3. Water as a Solvent
Bent (V-shaped) molecule
with 105° bond angle.
Covalent bonds – electron
sharing between oxygen and
hydrogen.
Unequal sharing of electrons
due to electronegativity
differences cause water to be
a polar molecule.
4. Water as a Solvent
Partial charges (δ) form on
the water molecule due to
the unequal sharing of
electrons.
Water’s polarity causes it to
be a solvent for ionic
compounds.
5. Polar Water Molecules Interact with the
Positive and Negative Ions of a Salt
Assisting in the Dissolving Process
6. Solubility
Solubility of ionic substances in water varies
greatly.
Solubility depends on the relative
attractions of the ions and water molecules
for each other.
Once dissolved, an ionic compound
becomes hydrated, therefore ions disperse
independent of one another.
7. Solubility
Water will dissolve non-ionic substances
depending upon their structure.
If a polar bond exists within the
structure, the molecule can be subject
to being water soluble.
In general, polar substances dissolve in
polar solvents and nonpolar substances
dissolve in nonpolar solvents.
8. An Ethanol Molecule Contains a Polar O-
H Bond Similar to Those in the Water
Molecule
9. The Polar Water Molecule Interacts
Strongly with the Polar-O-H bond in
Ethanol
11. Strong and Weak Electrolytes
A solution in which a substance is
dissolved in water; the substance is the
solute and the solvent is the water.
A common property for a solution is
electrical conductivity.
Its ability to conduct an electric
current.
12. Strong and Weak Electrolytes
Pure water is not an electrical conductor.
Strong electrolytes conduct current very
efficiently.
Weak electrolytes conduct only a small
current.
Nonelectrolytes do not allow current to
flow.
14. Svante Arrhenius
1859-1927
Studied the nature of solutions and
theorized that conductivity of solutions
arose from the presence of ions.
Proved that the strength of
conductivity is directly related to the
number of ions present in solution.
15. Strong Electrolytes
Strong electrolytes are substances
that are completely ionized when
they are dissolved in water.
Soluble salts
Strong acids
Strong bases
17. Acids
Arrhenius discovered in his studies of
solutions that when acids were
dissolved in water they behaved as
strong electrolytes.
This result was directly related to an
acid’s ability to ionize in water.
An Acid is a substance that produces
H+
ions when it is dissolved in water.
19. Acids
HCl
H2Ou xuuuu
→ Η +
(αθ) + Χλ−
(αθ)
HNO3
H2Ou xuuuu
→ Η +
(αθ) + ΝΟ−
3(αθ)
H2SO4
H2Ou xuuuu
→ Η +
(αθ) + ΗΣΟ−
4(αθ)
An Acid is a substance that produces H+
ions when it
is dissolved in water.
20. Acids
HCl
H2Ou xuuuu
→ Η +
(αθ) + Χλ−
(αθ)
HNO3
H2Ou xuuuu
→ Η +
(αθ) + ΝΟ−
3(αθ)
H2SO4
H2Ou xuuuu
→ Η +
(αθ) + ΗΣΟ−
4(αθ)
In conductivity studies, virtually every molecule
ionizes. Therefore, strong electrolytes are strong
acids.
21. Strong Acids
Sulfuric acid, nitric acid and hydrochloric
acid are aqueous solutions and should be
written in chemical equations as such.
A strong acid is one that completely
dissociates into its ions. In aqueous solutions,
the HCl molecule does not exist.
Sulfuric acid can produce two H+
ions per
molecule. Only the first H+
ion completely
dissociates. The anion HSO4
-
remains partially
intact.
22. Strong Bases
Bases are soluble ionic compounds
containing the hydroxide ion (OH-
).
Strong bases are strong electrolytes
and these compounds ionize
completely in water.
25. Weak Electrolytes
Weak electrolytes are substances that
exhibit a small degree of ionization in
water.
They produce relatively few ions when
dissolved in water
Most common weak electrolytes are
weak acids and weak bases.
26. Weak Acids
Formulas for acids are often written with
the acidic hydrogen atom or atoms
(the hydrogen atoms that will produce
H+
ions in solution) listed first. If any
nonacidic hydrogens are present they
are written later in the formula.
HC2 H3O2(aq) + Η 2Ο(λ)
→← Η3Ο(αθ)
+
+ Χ2 Η3Ο2(αθ)
−
27. Weak Acids
In Acetic acid, only 1% of its molecules
ionize.
The double arrow indicates that the
reaction can occur in either direction.
