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Chapter 4
Chemical Quantities and
Aqueous Reactions
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Stoichiometry (4.2)
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Amounts of Reactants and Products
• Answer the question: “How much?”
 How much product will be formed?
 How much reactant is needed?
• Stoichiometry - The quantitative study of reactants and
products in a chemical reaction.
• We need to learn one more “tool.”
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We will consider the following reaction:
2H2 + O2
2H2O
• This reaction says:
•2 molec H2 reacts with 1 molec O2 to give 2 molec water
•2 moles H2 reacts with 1 mole O2 to give 2 moles water
• We want to think of the coefficients in terms of moles.
•If 4 moles of H2 reacts with 2 moles O2, how many moles of
water would be formed?
•4 moles H2O
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• The coefficients in a balanced equation are
always used to convert between moles of
substances in a chemical reaction.
• Let us see how we use this as a unit factor.
4 mol H2 X
2 mol H2O = 4 mol H O
2
2 mol H2
numbers come from
balanced equation
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Practice  How many moles of water are made in the
combustion of 0.10 moles of glucose?
Given:
Find:
Conceptual
Plan:
Relationships:
0.10 moles C6H12O6
moles H2O
mol C6H12O6
mol H2O
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O 1 mol C6H12O6 : 6 mol H2O
Solution:
Check:
0.6 mol H2O = 0.60 mol H2O
because 6x moles of H2O as C6H12O6, the number makes
sense
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Amounts of Reactants and Products
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the
number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
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Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
molar mass
CH3OH
209 g CH3OH x
moles H2O
coefficients
chemical equation
grams H2O
molar mass
H 2O
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
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Predicting Amounts from Stoichiometry
• The amounts of any substance in a chemical reaction can be
determined from the amount of just one substance
• How much CO2 can be made from 22.0 moles of C8H18 in the
combustion of C8H18?
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
2 moles C8H18 : 16 moles CO2
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Example: Estimate the mass of CO2 produced in
2007 by the combustion of 3.5 x 1015 g gasolne
• Assuming that gasoline is octane, C8H18, the equation for
the reaction is
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
• The equation for the reaction gives the mole relationship
between amount of C8H18 and CO2, but we need to know
the mass relationship, so the conceptual plan will be
g C8H18
mol C8H18
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mol CO2
g CO2
Example: Estimate the mass of CO2 produced in
2007 by the combustion of 3.5 x 1015 g gasoline
Given: 3.4 x 1015 g C8H18
Find: g CO2
Conceptual g C8H18
Plan:
mol C8H18
mol CO2
g CO2
Relationships: 1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO2
Solution:
Check: because 8x moles of CO as C H , but the molar mass of
2
8 18
C8H18 is 3x CO2, the number makes sense
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The rapid decomposition of sodium azide, NaN3, to its
elements is one of the reactions used to inflate airbags:
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
How many grams of N2 are produced from 6.00 g of NaN3?
1.
2.
3.
4.
3.88 g
1.72 g
0.138 g
2.59 g
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Limiting Reagents and Yield (4.3)
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The Limiting Reactant
• For reactions with multiple reactants, it is likely that one of the
reactants will be completely used before the others
• When this reactant is used up, the reaction stops and no more product
is made
• The reactant that limits the amount of product is called the limiting
reactant
 sometimes called the limiting reagent
 the limiting reactant gets completely consumed
• Reactants not completely consumed are called excess reactants
• The amount of product that can be made from the limiting reactant is
called the theoretical yield
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Limiting Reagents
2NO + 2O2
2NO2
NO is the limiting reagent
O2 is the excess reagent
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Here’s a “non-chemical” example.
We want to build toy cars using wooden blocks and wheels.
Each car has one body (large block) and four wheels.
4 wheels + 1 block  1 car
We look in our toy box and find we have 10 large blocks and 24
wheels.
Which is our limiting “reagent”?
The wheels
So, how many cars can we build?
Six
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• The first step in doing a “Limiting Reagent”
problem is to identify it as one.
• When measured info is given for each of the
reactants (usually two,) you have a limiting
reactant problem.
