Chapter 4 Chemical Quantities and Aqueous Reactions

Report
Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Chapter 4
Chemical
Quantities
and Aqueous
Reactions
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright © 2011 Pearson Education, Inc.
Global Warming
• Scientists have measured an average 0.6 °C rise
in atmospheric temperature since 1860
• During the same period atmospheric CO2 levels
have risen 25%
• Are the two trends causal?
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The Sources of Increased CO2
• One source of CO2 is the combustion reactions of fossil
fuels we use to get energy
• Another source of CO2 is volcanic action
• How can we judge whether global warming is natural or
due to our use of fossil fuels?
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Quantities in Chemical Reactions
• The amount of every substance used and
made in a chemical reaction is related to
the amounts of all the other substances in
the reaction
Law of Conservation of Mass
Balancing equations by balancing atoms
• The study of the numerical relationship
between chemical quantities in a chemical
reaction is called stoichiometry
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Reaction Stoichiometry
• The coefficients in a balanced chemical
equation specify the relative amounts in moles
of each of the substances involved in the
reaction
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
2 molecules of C8H18 react with 25 molecules of O2
to form 16 molecules of CO2 and 18 molecules of H2O
2 moles of C8H18 react with 25 moles of O2
to form 16 moles of CO2 and 18 moles of H2O
2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O
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Making Pizza
• The number of pizzas you can make depends
on the amount of the ingredients you use
1 crust + 5 oz. tomato sauce + 2 cu cheese  1 pizza
• This relationship can be expressed mathematically
1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza
• If you want to make more or less than one pizza,
you can use the amount of cheese you have to
determine the number of pizzas you can make
 assuming you have enough crusts and tomato sauce
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Predicting Amounts from Stoichiometry
• The amounts of any other 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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Practice
• According to the following equation, how
many moles of water are made in the
combustion of 0.10 moles of glucose?
C6H12O6 + 6 O2  6 CO2 + 6 H2O
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Practice  How many moles of water are made in the
combustion of 0.10 moles of glucose?
Given:
Find:
0.10 moles C6H12O6
moles H2O
Conceptual
Plan:
Relationships:
mol C6H12O6
mol H2O
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O 1 mol C6H12O6 : 6 mol H2O
Solution:
0.6 mol H2O = 0.60 mol H2O
Check:
because 6x moles of H2O as C6H12O6, the number makes
sense
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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
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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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Which Produces More CO2;
Volcanoes or Fossil Fuel Combustion?
• Our calculation just showed that the world
produced 1.1 x 1016 g of CO2 just from
petroleum combustion in 2007
1.1 x 1013 kg CO2
• Estimates of volcanic CO2 production are
•
2 x 1011 kg/year
This means that volcanoes produce less
than 2% of the CO2 added to the air annually
2 . 0  10
1 . 1  10
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11 kg
yr
13 kg
 100 %  1 . 8 %
yr
12
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Example 4.1: How many grams of glucose can be
synthesized from 37.8 g of CO2 in photosynthesis?
Given:
Find:
37.8 g CO2, 6 CO2 + 6 H2O  C6H12O6+ 6 O2
g C6H12O6
Conceptual g CO2
mol CO2
mol C6H12O6
g C6H12O6
1mol
180.2 g
Plan:
1mol C6H12O6
44.01 g
Relationships:
6 mol CO2
1 mol
1 mol C6H12O6 = 180.2g, 1 mol CO2 = 44.01g, 1 mol C6H12O6 : 6 mol CO2
Solution:
1 mol C6H12O6 180.2 g C6H12O6
1 mol CO2



37.8 g CO2
44.01 g CO2
6 mol CO 2
1 mol C6H12O6
 25.8 g C6H12O 6
Check: because 6x moles of CO as C H O , but the molar mass
2
6 12 6
of C6H12O6 is 4x CO2, the number makes sense
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Practice — How many grams of O2 can be made
from the decomposition of 100.0 g of PbO2?