Acetic acid is a weak electrolyte and
therefore a weak acid because it
dissociates (ionizes) only to a slight extent
in aqueous solutions.
HC2 H3O2(aq) + Η 2Ο(λ)
→← Η3Ο(αθ)
+
+ Χ2 Η3Ο2(αθ)
−
29. Weak Bases
The most common weak base is NH3.
In an aqueous solution, ammonia results
in a basic solution.
NH3(aq) + Η2Ο(λ)
→← ΝΗ4(αθ)
+
+ ΟΗ(αθ)
−
33. Solutions
Most chemical reactions take place in the
environment of solutions. In order to perform
stoichiometric calculations in solutions, one
must know two things.
The nature of the reaction; which depends
on the exact forms the chemicals take
when dissolved.
The amounts of the chemicals present in
the solutions, usually expressed as
concentrations.
34. Concentration
Molarity (M) – is moles of solute
per volume of solution in liters:
Example: 1.0M= 1.0 molar =
1.0moles solute/1liter of solution
M = µολαριτψ=
µολεσοφ σολυτε
λιτερσοφ σολυτιον
35. Example
Calculate the molarity of a solution
prepared by dissolving 11.5g of solid
NaOH in enough water to make 1.50L
of solution.
.192 M NaOH
36. Example
Calculate the molarity of a solution
prepared by dissolving 1.56g of
gaseous HCl in enough water to make
26.8 ml of solution.
1.60M HCl
37. Example
Give the concentration of each type
of ion in 0.50M Co(NO3)2.
Co2+
= 0.50 M Co2+
NO3
-
= 1.0 M NO3
-
38. Example
Calculate the number of moles of Cl-
ions in 1.75L of 1.0 x 10-3
M ZnCl2.
3.5 x 10-3
mol Cl-
39. Dilution
The process of changing the molarity of a
solution from a more concentrated
solution to a lesser concentrated solution.
Moles of solute after dilution = moles of
solute before dilution.
M x V= moles
M1V1=M2V2
40. Example
What volume of 16M sulfuric acid must
be used to prepare 1.5L of 0.10 M
H2SO4 solution?
9.4 ml solution
43. Types of Solution Reactions
Most solution reactions can be
put into three types of reactions:
Precipitation Reactions
Acid-Base Reactions
Oxidation-Reduction Reactions
44. Precipitation Reactions
When two solutions are mixed, an
insoluble substance sometimes forms;
that is, a solid forms and separates
from the solution.
The solid that forms is called a
precipitate.
50. Solubility
Predicting the identity of a solid product in a
precipitation reaction requires knowledge of
the solubilities of common ionic substances.
Slightly soluble – the tiny amount of solid that
dissolves is not noticeable. The solid appears
insoluble to the naked eye.
Insoluble and slightly soluble are often used
interchangeably.
53. Precipitation Reactions
Precipitation reactions move forward
due to the decrease in energy state
of the compound. Bonds forming in
the compound increase stability and
push the reaction forward.
55. Formula Equation
Although the formula equation shows
the reactants and products of the
reaction, it does not give a correct
picture of what actually occurs in
solution.
Gives the overall reaction
stoichiometry but not necessarily the
actual forms of the reactants and
products.
56. Complete Ionic Equation
The complete ionic equation better
represents the actual forms of the
reactants and products in solution. All
substances that are strong electrolytes
are represented as ions.
Complete ionic equation shows all
ions in a reaction, even those that
do not participate in the reaction.
These ions are called spectator ions.
57. Net Ionic Equation
The net ionic equation includes only
those solution components that are
directly involved in the reaction.
Commonly used because it gives
the actual forms of the reactants
and products and includes only the
species that undergo a change.
Spectator ions are not included.
58. Example Problem
For the following reaction, write the formula
equation, the complete ionic equations, and
the net ionic equation. Aqueous potassium
chloride is added to aqueous silver nitrate to
form a silver chloride precipitate plus
aqueous potassium nitrate.
KCl(aq) + ΑγΝΟ3(αθ) → ΑγΧλ(σ) + ΚΝΟ3(αθ)
K(aq)
+
+ Χλ(αθ)
−
+ Αγ(αθ)
+
+ ΝΟ3(αθ)
−
→ ΑγΧλ(σ) + Κ(αθ)
+
+ ΝΟ3(αθ)
−
Cl(aq)
−
+ Αγ(αθ)
+
→ ΑγΧλ(σ)
60. Solution Stoichiometry
The rules of stoichiometry and limiting
reactant apply to chemical reactions in
solutions. But two rules need special
emphasis.
