Download Examples

AP Chemistry
The Ultimate Chemical
Equations Handbook
HOMEWORK
 Do
all exercises in this book on a
separate sheet of paper.
 DO
NOT WRITE IN THE BOOK.
Chapter 1
 Symbols
elements
 There
and Nomenclature of the
is interesting info where the
elements got their name, but nothing
we will cover.
Chapter 2 and 3 Naming
Binary Compounds
 First,
determine if you have an ionic
compound or a covalent compound.
 A metal and a nonmetal will form
an ionic bond.
 Compounds with Polyatomic ions
form ionic bonds.
 Nonmetals bonding together or
Nonmetals and a metalloid form
covalent bonds.
Covalent bonding is very
similar to ionic naming
 You
always name the one that is
least electronegative first (furthest
from fluorine)
 Most electronegative last, and gets
the suffix “-ide”.
Covalent bonding is very
different from ionic naming
 Ionic
names ignored the subscript
because there was only one
possible ratio of elements.
 Covalent gives several possibilities
so we have to indicate how many
of each atom is present in the
name
Prefixes you have to know
prefix
meaning
prefix
meaning
*mono-
1
hex-
6
di-
2
hept-
7
tri-
3
oct-
8
tetr-
4
non-
9
pent-
5
dec-
10
* the first atom named does not get the prefix
“mono-”, it just keeps its original name!
Examples
 CO
 carbon
monoxide
 CO2
 carbon
dioxide
 NI3
 nitrogen
triiodide
 P4O6
 tetraphosphorus
hexoxide
Continuing
 I4O9
 tetriodine
nonoxide
 S2F10
 disulfur
decafluoride
 IF7
 Iodine
heptafluoride
 Si2Cl6
 disilicon
hexachloride
Naming ionic compounds
 For
monoatomic anions only
 drop the ending and add “-ide”
 so F fluoride
 Cl-, O2-, C4 chloride, oxide and carbide
Continuing…
 cations
keep the name of the
element.
 When naming compounds always
name the positive (cation) first and
the negative (anion) last.
 so mixing ions of chlorine and
sodium give you

sodium chloride
 (positive) (negative)
Determining the formula of ions
 Ionic
compounds are neutral
 You need to find the lowest number
of each ion to make it neutral
 for example:
 Na+ and O2 2 sodium for every one oxygen
 Na2O
More examples
 Al3+
and O2-
 Al2O3
 K+ and
Cl-
 KCl
 the
subscripts don’t effect the
name if there is only one
possibility
 still (cation)(anion)
 Aluminum oxide
 Potassium chloride
Several atoms can form a
couple of different ions.
 All
of these are metals that are not in
group 1, 2 or aluminum.
 for example iron can form Fe2+ or
Fe3+
 These are said as iron (II) and iron
(III)
 Cu+ and Cu2+ is Copper (I) and
Copper (II)
Figuring out charge on these
elements
 If
the ion is named, the charge is in
the name.
 If you have the formula, use the
charges of the other ions present to
determine the charge.
 Remember
 Alkali will always be +1
 Alkaline Earth +2, Halogens -1,
oxygen group -2
 Aluminum will always be +3
Examples