2 PbO2(s) → 2 PbO(s) + O2(g)
(PbO2 = 239.2, O2 = 32.00)
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Practice — How many grams of O2 can be made from
the decomposition of 100.0 g of PbO2?
2 PbO2(s) → 2 PbO(s) + O2(g)
Given:
Find:
100.0 g PbO2, 2 PbO2 → 2 PbO + O2
g O2
Conceptual g PbO2
Plan:
Relationships:
mol PbO2
mol O2
g O2
1 mol O2 = 32.00g, 1 mol PbO2 = 239.2g, 1 mol O2 : 2 mol PbO2
Solution:
Check: because ½ moles of O as PbO , and the molar mass of
2
2
PbO2 is 7x O2, the number makes sense
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Stoichiometry Road Map
Pure
Substance
Solution
aAbB
% A(aq)
ppm A(aq)
M A(aq)
M B(aq)
% B(aq)
ppm B(aq)
M = moles
L
Moles A
MM
22.4 L
mass A
density
equation
MM
22.4 L
Volume A(g)
equation
volume A (l)
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Moles B
Volume B(g)
mass B
density
volume B(l)
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More Making Pizzas
• We know that
1 crust + 5 oz. tomato sauce + 2 cu cheese  1 pizza
• But what would happen if we had 4 crusts,
15 oz. tomato sauce, and 10 cu cheese?
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More Making Pizzas, Continued
• Each ingredient could potentially make a
•
•
different number of pizzas
But all the ingredients have to work together!
We only have enough tomato sauce to make
three pizzas, so once we make three pizzas,
the tomato sauce runs out no matter how much
of the other ingredients we have.
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More Making Pizzas, Continued
• The tomato sauce limits the amount of pizzas
we can make. In chemical reactions we call
this the limiting reactant.
 also known as the limiting reagent
• The maximum number of pizzas we can
make depends on this ingredient. In
chemical reactions, we call this the
theoretical yield.
 it also determines the amounts of the other
ingredients we will use!
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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 and Excess Reactants in the
Combustion of Methane
•
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Our balanced equation for the combustion of
methane implies that every one molecule of CH4
reacts with two molecules of O2
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Limiting and Excess Reactants in the
Combustion of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
• If we have five molecules of CH4 and eight molecules
of O2, which is the limiting reactant?
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since less CO2
can be made
from the O2
than the CH4,
so the O2 is
the limiting
reactant
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Practice — How many moles of Si3N4 can be
made from 1.20 moles of Si and 1.00 moles
of N2 in the reaction 3 Si + 2 N2  Si3N4?
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Practice — How many moles of Si3N4 can be made from
1.20 moles of Si and 1.00 moles of N2 in the reaction
3 Si + 2 N2  Si3N4?
Given:
Find:
Conceptual
Plan:
1.20 mol Si, 1.00 mol N2
mol Si3N4
mol Si
mol Si3N4
mol N2
mol Si3N4
Pick least
amount
Relationships:
Limiting
reactant and
theoretical
yield
2 mol N2 : 1 Si3N4; 3 mol Si : 1 Si3N4
Solution: Limiting
reactant
Theoretical
yield
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More Making Pizzas
• Let’s now assume that as we are making
•
pizzas, we burn a pizza, drop one on the
floor, or other uncontrollable events happen
so that we only make two pizzas. The actual
amount of product made in a chemical
reaction is called the actual yield.
We can determine the efficiency of making
pizzas by calculating the percentage of the
maximum number of pizzas we actually
make. In chemical reactions, we call this the
percent yield.
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Theoretical and Actual Yield
• As we did with the pizzas, in order to determine
•
the theoretical yield, we should use reaction
stoichiometry to determine the amount of product
each of our reactants could make
The theoretical yield will always be the least
possible amount of product
 the theoretical yield will always come from the limiting
reactant
• Because of both controllable and uncontrollable
factors, the actual yield of product will always be
less than the theoretical yield
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Example 4.4:
Finding limiting reactant,
theoretical yield, and
percent yield
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Example:
• When 28.6 kg of C are allowed to react with 88.2 kg
of TiO2 in the reaction below, 42.8 kg of Ti are
obtained. Find the limiting reactant, theoretical
yield, and percent yield.