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Procedure for Limiting Reagent Problems
• Calculate the moles of product formed for each reactant
given.
• The reactant which produces the least amount of product is
the limiting reagent.
• Using the information obtained by the limiting reagent,
finish the problem.
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Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed
OR
g Fe2O3
124 g Al x
mol Fe2O3
1 mol Al
27.0 g Al
x
mol Al needed
1 mol Fe2O3
2 mol Al
Start with 124 g Al
g Al needed
160. g Fe2O3
=
x
1 mol Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
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367 g Fe2O3
Use limiting reagent (Al) to calculate
amount of product that can be formed.
g Al
mol Al
mol Al2O3
2Al + Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
g Al2O3
Al2O3 + 2Fe
1 mol Al2O3
2 mol Al
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102. g Al2O3
=
x
1 mol Al2O3
234 g Al2O3
Reaction Yield
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
Theoretical Yield
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x 100
Nitroglycerin (C3H5N3O9) is a powerful explosive. Its decomposition may
be represented by:
4C3H5N3O9  6N2 + 12CO2 + 10 H2O + O2
What is the maximum amount of O2 in grams that can be obtained
from 2.00 x102 g of nitroglycerin? Calculate the percent yield in this
reaction if the amount of O2 generated is found to be 6.55 g.
molar mass of C3H5N3O9 = 227.1 g
2.00 x 102 g
x
1 mol C3H5N3O9
227.1 g C3H5N3O9
x
1 mol O2
x
4 mol C3H5N3O9
32.00 g O2
1 mol O2
= 7.05 g
Actual Yield
% Yield =
x 100
Theoretical Yield
= 6.55 g/7.05 g x 100
= 92.9%
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Ammonia is produced using the Haber process:
3 H2 + N2  2 NH3
What percent yield of ammonia produced from 15.0
kg each of H2 and N2, if 13.7 kg of product are
recovered? Assume the reaction goes to completion.
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7.53 x 10-2 %
1.50 x 10-1 %
75.3%
15.0%
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Solution Concentration (4.4)
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Solutions
• When table salt is mixed with water, it seems to disappear, or
become a liquid – the mixture is homogeneous
 the salt is still there, as you can tell from the taste, or simply
boiling away the water
• Homogeneous mixtures are called solutions
• The component of the solution that changes state is called the
solute
• The component that keeps its state is called the solvent
 if both components start in the same state, the major
component is the solvent
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Solution Concentration: Molarity
• Moles of solute per 1 liter of solution
• Used because it describes how many molecules of solute
in each liter of solution
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Preparing 1 L of a 1.00 M NaCl Solution
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Example 4.5: Find the molarity of a solution that has
25.5 g KBr dissolved in 1.75 L of solution
Given:
Find:
Conceptual
Plan:
25.5 g KBr, 1.75 L solution
molarity, M
g KBr
mol KBr
M
L sol’n
Relationships:
1 mol KBr = 119.00 g, M = moles/L
Solution:
Check:
because most solutions are between 0 and 18 M, the
answer makes sense
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Using Molarity in Calculations
• Molarity shows the relationship between the moles of
•
solute and liters of solution
If a sugar solution concentration is 2.0 M, then 1 liter of
solution contains 2.0 moles of sugar
 2 liters = 4.0 moles sugar
 0.5 liters = 1.0 mole sugar
• 1 L solution : 2 moles sugar
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Example 4.6: How many liters of 0.125 M NaOH
contain 0.255 mol NaOH?
Given: 0.125 M NaOH, 0.255 mol NaOH
Find: liters, L
Conceptual
Plan:
L sol’n
mol NaOH
Relationships: 0.125 mol NaOH = 1 L solution
Solution:
Check:
because each L has only 0.125 mol NaOH, it makes
sense that 0.255 mol should require a little more than 2 L
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Practice – How would you prepare 250.0 mL
of 0.150 M CaCl2?