Copper (II) chloride

Cobalt (III) sulfide

NiF2

TiS2
Polyatomic ions
 Polyatomic
ion
 You
Ions- many atoms in one
can NOT break these apart in
this section.
 the “ide” suffix only applies to
monoatomic anions
Common polyatomic ions
Ammonium
NH4+
Perchlorate
ClO4-
Sulfate
SO42-
Acetate
CH3CO2-
Chlorate
ClO3-
Sulfite
SO32-
Nitrate
NO3-
Chlorite
Nitrite
NO2-
Hypochlorite
ClO-
Dichromate
Cr2O72-
Cyanide
CN-
Iodate
IO3-
Chromate
CrO42-
Thiocyanate
SCN-
Bromate
BrO3-
Oxalate
C2O42-
Hydroxide
OH-
Silicate
SiO32-
HSO4-
Permanganate
MnO4-
Phosphate
PO43-
HSO3-
Thiosulfate
S2O32-
Arsenate
AsO43-
Hydrogen
carbonate
Hydrogen
sulfate
Hydrogen
sulfite
HCO3-
ClO2-
Carbonate
CO32-
YOU WILL HAVE TO
MEMORIZE THESE!
 This
is one of the big differences
from last year.
 We
will have a quiz just like the
elements quiz last year over these!
Determining the formula of ions
 Ionic
compounds are neutral
 Remember– don’t break a
polyatomic ion apart
 for example: Ammonium carbonate
 NH4+ and CO32 (NH4)2CO3
Chapter 4 acids and salts
 Oxyanions-
negative ions containing
oxygen.
 These have the suffix “-ate” or “-ite”
 “-ate” means it has more oxygen
atoms bonded, “-ite” has less
 For example
 SO42- sulfate
 SO32- sulfite
Oxyanions
 Oxyanions
may contain the prefix
“hypo-”, less than, or “per-”, more
than.
 For example
 ClO4Perchlorate
 ClO3Chlorate
 ClO2Chlorite
 ClOHypochlorite
Acids
 Certain
compounds produce H+ ions
in water, these are called acids.
 You can recognize them because the
neutral compound starts with “H”.
 For example HCl, H2SO4, and HNO3.
 Don’t confuse a polyatomic ion with a
neutral compound.
 HCO3- is hydrogen carbonate (or
bicarbonate), not an acid.
Naming acids
 Does
it contain oxygen?
 If it does not, it gets the prefix
“hydro-” and the suffix “-ic acid”
 HCl
 Hydrochloric acid
 HF
 Hydroflouric acid
 HCN
 Hydrocyanic acid
Naming Acids
 If
it does contain an oxyanion, then
replace the ending.
 If the ending was “–ate”, add “-ic
acid”
 If the ending was “–ite”, add “-ous
acid”
 H2SO4 Sulfuric Acid
 H2SO3 Sulfurous Acid
Examples
 HNO3
 HI
 H3AsO3
Chapter 5 Complex ions
 Complex
ion- transition metal ion with
attached ligands
 Fe(CN)63 Glance over this chapter. Skip the
problems, this material is out of the
test.
Ch 6 Organic
 Alkanes-
straight chain hydrocarbons
with all single bonds
 Alkenes- hydrocarbons with a double
bond
 Alkynes- hydrocarbons with a triple
bond
 Cyclic hydrocarbons- rings
Root words
Meth
# of C
atoms
1
Hex
# of C
atoms
6
Eth
2
Hept
7
Prop
3
Oct
8
But
4
Non
9
Pent
5
Dec
10
Name this molecule
And give its molecular formula
4 ethyl octane
C10H22
4 propyl decane
C13H28
Name and give the formula
Methyl cyclohexane
C7H14
Functional groups
 halogenated
*R means any carbon chain
Alcohols
R-OH
-al
-oic acid
Ketones
NOT at the edge
R-C-R
=O
Aldehydes
at the edge
R=O
R-C=O
-OH
-ol
Carboxylic Acids
-one
Predicting organic reactions
 Addition
reactions occur by
adding halogens or hydrogen to
alkene or alkynes.
 In the reaction, the new molecule
takes the place of the double or
triple bond.
 Cl2
+ CH3-CH=CH2  CH3-CClHCClH2
example
 1-
butene is reacted with fluorine
 C4
H8 + F2  C4H8F2
Predicting organic reactions
 Substitution
reactions occur by
adding halogens to an alkane.
 In the reaction, the new molecule
takes the place of a hydrogen.
 Cl2
+ CH3-CH3  CH3-CClH2 +
HCl
 Cl2 + C2H6  C2ClH5 + HCl
Predicting organic reactions
 Combustion
reactions occur when
an organic compound is burned in
oxygen.