TiO 2 (s)  2 C (s)  Ti(s)  2 C O (g)
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Example:
When 28.6 kg of C reacts with
88.2 kg of TiO2, 42.8 kg of Ti are
obtained. Find the limiting
reactant, theoretical yield, and
percent yield
TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
• Write down the given quantity and its units
Given:
28.6 kg C
88.2 kg TiO2
42.8 kg Ti produced
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Information
Example:
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find the limiting
reactant, theoretical
yield, and percent yield
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
• Write down the quantity to find and/or its units
Find: limiting reactant
theoretical yield
percent yield
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Example:
Find the limiting
reactant, theoretical
yield, and percent
yield
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
• Write a conceptual plan
kg
C
}
kg
TiO2
smallest
amount is
from
limiting
reactant
smallest
mol Ti
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Example:
Find the limiting
reactant, theoretical
yield, and percent
yield
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
• Collect needed relationships
1000 g = 1 kg
Molar Mass TiO2 = 79.87 g/mol
Molar Mass Ti = 47.87 g/mol
Molar Mass C = 12.01 g/mol
1 mole TiO2 : 1 mol Ti (from the chem. equation)
2 mole C : 1 mol Ti (from the chem. equation)
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Example:
Find the limiting
reactant,
theoretical yield,
and percent yield
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
• Apply the conceptual plan
limiting reactant
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smallest moles of Ti
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Example:
Find the limiting
reactant,
theoretical yield,
and percent yield
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
• Apply the conceptual plan
theoretical yield
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Example:
Find the limiting
reactant,
theoretical yield,
and percent yield
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
• Apply the conceptual plan
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Example:
Find the limiting
reactant,
theoretical yield,
and percent yield
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
• Check the solutions
limiting reactant = TiO2
theoretical yield = 52.9 kg
percent yield = 80.9%
Because Ti has lower molar mass than TiO2, the T.Y. makes
sense and the percent yield makes sense as it is less than 100%
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Practice — How many grams of N2(g) can be made
from 9.05 g of NH3 reacting with 45.2 g of CuO?
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
If 4.61 g of N2 are made, what is the percent yield?
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Practice — How many grams of N2(g) can be made from 9.05 g of NH3 reacting
with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
If 4.61 g of N2 are made, what is the percent yield?
Given:
Find:
9.05 g NH3, 45.2 g CuO
g N2
Conceptual g NH
3
Plan:
mol NH3
mol N2
Choose
smallest
g CuO
mol CuO
g N2
mol N2
Relationships: 1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g
2 mol NH3 : 1 mol N2, 3 mol CuO : 1 mol N2
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Practice — How many grams of N2(g) can be made from 9.05 g of NH3 reacting
with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
If 4.61 g of N2 are made, what is the percent yield?
Solution:
Theoretical
yield
Check:
because the percent yield is less than 100,
the answer makes sense
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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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Describing Solutions
• Because solutions are mixtures, the composition
can vary from one sample to another
pure substances have constant composition
saltwater samples from different seas or lakes have
different amounts of salt
• So to describe solutions accurately, we must
describe how much of each component is
present
we saw that with pure substances, we can describe
them with a single name because all samples are
identical
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Solution Concentration
• Qualitatively, solutions are
often described as dilute or
concentrated
• Dilute solutions have a
small amount of solute
compared to solvent
• Concentrated solutions
have a large amount of
solute compared to solvent
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Concentrations—Quantitative
Descriptions of Solutions
• A more precise method for describing a
•
solution is to quantify the amount of solute in a
given amount of solution
Concentration = amount of solute in a given
amount of solution
occasionally amount of 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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Practice — What Is the molarity of a solution containing
3.4 g of NH3 (MM 17.03) in 200.0 mL of solution?