Given: 250.0 mL solution
Find: mass CaCl2, g
Conceptual mL sol’n
L sol’n
Plan:
mol CaCl2
g CaCl2
Relationships: 1.00 L sol’n = 0.150 mol; 1 mL = 0.001L; 1 mol = 110.98 g
Solution:
Dissolve 4.16 g of CaCl2 in enough water to total 250.0 mL
Check: the unit is correct, the magnitude seems
reasonable as the volume is ¼ of a liter
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Dilution
• Often, solutions are stored as concentrated stock
solutions
• To make solutions of lower concentrations from these
stock solutions, more solvent is added
 the amount of solute doesn’t change, just the volume of solution
moles solute in solution 1 = moles solute in solution 2
• The concentrations and volumes of the stock and new
solutions are inversely proportional
M1∙V1 = M2∙V2
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Example 4.7: To what volume should you dilute
0.200 L of 15.0 M NaOH to make 3.00 M NaOH?
Given: V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M
Find: V2, L
Conceptual
Plan:
Relationships:
V1, M1, M2
V2
M1V1 = M2V2
Solution:
Check: because the solution is diluted by a factor of 5, the volume
should increase by a factor of 5, and it does
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Sulfuric acid is found in some types of batteries.
What volume of 3.50 M H2SO4 is required to
prepare 250.0 mL of 1.25 M H2SO4?
1.
2.
3.
4.
5.
17.5 mL
700. mL
89.3 mL
109 mL
None of the above
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Solution Stoichiometry
• Because molarity relates the moles of solute to the liters of
solution, it can be used to convert between amount of
reactants and/or products in a chemical reaction
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Example 4.8: What volume of 0.150 M KCl is required to
completely react with 0.150 L of 0.175 M Pb(NO3)2 in the
reaction 2 KCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2 KNO3(aq)?
Given:
Find:
Conceptual
Plan:
0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2
L KCl
L Pb(NO3)2
mol Pb(NO3)2
mol KCl
L KCl
Relationships: 1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 : 2 mol KCl
Solution:
Check:
because you need 2x moles of KCl as Pb(NO3)2, and the
molarity of Pb(NO3)2 > KCl, the volume of KCl should be
more than 2x the volume of Pb(NO3)2
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Types of Aqueous Solutions (4.5)
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What Happens When a Solute Dissolves?
• There are attractive forces between the solute particles
•
•
•
holding them together
There are also attractive forces between the solvent
molecules
When we mix the solute with the solvent, there are
attractive forces between the solute particles and the
solvent molecules
If the attractions between solute and solvent are strong
enough, the solute will dissolve
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Table Salt Dissolving in Water
Each ion is attracted to
the surrounding water
molecules and pulled off
and away from the crystal
When it enters the
solution, the ion is
surrounded by water
molecules, insulating it
from other ions
The result is a solution with
free moving charged
particles able to conduct
electricity
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An electrolyte is a substance that, when dissolved in water, results in a
solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved, results in a solution
that does not conduct electricity.
nonelectrolyte
weak electrolyte
strong electrolyte
To conduct electricity, there must be mobile ions. This occurs when a
compound breaks up (dissociates or ionizes) into its ions in water.
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The term “dissociation” is used for the breaking
up of a compound into cations and anions.
The term “ionization” is used to describe the
separation of acids and bases into ions.