 The products of a complete
combustion are water vapor and
carbon dioxide.
 C6H12O6
+ 6 O2  6 H2O+ 6 CO2
Predicting organic reactions
 Esterification
reactions
 Made by reacting carboxylic acids
with alcohols.
Carboxylic acid
+
H-O-R
R-C-O-R
alcohol
O=
O=
R-C-O-H
Ester
+ H-O-H
Examples
 Fluorine
 Ethanol
is added to 2 propene
is burned in oxygen
 Chlorine
is added to propane
 Ethanoic
acid is reacted with 1-
butanol
Ch 7 Predicting molecular
equations
 Synthesis
Reactions pg 42
 Metals and nonmetals combine to
form salts. CHECK CHARGES OF
IONS!!!!!
 Metal Oxides and water form bases,
bases have OH attached
 Nonmetal oxides and water form
acids
 Metal oxides and nonmetal oxides
form salts (you will make an
Examples
 Magnesium
 Calcium
burns in oxygen
reacts with chlorine gas
 Magnesiumvoxide
reacts with water
 Carbon
dioxide is bubbled through
 Lithium
oxide is added to carbon
water
Ch 7 Predicting molecular equations
 Decomposition
Reactions pg 42
 CHECK CHARGES OF IONS!!!!!
 Carbonates decompose into oxides and
CO2
 Chlorates decompose into chlorides and
O2
 Ammonium and a base makes ammonia
and water
 Hydrogen peroxide decomposes into
water and oxygen.
Three that always appear on the
AP test
 Hydrogen
peroxide decomposes
into water and oxygen
 Ammonium hydroxide
decomposes into ammonia and
water
 Carbonic acid decomposes into
carbon dioxide and water
Examples
 Titanium
 Copper
(IV) chlorate decomposes
(III) carbonate is heated
 Magnesium
 Carbonic
chloride is electrolyzed
acid is heated
 Hydrogen
peroxide decomposes
Ch 8 Single replacement reactions
A
+ BX  AX + B
 You will have a chart of activity series
 More active metals will replace less active
metals from their compound in a solution
 A less active element will have no reaction
when added to a more active element!
 Active metals replace hydrogen in water
 Active metals replace hydrogen in acids
 Active nonmetals replace less active
nonmetals from their compounds in
solutions
Activity
series
chart
Old chart
from test
 Zinc
Examples
is added to a solution of cobalt
(II) chloride
 Cadmium
is added to a solution of
barium iodide
 Lithium
is added to a solution of
copper (II) chlorate
 Chromium
is left in water
Examples
 Potassium
 Silver
is added to sulfuric acid
is added to hydrochloric acid
 Chlorine
gas is bubbled through a
solution of sodium bromide
Chapter 9 Double replacement
reactions
 AY
+ BX  AX + BY
 These reactions occur in solution
 Remember in solution the ions are
free floating. For a reaction to occur,
the ions have to come together and
leave their dissolved state.
 Formation of a precipitate
 Formation of a gas
 Formation of a molecular species
Solubility Rules
 Acids
are soluble.
 Compounds of: alkali metals,
ammonium, and nitrate are soluble.
 This
needs to be memorized.
Other soluble compounds
 All
acetates are soluble except Fe3+
 All chlorates are soluble.
 All binary compounds of the halogens
(other than F) with metals are soluble,
except those of Silver, Mercury(I), and
Lead. Fluorides are insoluble except
for rule 1 and 2.
 All sulfates are soluble, except those
of barium, strontium, calcium, lead,
silver, and mercury (I).
Insoluble compounds
 Carbonates,
oxalates, sulfites,
chromates, oxides, silicates, and
phosphates are insoluble.
 Hydroxides are insoluble except Ba,
Sr, and Ca
 Sulfides are insoluble except for
calcium, barium, strontium,
magnesium.
 The exception is with alkali
metals or ammonium.
Net Ionic Equation
 Determine