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Practice — What Is the molarity of a solution containing
3.4 g of NH3 (MM 17.03) in 200.0 mL of solution?
Given: 3.4
0.20gmol
NH3NH
, 200.0
mL solution
L solution
3, 0.2000
Find: M
Conceptual g NH3
mol NH3
Plan:
mL sol’n
L sol’n
M
Relationships: M = mol/L, 1 mol NH3 = 17.03 g, 1 mL = 0.001 L
Solve:
Check: the unit is correct, the number is reasonable because
the fraction of moles is less than the fraction of liters
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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 — Determine the mass of CaCl2
(MM = 110.98) in 1.75 L of 1.50 M solution
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Practice — Determine the mass of CaCl2
(MM = 110.98) in 1.75 L of 1.50 M solution
Given: 1.50 M CaCl2, 1.75 L
Find: mass CaCl2, g
Conceptual
Plan:
L sol’n
mol CaCl2
g CaCl2
Relationships: 1.50 mol CaCl2 = 1 L solution; 110.98 g CaCl2 = 1 mol
Solution:
Check:
because each L has 1.50 mol CaCl2, it makes sense
that 1.75 L should have almost 3 moles
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Example: How would you prepare 250.0 mL of a
1.00 M solution CuSO45 H2O(MM 249.69)?
Given: 250.0 mL solution
Find: mass CuSO4 5 H2O, g
mol CuSO4
Conceptual mL sol’n
L sol’n
Plan:
Relationships:
g CuSO4
1.00 L sol’n = 1.00 mol; 1 mL = 0.001 L; 1 mol = 249.69 g
Solution:
Dissolve 62.4 g of CuSO4∙5H2O 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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Practice – How would you prepare 250.0 mL
of 0.150 M CaCl2 (MM = 110.98)?
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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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Practice – What is the concentration of a
solution prepared by diluting 45.0 mL of
8.25 M HNO3 to 135.0 mL?
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Practice – What is the concentration of a solution prepared
by diluting 45.0 mL of 8.25 M HNO3 to 135.0 mL?
Given: V1 = 45.0 mL, M1 = 8.25 M, V2 = 135.0 mL
Find: M2, L
Conceptual
Plan:
Relationships:
V1, M1, V2
M2
M1V1 = M2V2
Solution:
Check:
because the solution is diluted by a factor of 3, the
molarity should decrease by a factor of 3, and it does
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Practice – How would you prepare 200.0 mL of
0.25 M NaCl solution from a 2.0 M solution?
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Practice – How would you prepare 200.0 mL of 0.25 M
NaCl solution from a 2.0 M solution?
Given: M1 = 2.0 M, M2 = 0.25 M, V2 = 200.0 mL
Find: V1, L
Conceptual
Plan:
Relationships:
M1, M2, V2
V1
M1V1 = M2V2
Solution:
Dilute 25 mL of 2.0 M solution up to 200.0 mL
Check: because the solution is diluted by a factor of 8, the volume
should increase by a factor of 8, and it does
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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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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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When Will a Salt Dissolve?
• Predicting whether a compound will dissolve in
•
water is not easy
The best way to do it is to do some
experiments to test whether a compound will
dissolve in water, then develop some rules
based on those experimental results
we call this method the empirical method
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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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Electrolytes and Nonelectrolytes
• Materials that dissolve
in water to form a
solution that will
conduct electricity are
called electrolytes
• Materials that dissolve
in water to form a
solution that will not
conduct electricity are
called nonelectrolytes
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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
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do not dissociate when
they dissolve
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Acids
• Acids are molecular compounds that ionize when
they dissolve in water
 the molecules are pulled apart by their attraction for the
water
 when acids ionize, they form H+ cations and also anions
• The percentage of molecules that ionize varies
•
•
from one acid to another
Acids that ionize virtually 100% are called strong
acids
HCl(aq)  H+(aq) + Cl−(aq)
Acids that only ionize a small percentage are called
weak acids
HF(aq)  H+(aq) + F−(aq)
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Strong and Weak Electrolytes
• Strong electrolytes are materials that dissolve
completely as ions
 ionic compounds and strong acids
 their solutions conduct electricity well
• Weak electrolytes are materials that dissolve
mostly as molecules, but partially as ions
 weak acids
 their solutions conduct electricity, but not well
• When compounds containing a polyatomic ion
dissolve, the polyatomic ion stays together
HC2H3O2(aq)  H+(aq) + C2H3O2−(aq)
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Dissociation and Ionization
•
•
•
When ionic compounds dissolve in water, the anions and
cations are separated from each other. This is called
dissociation.
