Indiana University CPurdue University Fort Wayne IPFW

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					     Chapter 4
I. Types of chemical reactions
A) There are millions of reactions. We can’t
remember them all so what to we do?
 B) Fall into several categories.
C) Your text divides them into three types.
1. Precipitation reactions – (metathesis
reactions - 2 compounds exchange parts -
generally take place in water solutions of
two ionic solids and a solid (ppt) forms.)
 The general form of the equation for
 such a reaction is: AB + CD  AD + CB
2) Acid-Base reactions. An acid substance
react with a substance called a base. Such
reactions involve the transfer of a proton (H+)
between reactants.
3) Oxidation-reduction reactions. These
involve the transfer of electrons between
D) Ionic Theory of Solutions and Solubility
1) A solution is a homogeneous mixture of
particles less than one nanometer in diameter
distributed evenly throughout.
3) We will focus on solids dissolved in water,
but there are other solvents than water, and
solutions other than liquids - gaseous (clean
air), solid solutions (alloys like steel, solder,
brass, bronze.
4) Solutes which dissolve are either
electrolytes or nonelectrolytes.
a) electrolytes are substances which when
dissolved in water result in a solution which
conducts electricity. Moving charged particles
must be there. These charged particles are
called ions.
There are strong and weak electrolytes.

b) nonelectrolytes are substances which
when dissolved in water do not conduct
electricity - uncharged molecules are the
particles in solution.
c) Soluble Salts - substances which result
from the neutralization of an acid by a
base and are soluble in water are of
interest - since they dissociate in water to
give separate ions
      NaCl(aq)  Na+(aq) + Cl- (aq)
      CaCl2(aq)  Ca2+(aq) + 2 Cl- (aq)
      C12H22O11(aq)  C12H22O11(aq)
  molecules stay intact - nonelectrolyte
5) The formation of an insoluble solid (a
precipitate) drives a chemical reaction.
6) For these reactions we can write three
kinds of equations.
a) Molecular Equations-complete
formulas are written for all the reactants
and products, no ions are written.
b) Ionic equations-all strongly soluble
electrolytes are written in their
dissociated (ionized) forms.
c) Net Ionic equations-only involve those
chemical species which are involved in a
chemical reaction. All spectator ions are
Spectator ions-those ions which do not
participate in the chemical reaction but
are present in the reaction mixture.
d) Examples
Write the molecular, ionic, and net ionic
equations for the reaction of an aqueous
solution of CaCl2 and an aqueous solution
of Na2CO3.
1) the molecular equation is:
CaCl2(aq) + Na2CO3(aq) CaCO3(s)+2NaCl(aq

2) the ionic equation is:
Ca2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + CO32-(aq) 
CaCO3(s) + 2 Na+(aq) + 2 Cl-(aq)
3) the net ionic equation is:
Ca2+(aq) + CO32- (aq)  CaCO3(s)
The solubility rules are used to determine
whether precipitation reactions occur or
1) you need to know whether the
predicted products are soluble or
insoluble in water. A precipitate is an
insoluble compound formed during a
chemical reaction in solution.
9)The solubility rules are given on page
128 - table 4.1
You do not have to memorize these rules,
I will give them to you on a quiz or exam.
You must know how to use the chart in
order to predict whether or not a
precipitate will occur and the formula of
the precipitate.
If we add a solution of KCl to a AgNO3
solution will a precipitate form? We first
write a predicted metathesis reaction.
KCl(aq) + AgNO3(aq)  KNO3 + AgCl
Then we look at the solubility table to see
if any of the products are insoluble in
We see that the table indicates that AgCl
is insoluble - most chlorides are soluble
except for Ag+, Hg2+, Hg22+ and Pb2+
The molecular equation then becomes:
KCl(aq) + AgNO3(aq)  KNO3(aq)+ AgCl(s)
The net ionic equation is:
         Cl-(aq) + Ag+(aq)  AgCl(s)
If we add a solution of NaNO3 to an
NH4Cl solution will a precipitate form?
The predicted equation then becomes:
NaNO3(aq) + NH4Cl(aq)  NaCl + NH4NO3
We see that the table indicates that both
compounds are soluble - most chlorides
are soluble, NaCl is soluble, and nitrates
are soluble as well as NH4+ compounds, so
mixing these two solutions gives no
precipitates, no reaction results. After the
 we would write N. R. for no reaction.
What will result if we add a solution of
Pb(NO3)2 to a solution of KI?

A) There are three definitions of acids
and bases.
1) The Arrhenius definition - which is
based on properties of these substances:
in water solution, acids taste sour, turn
blue litmus to red, react with active
metals to give off hydrogen gas,
bases have a bitter taste, turn red litmus
to blue, feel soapy (slippery) and
neutralize acids.
 The result of these observations led
 Arrhenius to the theory that acids ionize
 in water to give H+ ions and bases give
 OH- ions.
HCl(g) dissolves in water to give H+(aq) and
Cl-(aq) ions. Therefore an Arrhenius acid.
NaOH(s) dissolves in water to give Na+(aq)
ions and OH- (aq) ions. Therefore an
Arrhenius base.
The neutralization reaction then
HCl(aq) + NaOH (aq)  H2O(l) + NaCl(aq)
Sodium hydroxide and Nitric acid
Potassium hydroxide and Sulfuric acid
2) The Brønsted - Lowry Theory of acids
and bases.
A B-L acid is a species which donates a
proton (H+ ion) to another species.
A base is a species which accepts a proton
from another species. You must have a
reaction in order to name a species as a
B-L acid or base.
   HCl(aq) + H2O (l)  H3O+(aq) + Cl-(aq)
HCl is the B-L Acid, it donates the proton
to the H2O which is the B-L Base since it
accepts the proton (H+ion)
  NH3(g) + H2O(l)  NH4+(aq) + OH-(aq)