what dissolved and precipitated
Mg(NO3)2(aq)+Na2CO3(aq) MgCO3(s) + 2NaNO3(aq)
 Dissociate
everything that is aqueous, not
solid
 Mg++ + 2NO3- + 2Na+ + CO3-–  MgCO3(s) + 2 Na+ + 2NO3-
 Now
cancel out everything that is the same
on both sides of the equation
 These are called spectator ions
The remaining
part is the net
Mg++ + CO3--  MgCO3(s) ionic equation
Examples
 Hydrochloric
nitrate
acid reacts with silver
 Potassium
carbonate reacts with
calcium chlorate
 Sodium
chloride reacts with
ammonium oxalate
 Scandium
chromate
acetate reacts with lithium
 H2S
Gases
(hydrogen sulfide) is formed from
any sulfide reacting with an acid
 CO2 (Carbon dioxide) is formed from
any carbonate reacting with an acid,
water is also produced
 SO2 (sulfur dioxide) is formed from any
sulfite reacting with an acid, water is
also produced
 NH3 (ammonia) is formed from
ammonium reacting with a soluble
hydroxide
Examples
 Ammonium
chloride reacts with
calcium hydroxide
 Sodium
sulfide is combined with
nitric acid
 Ammonium
carbonate is combined
with barium chlorate
 Lithium
acid
sulfite reacts with phosphoric
Formation of a molecular
species
 It
is the same as precipitates or
gases except a liquid is formed.
 Acid base neutralization reactions will
produce water.
 NaOH + HNO3  H2O (l) + NaNO3 (aq)
Strong acids
Acid
formula Acid
Hydrochloric HCl
acid
Hydrobromic HBr
acid
Hydriodic
HI
acid
Formula
Sulfuric
Acid
Nitric Acid
H2SO4
Perchloric
Acid
HClO4
Chloric
Acid
HClO3
HNO3
Strong Bases
these make a lightning bolt
on the periodic table!
Name
Formula
Name
Formula
Sodium
NaOH
Hydroxide
Calcium
Ca(OH)2
Hydroxide
Potassium KOH
Hydroxide
Strontium Sr(OH)2
Hydroxide
Barium
Ba(OH)2
Hydroxide
Strong acids and bases
 Strong
acids and bases are not at
equilibrium, there is no reverse
reaction.
 Strong acids and bases will never be
formed in a net ionic equation.
 All other acids/bases can be formed,
and will be formed by reacting the
appropriate ion with a strong
acid/base.
 *Most other bases are insoluble
Examples
 Calcium
hydroxide reacts with chloric
acid
 Hydrochloric acid reacts with calcium
nitrite
 Nitric acid reacts with sodium chlorite
 Sodium chloride is mixed with
sulfuric acid
Chapter 10 Aqueous Solutions
and Ionic Equations
 This
chapter has already been
covered
 *Only dissociate soluble ionic
compounds
 Molecular equation
 Na2S + CrCl2  CrS + NaCl
 Full Ionic Equation
 2Na++S2-+Cr2++2Cl- CrS +Na++Cl Net Ionic Equation
Chapter 11 Redox Reactions
 Redox
or oxidation-reduction reactions
are reactions that involve a transfer of
electrons.
 Oxidation is the loss of electrons.
 Reduction is the gain of electrons.
 (think of the charge, OIL RIG)
 4 K + O2 → 4 K+ + 2 O2 Potassium get oxidized, oxygen get reduced
Using oxidation states
 In
the reaction…
 2 Na +2 H2O →2 NaOH + H2