Na2S(aq)  2 Na+(aq) + S2-(aq)
When compounds containing polyatomic ions dissociate,
the polyatomic group stays together as one ion
Na2SO4(aq)  2 Na+(aq) + SO42−(aq)
When strong acids dissolve in water, the molecule
ionizes into H+ and anions
H2SO4(aq)  H+(aq) + HSO4−(aq)
HSO4−(aq)  H+(aq) + SO42− (aq)
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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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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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No Precipitate Formation =
No Reaction
KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq)
all ions still present,  no reaction
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Process for Predicting the Products of
a Precipitation Reaction
1. Determine what ions each aqueous reactant has
2. Determine formulas of possible products
 exchange ions

(+) ion from one reactant with (-) ion from other
 balance charges of combined ions to get formula of each
product
3. Determine solubility of each product in water
 use the solubility rules
 if product is insoluble or slightly soluble, it will precipitate
4. If neither product will precipitate, write no reaction
after the arrow
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Process for Predicting the Products of
a Precipitation Reaction
5. If any of the possible products are insoluble,
write their formulas as the products of the
reaction using (s) after the formula to indicate
solid. Write any soluble products with (aq)
after the formula to indicate aqueous.
6. Balance the equation
 remember to only change coefficients, not
subscripts
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Example 4.10: Write the equation for the
precipitation reaction between an aqueous
solution of potassium carbonate and an aqueous
solution of nickel(II) chloride
1. Write the formulas of the reactants
K2CO3(aq) + NiCl2(aq) 
2. Determine the possible products
a) determine the ions present
(K+ + CO32−) + (Ni2+ + Cl−) 
b) exchange the Ions
(K+ + CO32−) + (Ni2+ + Cl−)  (K+ + Cl−) + (Ni2+ + CO32−)
c) write the formulas of the products

balance charges
K2CO3(aq) + NiCl2(aq)  KCl + NiCO3
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Example 4.10: Write the equation for the
precipitation reaction between an aqueous
solution of potassium carbonate and an
aqueous solution of nickel(II) chloride
3. Determine the solubility of each product
KCl is soluble
NiCO3 is insoluble
4. If both products are soluble, write no
reaction
does not apply because NiCO3 is insoluble
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Example 4.10: Write the equation for the
precipitation reaction between an aqueous
solution of potassium carbonate and an
aqueous solution of nickel(II) chloride
5. Write (aq) next to soluble products and (s)
next to insoluble products
K2CO3(aq) + NiCl2(aq)  KCl(aq) + NiCO3(s)
6. Balance the equation
K2CO3(aq) + NiCl2(aq)  2 KCl(aq) + NiCO3(s)
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Practice – Predict the products and
balance the equation
KCl(aq) + AgNO3(aq) 
(K+ + Cl−) + (Ag+ + NO3−) → (K+ + NO3−) + (Ag+ + Cl−)
KCl(aq) + AgNO3(aq) → KNO3 + AgCl
KCl(aq) + AgNO3(aq)  KNO3(aq) + AgCl(s)
Na2S(aq) + CaCl2(aq) 
(Na+ + S2−) + (Ca2+ + Cl−) → (Na+ + Cl−) + (Ca2+ + S2−)
Na2S(aq) + CaCl2(aq) → NaCl + CaS
Na2S(aq) + CaCl2(aq)  NaCl(aq) + CaS(aq)
No reaction
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Practice – Write an equation for the
reaction that takes place when an
aqueous solution of (NH4)2SO4 is mixed
with an aqueous solution of Pb(C2H3O2)2.