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Molecular View of
Electrolytes and Nonelectrolytes
• To conduct electricity, a material must have charged
particles that are able to flow
• Electrolyte solutions all contain ions dissolved in the water
 ionic compounds are electrolytes because they dissociate into their
ions when they dissolve
• Nonelectrolyte solutions contain whole molecules
dissolved in the water
 generally, molecular compounds do not ionize when they dissolve
in water
 the notable exception being molecular acids
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Salt vs. Sugar Dissolved in Water
ionic compounds
dissociate into ions when
they dissolve
molecular compounds
do not dissociate when
they dissolve
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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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Practice – Write the equation for the process that
occurs when the following strong electrolytes
dissolve in water
CaCl2
CaCl2(aq)  Ca2+(aq) + 2 Cl−(aq)
HNO3
HNO3(aq)  H+(aq) + NO3−(aq)
(NH4)2CO3 (NH4)2CO3(aq)  2 NH4+(aq) + CO32−(aq)
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Solubility of Ionic Compounds
• Some ionic compounds, such as NaCl, dissolve very well in
water at room temperature
• Other ionic compounds, such as AgCl, dissolve hardly at all
in water at room temperature
• Compounds that dissolve in a solvent are said to be soluble,
where as those that do not are said to be insoluble
 NaCl is soluble in water, AgCl is insoluble in water
 the degree of solubility depends on the temperature
 even insoluble compounds dissolve, just not enough to be meaningful
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Solubility Rules Compounds that Are
Generally Soluble in Water
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Solubility Rules Compounds that Are
Generally Insoluble in Water
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Practice – Determine if each of the following is
soluble in water
KOH
KOH is soluble because it contains K+
AgBr
AgBr is insoluble; most bromides are soluble,
but AgBr is an exception
CaCl2
CaCl2 is soluble; most chlorides are soluble,
and CaCl2 is not an exception
Pb(NO3)2
Pb(NO3)2 is soluble because it contains NO3−
PbSO4
PbSO4 is insoluble; most sulfates are soluble,
but PbSO4 is an exception
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Which of the following compounds will be insoluble in
water?
1.
2.
3.
4.
5.
Ca(OH)2
Na3PO4
CaS
Hg2Cl2
LiF
Types of Reactions in Aqueous Solution
Precipitation Reactions (4.6)
Acid - Base Reactions (4.7)
Oxidation-Reduction Reactions (4.8)
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Precipitation Reactions
• Precipitation reactions are reactions in which a solid
forms when we mix two solutions
 reactions between aqueous solutions of ionic compounds
 produce an ionic compound that is insoluble in water
 the insoluble product is called a precipitate
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2 KI(aq) + Pb(NO3)2(aq)  PbI2(s) + 2 KNO3(aq)
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Predicting a precipitation reaction and writing
the chemical equation
1 Write the reactants for the reaction
2 Show the reactants in their ionic form, i.e., if they are
soluble, write their dissociated form
3 Determine which cation-anion combination yields an
insoluble compound
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4 Write the products of the reaction.
– The insoluble product is written in its combined form with
an (s) after it.
– The remaining ions are written as aqueous ions.
– This reaction equation is called the ionic equation.
5 Cancel the “spectator ions,” which are not involved in the
overall reaction.
6 What remains is the “net ionic equation,” which shows
only the species that actually take part in the reaction.
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Writing Net Ionic Equations -Example
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-
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AgCl (s)
Practice – Write the ionic and net ionic
equation for each
K2SO4(aq) + 2 AgNO3(aq)  2 KNO3(aq) + Ag2SO4(s)
2K+(aq) + SO42−(aq) + 2Ag+(aq) + 2NO3−(aq)
 2K+(aq) + 2NO3−(aq) + Ag2SO4(s)
2 Ag+(aq) + SO42−(aq)  Ag2SO4(s)
Na 2CO 3(aq) + 2 HCl(aq)  2 NaCl(aq) + CO 2(g) + H 2O(l)
2Na+(aq) + CO32−(aq) + 2H+(aq) + 2Cl−(aq)
 2Na+(aq) + 2Cl−(aq) + CO2(g) + H2O(l)
CO32−(aq) + 2 H+(aq)  CO2(g) + H2O(l)
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Types of Reactions in Aqueous Solution
Acid - Base Reactions (4.7)
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Acid-Base Reactions
• Also called neutralization reactions because the acid and
base neutralize each other’s properties
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
• The net ionic equation for an acid-base reaction is
H+(aq) + OH(aq)  H2O(l)
 as long as the salt that forms is soluble in water
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Acids and Bases in Solution
• Acids ionize in water to form H+ ions
 more precisely, the H from the acid molecule is donated to a water
molecule to form hydronium ion, H3O+
 most chemists use H+ and H3O+ interchangeably
• Bases dissociate in water to form OH ions
 bases, such as NH3, that do not contain OH ions, produce OH by
pulling H off water molecules
• In the reaction of an acid with a base, the H+ from the acid
combines with the OH from the base to make water
• The cation from the base combines with the anion from the
acid to make the salt
acid + base salt + water
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Some Additional Examples
HF + NaOH 
NaF + H2O
2HNO3 + Ba(OH)2 
Ba(NO3)2
H2SO4 + 2LiOH 
+ 2H2O
Li2SO4 + 2H2O
64
Titration
• Often in the lab, a solution’s concentration is
determined by reacting it with another material
and using stoichiometry – this process is called
titration
• In the titration, the unknown solution is added to a
known amount of another reactant until the
reaction is just completed. At this point, called the
endpoint, the reactants are in their stoichiometric
ratio.