0
+1 -2
+1 -2 +1 0
 Note the changes
 Sodium went from 0 to 1
 2 of the hydrogen atoms went from +1 to 0
(the other two were unchanged)
Breaking into two half reactions
 Sodium
must have lost 2 electrons
 2 Na → 2Na+ + 2 e And Hydrogen gained two electrons
 2 H2O +2 e-→ 2 OH- + H2
 Sodium is oxidized, hydrogen is reduced in
this reaction
 Oxidation is an increase in oxidation state
 Reduction is a decrease in oxidation state
Balancing Redox Equations
 by
Half Reactions Method or
oxidation state method
 The book does not separate these
into half reactions, although it adds
another step I think it makes it
easier
Half reactions
 Ce4+
+ Sn2+ → Ce3+ + Sn4+
 Half reactions
 Ce4+ + e- → Ce3+
 Sn2+ → 2e- + Sn4+
 Electrons lost must equal
electrons gained!
 2 Ce4+ +2 e- →2 Ce3+
 Merge the two half reactions
 2 Ce4+ + Sn2+ → 2 Ce3+ + Sn4+
Redox reactions in acidic
solutions
 It
will be noted in the problem
 Balance all elements except
hydrogen and oxygen.
 Balance oxygen by adding H2O
(which is always prevalent in an
acidic solution)
 Balance hydrogen by adding H+
 Then balance the charge adding
electrons and proceed as normal.
Example
 In
an acidic solution
 Cr2O7 2- + Cl- → Cr3+ + Cl2
 Half reactions
 Cr2O7 2- → Cr3+
 Cl- → Cl2
Reduction side
 Cr2O7 2 Cr2O7 2 Cr2O7 2-
 Cr2O7 2 Cr2O7 2-
H2O
→ Cr3+
→ 2 Cr3+
→ 2 Cr3+ + 7 H2O
+ 14 H+→ 2 Cr3+ + 7 H2O
+ 14 H++ 6 e- →2Cr3++7
Oxidation side
 Cl-
→ Cl2
 2 Cl- → Cl2
 2 Cl- → Cl2 + 2 e I have to equal 6 e- so multiply by 3
 6 Cl- → 3 Cl2 + 6 e-
Combine my half reactions
 Cr2O7 2-
+ 14 H++ 6 e- → 2 Cr3+ + 7
H2O
 6 Cl- → 3 Cl2 + 6 e And you get
 Cr2O7 2-+14 H++6Cl-→2Cr3++3
Cl2+7H2O
 The electrons cancel out .
Example
 In
an acidic solution
 MnO4- + H2O2 → Mn2+ + O2
Balancing Redox Equations in a
basic solution
 Look
for the words basic or alkaline
 Follow all rules for an acidic solution.
 After you have completed the acidic
reaction add OH- to each side to
neutralize any H+.
 Combine OH- and H+ to make H2O.
 Cancel out any extra waters from
both sides of the equation.
Example
 We
will use the same equation as
before
 In a basic solution
 MnO4- + H2O2 → Mn2+ + O2
2
MnO4- + 6 H++ 5 H2O2
→ 2 Mn2+ + 5 O2 + 8 H2O
Another example
 In
a basic solution
 MnO4 − + SO32-→MnO4 2− + SO42 Half reactions
 MnO4 − → MnO4 2−
 SO32-→ SO42-
Predicting redox reactions
 Single
replacement reactions we
went over a few chapters ago are all
redox reactions.
 Here are the same examples, this
time write out the net ionic equation
 The change will be the charge!!
Examples
 Zinc
is added to a solution of cobalt (II)
chloride
 Cadmium
is added to a solution of
barium iodide
 Lithium
is added to a solution of copper
(II) chlorate
 Chromium
is left in water
Examples
 Potassium
 Silver
is added to sulfuric acid
is added to hydrochloric acid
 Chlorine
gas is bubbled through a
solution of sodium bromide
Common Oxidizing/reducing agents
Oxidizing agent
Product
Reducing Agent Product
MnO4- in acid
Mn2+
H2O2
O2
MnO2 in acid
Mn2+
MnO4- in base
MnO2
Halogens
(dilute basic)
Hypohalite ion
(hypochlorite)
CrO42- in acid
Cr3+
Cr2O72- in acid
Cr3+
halogens
(conc basic)
Halate ion
(chlorate)
HClO4
Cl-
Free metals
Metal ions
Na2O2
NaOH
H2O2
H2O
C2O42-
CO2
H2SO4 conc.
SO2
Sulfite or SO2
SO4 2-
Free halogens
Halide ion
Halide ion
Free halogen
HNO3 conc.
NO2
NO2-
NO3-
HNO3 dilute
NO
Metal ions
Lower oxidation
number
Metal ion
Higher oxidation
number
IMPORTANT
 For
every redox equation, you have
to have both an oxidizing agent and
a reducing agent!!
 Otherwise it doesn’t work.
Examples
 Bromate
reacts with bromide in an
acidic solution
 Permanganate reacts with oxalate in
an acidic solution
 Calcium metal reacts with
permanganate in a sodium hydroxide
solution
Example
 Chromate
reacts with chloride in an
acidic solution
 Chlorine reacts with permanganate in
a concentrated solution of sodium
hydroxide
Ch 12 Electrolysis in water
 Electrolysis
is a fairly simple process.
 There are two plates in a solution, and
an electric current is sent through.
 The plates are the cathode, where
reduction takes place, and the anode,
where oxidation takes place.
 cathode-reduction
anodeoxidation
Rules for cathode reaction
A
cation may be reduced to a metal
 Cu+ + 1 e-  Cu
 Or water way be reduce to hydrogen
 2 H2O + 2 e-  H2 + 2 OH Transition metals tend to reduce
before water, main group metals
tend to reduce after
Rules for anode reactions
 An
anion nonmetal may be oxidized
to a nonmetal
 2 Cl-  Cl2+ 2 e Water may be oxidized to oxygen
 2 H2O  O2 + 4 H+ + 4 e Chlorine, bromine and iodine will
oxidize before oxygen. That is it.
Rules for molten binary salts
 Molten
means melted, with no water.
 These are straightforward and easy!
 Molten magnesium chloride is
