
– Endpoint is where the titration is stopped.
nacid = nbase
– An indicator is used to signal the endpoint.
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Calculate the molarity of 25.0 mL of a monoprotic acid if
it took 45.50 mL of 0.25 M KOH to neutralize the acid.
nacid = nbase
For monoprotic acid and base:
(MV)acid = (MV)base
0.01338 mol acid
 0.455 M
0.0250 L
Calculate the concentration of the acid or base remaining
in solution when 10.7 mL of 0.211M HNO3 is added to
16.3 mL of 0.258 M NaOH.
A) 7.21 x10-2 M NaOH