 the unknown solution is added slowly from an
instrument called a burette
65
Acid-Base Titrations
• The difficulty is determining when there has been just
enough titrant added to complete the reaction
 the titrant is the solution in the burette
• In acid-base titrations, because both the reactant and
product solutions are colorless, a chemical is added that
changes color when the solution undergoes large changes
in acidity/alkalinity
 the chemical is called an indicator
• At the endpoint of an acid-base titration, the number of
moles of H+ equals the number of moles of OH
 also known as the equivalence point
66
Titration
The titrant is the base
solution in the burette.
As the titrant is added to
the flask, the H+ reacts with the
OH– to form water. But there is
still excess acid present so the
color does not change.
At the titration’s endpoint,
just enough base has been
added to neutralize all the acid.
At this point the indicator
changes color.
67
Color Change of Phenolphthalein in a Titration
68
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the end
point. What is the concentration of
the unknown HCl solution?
• Write down the given quantity and its units
Given:
10.00 mL HCl
12.54 mL of 0.100 M NaOH
69
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the end
point. What is the concentration of
the unknown HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
• Write down the quantity to find, and/or its units
Find:
concentration HCl, M
70
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the end
point. What is the concentration of
the unknown HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find: M HCl
• Collect needed equations and conversion factors
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
 1 mole HCl = 1 mole NaOH
0.100 M NaOH 0.100 mol NaOH  1 L sol’n
71
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the end
point. What is the concentration of
the unknown HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find: M HCl
Rel:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
• Write a conceptual plan
mL
NaOH
L
NaOH
mL
HCl
L
HCl
mol
NaOH
72
mol
HCl
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown
concentration requires 12.54
mL of 0.100 M NaOH solution
to reach the end point. What is
the concentration of the
unknown HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find: M HCl
Rel:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
• Apply the conceptual plan
= 1.25 x 103 mol HCl
73
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown
concentration requires 12.54
mL of 0.100 M NaOH solution
to reach the end point. What is
the concentration of the
unknown HCl solution?
• Apply the conceptual plan
74
Information
Given: 10.00 mL HCl
12.54 mL NaOH
Find: M HCl
Rel:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown
concentration requires 12.54
mL of 0.100 M NaOH solution
to reach the end point. What is
the concentration of the
unknown HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL NaOH
Find: M HCl
Rel:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
• Check the solution
HCl solution = 0.125 M
The units of the answer, M, are correct.
The magnitude of the answer makes sense because
the neutralization takes less HCl solution than
NaOH solution, so the HCl should be more concentrated.
75
Gas-Evolving Reactions
• Some reactions form a gas directly from the ion exchange
K2S(aq) + H2SO4(aq)  K2SO4(aq) + H2S(g)
• Other reactions form a gas by the decomposition of one of
the ion exchange products into a gas and water
K2SO3(aq) + H2SO4(aq)  K2SO4(aq) + H2SO3(aq)
H2SO3  H2O(l) + SO2(g)
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Types of Reactions in Aqueous Solution
Oxidation-Reduction Reactions (4.8)
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Other Patterns in Reactions
• The precipitation, acid-base, and gas-evolving reactions all
involve exchanging the ions in the solution
• Other kinds of reactions involve transferring electrons from
one atom to another – these are called oxidationreduction reactions
 also known as redox reactions
 many involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
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Oxidation-Reduction Reactions
Also called redox reactions
electron-transfer reactions
2Na(s) + Cl2(g)  2NaCl(s)
These are called half
reactions. One cannot occur
Na becomes Na+ ; it loses an electron.
without the other.