electrolyzed
 MgCl2  Mg + Cl2
examples
 Aqueous
calcium bromide is
electrolyzed
 Aqueous chromium (III) nitrate is
electrolyzed
 Aqueous cobalt (II) bromide
electrolyzed
 Molten sodium chloride is
electrolyzed
Ch 13 Complex ion reactions
 Formation
of complex ions
 Common complex ions metals
 Fe Co Ni Cr Cu Zn Ag Al
 Common ligands
 NH3
CN- OHSCN General rule: the number of ligands
will be twice the charge of the metal
ion
Example
 Iron
(III) chloride reacts with
potassium cyanide
 Fe3+ + CN-  Fe(CN)63 How did I get the charge? Iron is 3+
, 6 cyanides at 1-
Examples
 Zinc
(I) fluoride reacts with sodium
thiocyanate
 Concentrated ammonia is reacted
with cobalt (III) iodide
 Barium hydroxide reacts with nickel
(II) nitrate
Solution Stoichiometry
Ch 4
Homework
 Due
with test
 Pg 171 Chapter 4 review
 1-87 odd
Water and the Nature of Aqueous
Solutions
 Water
is the foundation of all life on
Earth.
 Each O-H covalent bond is highly polar
because the electronegativity of oxygen
is so much greater than hydrogen.
 The dipole moment creates a slightly
negative O and a slightly positive H as the
electrons are pulled strongly toward O.
Water is a bent molecule with
o
a bond angle of 109.5
109.5o
Hydrogen bond
Polar
molecules are attracted to one
another by dipole attractions.
In water, it is hydrogen bonding.
The molecules can’t easily slide past one
another, they are held in place.
The strength of the hydrogen bonds
accounts for water’s high surface tension,
its low vapor pressure, its high specific
heat, its high heat of vaporization, and its
high boiling point.
Hydrogen bonding in water
Solvents and Solutes
Chemically
pure water never exists in
nature because water dissolves so many
substances
* An aqueous solution is one in which
water samples contain dissolved
substances.
Components of a solution.
A solvent is the dissolving medium.
A solute is the dissolved particles.
Solutions
Solutions
are homogeneous mixtures in
which solute particles are usually less than
1 nm in diameter. The solute and solvent
are not capable of being separated by
filtration.
Substances that dissolve most readily in
water include ionic compounds and polar
covalent molecules.
Nonpolar molecules do not dissolve in
water (a polar molecule).
 Rule: Like dissolves like
The Solution Process
Water
molecules are in constant motion as
a result of their kinetic energy.
The molecules collide with solute particles.
The solvent molecules attract the solute
ions.
The ionic crystal breaks apart by the
action of the solvent.
Solvation is the process that occurs when
a solute dissolves.
http://mw2.concord.org/public/student/solution/dissolve.cml
Insoluble
In
some ionic compounds, the
attractions between the ions in the
crystal are stronger than the
attractions exerted by water.
* These compounds are insoluble.
 Nonpolar substances form a solution
because there are no repulsive forces
between them, not because the solute
and solvent are attracted.
Electrolytes and
Nonelectrolytes
Compounds that conduct an electric
current in aqueous solution or the molten
state are called electrolytes.
All ionic compounds are electrolytes.
 Soluble ionic compounds conduct
electricity in both a solution and in the
molten state.
Insoluble ionic compounds only conduct
electricity in the molten state.
Nonelectrolytes
Compounds
that do not conduct an electric
current in either aqueous solution or the molten
state are called nonelectrolytes.
Many molecular compounds are nonelectrolytes
because they do not contain ions.
 Most compounds of carbon are nonelectrolytes.
Some very polar molecular compounds are
nonelectrolytes in the pure state, but are
electrolytes when they dissolve in water.
This occurs because such compounds ionize in
solution.
Weak Electrolytes
Not
all electrolytes conduct electricity
to the same degree.
A weak electrolyte conducts electricity
poorly because only a fraction of the
solute exists as ions.
* Most of the compound is in the
original form.
* The most common weak electrolytes
are weak acids and weak bases.
Strong Electrolytes
A
strong electrolyte conducts
electricity very well because almost all
of the solute exists as separated ions.
* Very little of the original compound
remains intact.
* Classes of electrolytes include
soluble salts, strong acids and strong
bases.
The Composition of Solutions
 To
perform stoichiometric calculations
when two solutions are mixed, two things
must be known.
 The nature of the reaction, which
depends on the exact forms the chemical
takes when dissolved.
 The amounts of chemical present in
solution, usually expressed as
concentrations.
Molarity
Molarity
(M) is defined as the moles of
solute per volume of solution in liters.