Always write a balanced equation of the
reaction and give special attention to the
products that are formed and the true
form of the ions in solution.
Moles must still be calculated, but molarity
and volume must be used in the
calculation.
62. Example Problem
Determine the mass of solid NaCl that must be
added to 1.50L of a 0.100 M AgNO3 solution to
precipitate all the Ag+
ions in the form of AgCl.
Ag(aq)
+
+ Χλ(αθ)
−
→ ΑγΧλ(σ)
Determine the amount of Ag+
ions in solution.
0.100M Ag+
1Λ
ξ
1.50Λ
= .15µολεσΑγ+
Determine the amount of Cl-
needed to react with Ag+
.15molesAg+
ξ
1Χλ−
1Αγ+
ξ
1ΝαΧλ
1Χλ−
ξ
58.44γ
1µολε
= 8.77γΝαΧλ
63. Steps to Solution Stoichiometry
1. Identify the species present in the combined solution,
and determine what reaction occurs.
2. Write the balanced net ionic equation for the
reaction.
3. Calculate the moles of reactants.
4. Determine which reactant is limiting.
5. Calculate the moles of product or products, as
required.
6. Convert to grams or other units, as required.
65. Acids
Arrhenius’s concept of acids and bases:
An acid is a substance that produces H+
ions when dissolved in water and a base
is a substance that produces OH-
ions.
Fundamentally correct but doesn’t
include all bases.
66. Brønsted-Lowry Acids
Johannes BrØnsted (1879-1947) and
Lowry (1874-1936) gives a more
general definition of a base that
includes substances that do not
contain OH-
.
An acid is a proton donor.
A base is a proton acceptor.
67. Acid-Base Reactions
An Acid-Base reaction often forms two
things:
A salt (sometimes soluble)
Water
Since water is a nonelectrolyte, large
quantities of H+
and OH-
cannot coexist in
solution.
The net ionic equations is:
H(aq)
+
+ ΟΗ(αθ)
−
→ Η2Ο(λ)
68. Acid-Base Reactions
HCl(aq) + ΝαΟΗ(αθ) → ΝαΧλ(αθ) + Η2Ο(λ)
Η(αθ)
+
+ Χλ(αθ)
−
+ Να(αθ)
+
+ Χλ(αθ)
−
→ Να(αθ)
+
+ Χλ(αθ)
−
+ Η2Ο(λ)
Η(αθ)
+
+ ΟΗ(αθ)
−
→ Η2Ο(λ)
When a strong acid and a strong base
react, we expect both substances to
completely ionize. We then check to
see what will form that is soluble.
In this case, the salt is soluble and remains as ions. But
water, a nonelectrolyte, will form since H+
and OH-
have a
strong attraction for each other and do not ionize.
69. Acid-Base Reactions
HC2 H3O2(aq) + ΚΟΗ(αθ) →
ΟΗ(αθ)
−
+ ΗΧ2 Η3Ο2(αθ) → Η2Ο(λ) + Χ2 Η3Ο2(αθ)
−
When a weak acid and a strong base react, the
weak acid usually doesn’t ionize. However, the
hydroxide ion is such a strong base that for the
purposes of stoichiometric calculations it can be
assumed to react completely with any weak acid.
70. Performing Calculations for
Acid-Base Reactions
1. List the substances present in the combined
solution before any reaction occurs and
decide what reaction will occur.
2. Write the balanced net ionic equation for
this reaction.
3. Calculate the moles of reactants. For
reactions in solution, use the volumes of the
original solutions and their molarities.
71. Performing Calculations for
Acid-Base Reactions
4. Determine the limiting reactant where
appropriate.
5. Calculate the moles of the required
reactant or product.
6. Convert to grams or volume (of solution), as
required.
72. Acid-Base Reaction
An acid-base reaction is often called
a neutralization reaction.
When just enough base is added to
react exactly with the acid in a
solution, we say the acid has been
neutralized.
73. Step by Step Example
What volume of a .100 M HCl solution is
needed to neutralize 25.0ml of .350 M NaOH?
What ions are present? What are the possible reactions?
Na(aq)
+
+ Χλ(αθ)
−
→ ΝαΧλ(σ)
Η(αθ)
+
+ ΟΗ(αθ)
−
→ Η2Ο(λ)
74. Example
What volume of a .100 M HCl solution is
needed to neutralize 25.0ml of .350 M NaOH?