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) 
(NH4+ + SO42−) + (Pb2+ + C2H3O2−) → (NH4+ + C2H3O2−) + (Pb2+ + SO42−)
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq)  NH4C2H3O2 + PbSO4
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) 
NH4C2H3O2(aq) + PbSO4(s)
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) 
2 NH4C2H3O2(aq) + PbSO4(s)
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Ionic Equations
• Equations that describe the chemicals put into the
water and the product molecules are called molecular
equations
2 KOH(aq) + Mg(NO3)2(aq)  2 KNO3(aq) + Mg(OH)2(s)
• Equations that describe the material’s structure when
dissolved are called complete ionic equations
 aqueous strong electrolytes are written as ions
 soluble salts, strong acids, strong bases
 insoluble substances, weak electrolytes, and nonelectrolytes
are written in molecule form
 solids, liquids, and gases are not dissolved, therefore molecule form
2K+(aq) + 2OH−(aq) + Mg2+(aq) + 2NO3−(aq)  2K+(aq) + 2NO3−(aq) + Mg(OH)2(s)
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Ionic Equations
• Ions that are both reactants and products are called
spectator ions
2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq)  2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s)
 An ionic equation in which the spectator ions are
removed is called a net ionic equation
2 OH−(aq) + Mg2+(aq)  Mg(OH)2(s)
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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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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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Common Acids
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Common Bases
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HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
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Example: Write the molecular, ionic, and netionic equation for the reaction of aqueous
nitric acid with aqueous calcium hydroxide
1.
Write the formulas of the reactants
HNO3(aq) + Ca(OH)2(aq) 
2.
Determine the possible products
a) determine the ions present when each reactant
dissociates or ionizes
(H+ + NO3−) + (Ca2+ + OH−) 
b) exchange the ions, H+ combines with OH− to make H2O(l)
(H+ + NO3−) + (Ca2+ + OH−)  (Ca2+ + NO3−) + H2O(l)
c
write the formula of the salt
(H+ + NO3−) + (Ca2+ + OH−)  Ca(NO3)2 + H2O(l)
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Example: Write the molecular, ionic, and netionic equation for the reaction of aqueous
nitric acid with aqueous calcium hydroxide
3. Determine the solubility of the salt
Ca(NO3)2 is soluble
4. Write an (s) after the insoluble products and an
(aq) after the soluble products
HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + H2O(l)
5. Balance the equation
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
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Example: Write the molecular, ionic, and netionic equation for the reaction of aqueous
nitric acid with aqueous calcium hydroxide
6. Dissociate all aqueous strong electrolytes to get
complete ionic equation
 not H2O
2 H+(aq) + 2 NO3−(aq) + Ca2+(aq) + 2 OH−(aq) 
Ca2+(aq) + 2 NO3−(aq) + H2O(l)
7. Eliminate spectator ions to get net-ionic equation
2 H+(aq) + 2 OH−(aq)  2 H2O(l)
H+(aq) + OH−(aq)  H2O(l)
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Practice – Predict the products and
balance the equation
HCl(aq) + Ba(OH)2(aq) 
(H+ + Cl−) + (Ba2+ + OH−) → (H+ + OH−) + (Ba2+ + Cl−)
HCl(aq) + Ba(OH)2(aq) → H2O(l) + BaCl2
2 HCl(aq) + Ba(OH)2(aq)  2 H2O(l) + BaCl2(aq)
H2SO4(aq) + Sr(OH)2(aq) 
(H+ + SO42−) + (Sr2+ + OH−) → (H+ + OH−) + (Sr2+ + SO42−)
H2SO4(aq) + Sr(OH)2(aq) → H2O(l) + SrSO4
H2SO4(aq) + Sr(OH)2(aq) → 2 H2O(l) + SrSO4
H2SO4(aq) + Sr(OH)2(aq)  2 H2O(l) + SrSO4(s)
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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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NaHCO3(aq) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