2Na  2Na+ + 2eThis is the oxidation half
reaction: loss of electrons.
Cl2 becomes 2Cl ; it gains electrons.
This is the reduction half
Cl2 + 2e  2Clreaction: gain of electrons.
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Redox Reactions
Loss of electrons – oxidation (LEO)
Gain of electrons – reduction (GER)
Reduction is easily remembered because it is
accompanied by a reduction in the charge.
Oxidizing agent - the substance being reduced (causes oxidation)
Reducing agent - the substance being oxidized (causes reduction)
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Zn (s) + CuSO4 (aq)
Zn
ZnSO4 (aq) + Cu (s)
Zn2+ + 2e- Zn is oxidized
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
Which of the following is a redox reaction?
1.
2.
3.
4.
5.
HBr (aq) + KOH (aq) → H2O (l) + KBr (aq)
SO3 (g) + H2O (l) → H2SO4 (aq)
HBr (aq) + Na2S (aq) → NaBr (aq) + H2S (g)
NH4+ (aq) + 2 O2 (g) → H2O (l) + NO3– (aq) + 2 H+ (aq)
None of the above
Oxidation Number
To understand more complex redox reactions, we must
know the oxidation number to determine what is oxidized
and what is reduced.
Oxidation number - the number of charges an atom would
have if electrons were transferred completely.
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Rules for assigning oxidation numbers to atoms
1 In free elements, each atom has an oxidation
number of zero.
 H2, Br2 S8, K, Al
ON = 0
2 For monatomic ions, the oxidation number is
the same as the charge of the ion.
 Li+ ON = +1
S2- ON = -2
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Rules (continued)
3 The ON of oxygen is usually -2. A common
exception is in hydrogen peroxide H2O2 and
other peroxides (O22-) in which the is ON is -1.
4 The ON of hydrogen is usually +1, except
when it is bonded to metals in binary
compounds, in which case it is -1.
 H2O ON of H = +1
 LiH
ON of H = -1
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Rules (continued)
5 Fluorine has an ON of -1. Other halogens can
be -1, if they are the halide anions, or they
can have various positive ONs, depending on
what they are combined with.
 NaF
ON of F = -1
 NaCl
ON of Cl = -1
 HClO4 ON of Cl  -1 revisit later
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Rules (continued)
6 In a neutral molecule, the sum of the ONs of
all the atoms must be zero. In a polyatomic
ion, the sum of the ONs of all the elements
must equal the net charge of the ion.
7 Oxidation numbers do not have to be
integers. e.g., O2- ON = -½
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The oxidation numbers of elements in their compounds
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IF7
Oxidation
numbers of all the
elements in the
following ?
F = -1
7x(-1) + ? = 0
I = +7
K2Cr2O7
NaIO3
O = -2
Na = +1 O = -2
3x(-2) + 1 + ? = 0
K = +1
7x(-2) + 2x(+1) + 2x(?) = 0
I = +5
Cr = +6
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Determine the oxidation number of the red element in
each of the following compounds:
H2PO4–
1.
2.
3.
4.
5.
+6
+5
+5
+5
+6
+6
+2
+4
+4
+4
SO32–
+4
+4
+4
+8
+4
N2O4
Oxidation and Reduction: Another Definition
• Oxidation occurs when an atom’s oxidation state increases
•
during a reaction
Reduction occurs when an atom’s oxidation state
decreases during a reaction
CH4 + 2 O2 → CO2 + 2 H2O
−4 +1
+4 –2
0
oxidation
reduction
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+1 −2
Assign oxidation states,
determine the element oxidized and reduced,
and determine the oxidizing agent and
reducing agent in the following reactions:
Sn4+ + Ca → Sn2+ + Ca2+
+4
0
+2
+2
Ca is oxidized, Sn4+ is reduced
Ca is the reducing agent, Sn4+ is the oxidizing agent
F2 + S → SF4
0
0
+4−1
S is oxidized, F is reduced
S is the reducing agent, F2 is the oxidizing agent
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