M = molarity = moles of solute

liters of solution

Calculate the molarity of a solution
prepared by dissolving 11.5 g of solid
NaOH in enough water to make 1.50 L of
solution.

Calculation of Molarity II
Calculate
the molarity of a solution
prepared by dissolving 1.56 g of gaseous
HCl in enough water to make 26.8 mL of
solution.
Concentrations of Ions
Give
the concentration of each type of
ion in the following solutions:

a. 0.50 M Co(NO3)2


b. 1 M Fe(ClO4)3
Concentrations of Ions II
Calculate
the number of moles of Clions in 1.75 L of 1.0 x 10-3M ZnCl2.

Concentration and Volume I.
Typical blood serum is about 0.14 M
NaCl. What volume of blood contains 1.0
mg NaCl?

Solutions of Known Concentration
.To
analyze the alcohol content of a
certain wine, a chemist needs 1.00 L of
an aqueous 0.200 M K2Cr2O7 (potassium
dichromate) solution. How much solid
K2Cr2O7 must be weighed out to make
this solution?
Dilution
.To
save money and space in a
laboratory, solutions are often in
concentrated form. Water is added to
achieve the molarity desired for a
particular solution.
 This process is called dilution.
 Dilution with water does not alter the
numbers of moles of solute present.
Concentration and Volume
What
volume of 16M sulfuric acid must
be used to prepare 1.5 L of a 0.10 M
H2SO4 solution?