NaCl is soluble. Na+
and Cl-
are spectators.
Write a balanced net ionic equation.
H(aq)
+
+ ΟΗ(αθ)
−
→ Η2Ο(λ)
75. Example
What volume of a .100 M HCl solution is
needed to neutralize 25.0ml of .350 M NaOH?
What are the moles of reactant present in solution?
25.0ml NaOH
x
1L
1000ml
x
0.350mol OH −
ΛΝαΟΗ
= 8.75ξ10−3
µολΟΗ −
76. Example
What volume of a .100 M HCl solution is
needed to neutralize 25.0ml of .350 M NaOH?
How many moles of H+
are needed?
= 8.75x10−3
molH +
The mole ratio is 1:1
77. Example
What volume of a .100 M HCl solution is
needed to neutralize 25.0ml of .350 M NaOH?
What volume of HCl is required?
Vx
0.100molH +
Λ
= 8.75ξ10−3
µολΗ +
ς = 8.75ξ10−2
Λ
78. Example
In a certain experiment, 28.0ml of 0.250 M HNO3
and 53.0ml of .320 M KOH are mixed. Calculate
the amount of water formed in the resulting
reaction. What is the concentration of H+
or OH-
ions in excess after the reaction goes to
completion?
H+
is limiting
7.00x10-3
mol
H2O
.123 M OH-
excess
79. Acid-Base Titrations
Volumetric analysis is a technique for
determining the amount of a certain
substance by doing a titration.
A titration involves delivery (from a buret)
of a measured volume of a solution of
known concentration (the titrant) into a
solution containing the substance being
analyzed (the analyte).
80. Acid-Base Titrations
The point in the titration where enough
titrant has been added to react exactly
with the analyte is called the equivalence
point or the stoichiometric point.
This point is often marked with an indicator,
a substance added at the beginning of the
titration that changes color at the
equivalence point.
81. Acid-Base Titrations
The point at which the indicator actually
changes color is called the endpoint of
the titration.
The procedure for determining
accurately the concentration of a
solution is called standardizing the
solution.
82. Indicators
A common indicator for acid-base
titrations is phenolphthalein, which is
colorless in an acidic solution and pink
in a basic solution.
83. Example
A student carries out an experiment to standardize a
sodium hydroxide solution. To do this, the student weighs
out a 1.3009g sample of potassium hydrogen phthalate
(KHC8H4O4 or KHP). KHP (molar mass 204.22g/mol) has one
acidic hydrogen. The student dissolves the KHP is distilled
water, add phenolphthalein as an indicator, and titrates
the resulting solution with the sodium hydroxide solution to
the phenolphthalein endpoint. The difference between
the final and initial buret readings indicate that 41.20 ml of
the sodium hydroxide solution is required to react exactly
with the 1.3009g KHP. Calculate the concentration of the
sodium hydroxide solution.
84. Example
1.3009g KHP, molar mass = 204.22g/mol
41.20 ml NaOH solution to neutralize KHP
Calculate concentration of NaOH
KHP has 1 acidic hydrogen so it should react in a 1 to 1
ratio:
KHC8H4O4(aq) + ΝαΟΗ(αθ) → Η2Ο(λ) +Χ8Η4Ο4(αθ)
2−
+Κ(αθ)
+
+ Να(αθ)
+
.1546 M
86. Oxidation and Reduction
Oxidation Reduction reactions transfer
one or more electrons from one
species to another.
Called Redox reactions.
Common and important type of
reaction: photosynthesis, energy
production and combustion
87. Oxidation States
Also called oxidation numbers provides a
way to keep track of electrons in oxidation-
reduction reactions.
In a covalent compound when electrons
are shared, oxidation numbers are based
on the relative electron affinity of the
elements involved.
88. Oxidation States
In a covalent compound:
If the bond is between two identical
atoms, the electrons are divided
equally.
If the bond is between different
atoms, the electrons are divided
based on electron attraction.
89. Oxidation State Example
H2O
Total electrons: 4
O has more electronegativity and
maintains a δ-
.
Therefore O is assumed to have taken
both electrons, one from each of the
Hydrogens.
H has an oxidation state of +1 (each)
O has an oxidation state of -2
90. Oxidation States
The sum of the oxidation states must
be zero for an electrically neutral
compound.
The sum of the oxidation states must
equal the charge of the ion.