H2O is the B-L Acid, it donates the proton
to the NH3 which is the B-L Base since it
accepts the proton (H+ion).
We will discuss the third theory of acids
and bases, the Lewis theory, later.
 B) Strong and weak acids and bases
1) A strong acid is an acid that ionizes
completely in water; it is a strong
  HCl(aq) + H2O (l)  H3O+(aq) + Cl-(aq)
At this time you should learn 6 strong
acids and bases:
2) A weak acid is an acid that only
partially ionizes in water; it is a weak
The hydrogen cyanide molecule, HCN,
reacts with water to produce a small
percentage of ions in solution.
HCN(aq) + H2O(l)  CN-(aq) + H3O+(aq)
Table 4.1 lists some common weak acids,
remember, if you memorize the strong
ones, all others are weak.
3) A strong base is a base that is present
entirely as ions, one of which is OH-.
4) Neutralization reactions involving
weak acids.
Since KCN is a strong electrolyte, the net
ionic equation is

5) There are acid-base reactions which
give a gas as a product.
The net ionic equation would be:
   SO32-(aq) + 2H+(aq)  H2O(l) + SO2(g)
A) When electrons are transferred from
one compound to another,an oxidation-
reduction is said to occur.
B) The species which loses electrons is
said to be oxidized. The oxidation
number of the species increases. This is
an oxidation reaction.
C) The species which gains electrons is
said to be reduced. The oxidation
number of the species decreases. This is
a reduction reaction.
Oxidizing agents are themselves reduced.
They oxidize the other reactant.
Reducing agents are therefore oxidized.
They reduce the other reactant.
Simply a bookkeeping method for
studying redox reactions.
E) Rules for determining oxidation
1. The oxidation number of any free
element is zero, regardless of how
complex the allotrope may be.
2. The oxidation number for any simple,
monatomic ion is equal to the charge of
the ion.
3. The sum of all the oxidation numbers
of the atoms in a molecule or polyatomic
ion must be equal to the charge of the
 4. In its compounds, fluorine always has
 an oxidation number of –1.
5. In its compounds, hydrogen has an
oxidation number of +1 except for
hydrides, where the oxidation number is
6. In its compounds, oxygen has an
oxidation number of -2 except for the
peroxides and superoxides, which have
oxidation numbers of -1 and -1/2
7. Examples: Find the oxidation
numbers of all of the species present in
the following compounds.
F. An example:
         4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
Feo  Fe3+ + 3 e- (loss of electrons is
4 e + O2o  2 O2- (gain of electrons is
Fe(s) is oxidized - it gives electrons - it is
the reducing agent.
O2(g) is reduced - it takes electrons - it is
the oxidizing agent.
G. Most oxidation-reduction reactions
fall into one of the following simple
1. Combination

2. Decomposition
3. Displacement

 4. Combustion
Balancing Redox equations on separate

 IV. Solutions

A) to describe the relative amounts of
solute (that which is in the smaller
amount in the solution) and the solvent
(greater amount) the general term used is
B) Chemists like numbers, so we have
quantitative descriptions of
concentrations of solutions, one of which
is MOLARITY which MUST be related
to moles.

The important idea here is that when we
measure out a certain volume of solution
we know how many moles of solute we
are getting.
The number of grams of solute we are
pouring out of the container, is related to
we have in our new container.
 2) Examples: If we have 1.00 L of a 3.00
 M solution of HCl, how many moles of
 HCl are in the liter of solution?

How many grams of HCl are in 1.00 L
of 3.00 M HCl?
If we have 500. mL of 6.00 M HCl
solution, how many moles do we have?

How many moles of HCl are in 25.0 mL of
6.00 M HCl solution?
3) Preparing solutions of a desired M
a) Calculate the amount of solute needed
and weigh out that amount.
b) Obtain the correct volumetric flask.
c) Put some water into the flask, add the
solute and dissolve.
d) Fill the flask with solvent to the
scratch on the flask and shake to
4) How would you prepare 250.00 mL of
6.00 M aqueous solution of NaOH from
solid NaOH?
a) The calculation of the amount of
NaOH required is found as follows:
b) Obtain the 250.00 mL volumetric flask.
c) Add a bit of water, add 60.0 g of NaOH,
d) Fill to scratch with water and shake to
C) Many times we start with
concentrated solutions to make dilute
1) We must calculate the volume of the
concentrated solution required to make
the requested amount of dilute solution
and measure its volume.
2) Obtain the correct volumetric flask.
3) put some water in the flask and a
measured amount of concentrated
4) add solvent to the scratch and shake to
D) To calculate the amount of
concentrated solution required, the major
concept to keep in mind is that the moles
of solute obtained from the concentrated
solution are equal to the moles of solute in
the dilute solution.
Upon dilution, the moles of solute do not
change only the amount of solvent
moles of solute in the conc. solution =
moles of solute in the dilute solution.

            Mc X Lc or Mc X Vc
         Md X Ld or Md X Vd

          Mc X Vc = Md X Vd
How would you prepare 250.0 mL of
0.100M H2SO4 from 6.00 M H2SO4?
1) first calculate the amount of
concentrated sulfuric acid (6.00 M) we

2) obtain a 250.00 mL volumetric flask.
3) put some water in the flask and add the
4.17 mL H2SO4 from 6.00 M bottle.
4) fill to the scratch with water and