Another
way to express the dilution
process is:
M1V1 = M2V2
What
volume of 14.8M ammonia must be
used to prepare .75 L of a 1.5 M H2SO4
solution?
Stoichiometry of Precipitation
Reactions
 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 reactants or
products, as required.
 6. Convert to grams or other units, as
required.
Determining the Mass of
Products
Formed
 Calculate the mass of solid NaCl that
must be added to 1.50 L of a 0.100 M
AgNO3 solution to precipitate all the Ag+
ions in the form of AgCl.
Stoichiometry of Precipitation
Reactions
 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 reactants or
products, as required.
 6. Convert to grams or other units, as
required.
Determining the Mass of
Products
Formed
 Calculate the mass of solid NaCl that
must be added to 1.50 L of a 0.100 M
AgNO3 solution to precipitate all the Ag+
ions in the form of AgCl.
Acid-Base Reactions
The
Bronsted-Lowry Definitions:
An acid is a proton donor.
A base is a proton acceptor.
Arrhenius definitions:
Acid generate protons (hydronium)
Bases generate hydroxide
Lewis definitions
Acids are electron pair acceptors
Bases are electron pair donors
Performing Calculations for AcidBase Reactions
1.List
the species present in the combined
solution before any reaction occurs, and
decide what reaction will occur.
2. Write the balance 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.
4. Determine the limiting reactant.
 5. Calculate the moles of the required
reactant and product.
6. Convert to grams or volume (of solution),
as required.
Example Problem
What
volume of a 0.100 M HCl solution is
needed to neutralize 25.0 mL of 0.350 M
NaOH?
In a certain experiment, 28.0 mL of 0.250
M HNO3 and 53.0 mL of 0.320 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?
Acid-Base Titrations
Volumetric
analysis is a technique for
determining the amount of a certain substance
by doing a titration.
A titration is a process in which a solution of
known concentration is used to determine the
concentration of another solution through a
monitored reaction.
The equivalence point is reached when all of
the moles of H+ ions present in the original
volume of acid solution have reacted with an
equivalent number of moles of OH- ions added.
This is also known as the stoichiometric point.
Titration
An
acid-base indicator is a chemical that
changes color once the equivalence point is
reached.
 The endpoint of a titration is the point in a
titration where the indicator actually changes
color.
 Three requirements for a titration
1. The exact reaction between titrant
(known solution) and analyte (unknown
solution or one you are analyzing) must be
known (and rapid).
Three requirement (cont.)
2.
The stoichiometric (equivalence)
point must be marked accurately.
3. The volume of titrant required to
reach the stoichiometric point must be
known accurately.
Problem solving
 The
first step in the analysis of a
complex solution is to write down the
components and focus on the chemistry
of each one.
 Take a big problem and look at the small
problems within.
Neutralization Titration
A student carries out an experiment to standardize
(determine the exact concentration of) a sodium
hydroxide solution. To do this, the student weighs out
a 1.3009-g sample of potassium hydrogen phthalate
(KHC8H4O4, often abbreviated KHP. KHP (molar mass
204.22 g/mol) has one acidic hydrogen. The student
dissolves the KHP in distilled water, adds
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 indicates that 41.20
mL of the sodium hydroxide solution is required to
react exactly with the 1.3009 g KHP. Calculate the
concentration of the sodium hydroxide solution.
Neutralization Analysis
An
environmental chemist analyzed the effluent
(the released waste material) from an industrial
process known to produce the compounds carbon
tetrachloride (CCl4) and benzoic acid (HC7H5O2), a
weak acid that has one acidic hydrogen atom per
molecule. A sample of this effluent weighing
0.3518 g was shaken with water, and the resulting
aqueous solution required 10.59 mL of 0.1546 M
NaOH for neutralization. Calculate the mass
percent of HC7H5O2 in the original sample.