Ion charges are written as n+ or n-,
while oxidation numbers are written
+n or -n
93. Non- Integer Example
Fe3O4
Oxygen is assigned first, -2
Giving Fe a +8/3 state
Acceptable because all Fe
is assumed to have the
same charge within the
compound. But the
compound actually
contains two Fe3+
ions and
one Fe2+
ion.
94. Redox Reactions
Characterized by a transfer of electrons
2Na(s) + Χλ2(γ ) → 2ΝαΧλ(σ)
CH4(g) + 2Ο2(γ ) → ΧΟ2(γ ) + 2Η2Ο(λ)
95. Redox Reactions
C -4 C +4
Carbon loses 8 electrons
Increase in oxidation state is Oxidation
CH4(g) + 2Ο2(γ ) → ΧΟ2(γ ) + 2Η2Ο(λ)
O 0 O -2
Oxygen gains 8 electrons: 4 (-2) = -8
Decrease in oxidation state is Reduction
96. Redox Reactions
C -4 C +4
Carbon is Oxidized
Oxygen gas is the oxidizing agent
CH4(g) + 2Ο2(γ ) → ΧΟ2(γ ) + 2Η2Ο(λ)
O 0 O -2
Oxygen is Reduced
Methane is the reducing agent.
97. Example
Metallurgy, the process of producing a metal from its
ore, always involves oxidation-reduction reactions. In
the metallurgy of galena (PbS), the principal lead-
containing ore, the first step is the conversion of lead
sulfide to its oxide (a process called roasting):
The oxide is then treated with carbon monoxide to
produce the free metal:
For each reaction, identify the atoms that are oxidized
and reduced, and specify the oxidizing and reducing
agents.
2PbS(s) + 3Ο2(γ ) → 2ΠβΟ(σ) + 2ΣΟ2(γ )
PbO(s) + ΧΟ(γ ) → Πβ(σ) + ΧΟ2(γ )
98. Example
Pb +2 Pb +2
S -2 S +4
O 0 O -2
2PbS(s) + 3Ο2(γ ) → 2ΠβΟ(σ) + 2ΣΟ2(γ )
Sulfur is oxidized and oxygen is reduced.
Oxygen gas is the oxidizing agent and lead
sulfide is the reducing agent.
99. Example
Pb +2 Pb 0
O -2 O -2
C +2 C +4
PbO(s) + ΧΟ(γ ) → Πβ(σ) + ΧΟ2(γ )
Lead is reduced and carbon is oxidized.
PbO is the oxidizing agent, and CO is the
reducing agent.
101. Balancing Redox Reactions
Two methods are normally used:
1. Balancing of Oxidation states
2. Separation of the reaction into two half-
reactions
Normally used for more complex
reactions
102. Oxidation States Balancing
Method
We know that in a redox reaction we
must ultimately have equal numbers
of electrons gained and lost, and we
can use this principle to balance
redox equations.
103. Balancing Redox Steps
1. Write the unbalanced equation.
2. Determine the oxidation states of all atoms.
3. Show electrons gained and lost.
4. Use coefficients to equalize the electrons
gained and lost.
5. Balance the rest of the equations by inspection.
6. Add appropriate states.
104. Example
H(aq)
+
+ Χλ(αθ)
−
+ Σν(σ) + ΝΟ3(αθ)
−
→ ΣνΧλ6(αθ)
2−
+ ΝΟ2(γ ) + Η2Ο(λ)
+1 -1 0 +5 -2 +4 -1 +4 -2 +1 -2
Note that hydrogen, chlorine, and oxygen do not change
oxidation states and are not involved in electron exchange.
Sn(s) + ΝΟ3(αθ)
−
→ ΣνΧλ6(αθ)
2−
+ ΝΟ2(γ )
0 +5 -2 +4 -1 +4 -2
105. Example
Sn(s) + ΝΟ3(αθ)
−
→ ΣνΧλ6(αθ)
2−
+ ΝΟ2(γ )
0 +5 -2 +4 -1 +4 -2
Tin lost 4 electrons and each Nitrogen gained 1
electron. Therefore each nitrogen must have a
coefficient of 4.
H +
+ Χλ−
+ Σν + 4ΝΟ3 → ΣνΧλ6 + 4ΝΟ2 + Η2Ο
Balance the rest as usual
8H(aq)
+
+ 6Χλ(αθ)
−
+ Σν(σ) + 4ΝΟ3(αθ)
−
→ ΣνΧλ6(αθ)
2−
+ 4ΝΟ2(γ ) + 4Η2Ο(λ)