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Compounds that Undergo
Gas-Evolving Reactions
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Example 4.14: When an aqueous solution of
sodium carbonate is added to an aqueous
solution of nitric acid, a gas evolves
1. Write the formulas of the reactants
Na2CO3(aq) + HNO3(aq) 
2. Determine the possible products
a) determine the ions present when each reactant
dissociates or ionizes
(Na+ + CO32−) + (H+ + NO3−) 
b) exchange the anions
(Na+ + CO32−) + (H+ + NO3−)  (Na+ + NO3−) + (H+ + CO32−)
c) write the formula of compounds
 balance the charges
Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3
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Example 4.14: When an aqueous solution of
sodium carbonate is added to an aqueous
solution of nitric acid, a gas evolves
3. Check to see if either product is H2S - No
4. Check to see if either product decomposes –
Yes
 H2CO3 decomposes into CO2(g) + H2O(l)
Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l)
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Example 4.14: When an aqueous solution of
sodium carbonate is added to an aqueous
solution of nitric acid, a gas evolves
5. Determine the solubility of other product
NaNO3 is soluble
6. Write an (s) after the insoluble products and an
(aq) after the soluble products
Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + CO2(g) + H2O(l)
7. Balance the equation
Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + CO2(g) + H2O(l)
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Practice – Predict the products and
balance the equation
HCl(aq) + Na2SO3(aq) 
(H+ + Cl−) + (Na+ + SO32−) → (H+ + SO32−) + (Na+ + Cl−)
HCl(aq) + Na2SO3(aq) → H2SO3 + NaCl
HCl(aq) + Na2SO3(aq) → H2O(l) + SO2(g) + NaCl
2 HCl(aq) + Na2SO3(aq)  H2O(l) + SO2(g) + 2 NaCl(aq)
H2SO4(aq) + CaS(aq) 
(H+ + SO42−) + (Ca2+ + S2−) → (H+ + S2−) + (Ca2+ + SO42−)
H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4
H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4(s)
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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 oxidation-reduction 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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Combustion as Redox
2 H2(g) + O2(g)  2 H2O(g)
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Redox without Combustion
2 Na(s) + Cl2(g)  2 NaCl(s)
2 Na  2 Na+ + 2 e
Cl2 + 2 e  2 Cl
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Reactions of Metals with
Nonmetals
• Consider the following reactions:
•
•
4 Na(s) + O2(g) → 2 Na2O(s)
2 Na(s) + Cl2(g) → 2 NaCl(s)
The reactions involve a metal reacting with a
nonmetal
In addition, both reactions involve the conversion
of free elements into ions
4 Na(s) + O2(g) → 2 Na+2O2– (s)
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
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Oxidation and Reduction
• To convert a free element into an ion, the
atoms must gain or lose electrons
 of course, if one atom loses electrons, another must
accept them
• Reactions where electrons are transferred
•
from one atom to another are redox reactions
Atoms that lose electrons are being oxidized,
atoms that gain electrons are being reduced
Ger
Na+Cl–(s)
2 Na(s) + Cl2(g) → 2
Na → Na+ + 1 e– oxidation
Cl2 + 2 e– → 2 Cl– reduction
Leo
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Electron Bookkeeping
• For reactions that are not metal + nonmetal, or
•
do not involve O2, we need a method for
determining how the electrons are transferred
Chemists assign a number to each element in
a reaction called an oxidation state that
allows them to determine the electron flow in
the reaction
 even though they look like them, oxidation states
are not ion charges!
oxidation states are imaginary charges assigned
based on a set of rules
ion charges are real, measurable charges
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Rules for Assigning Oxidation States
• Rules are in order of priority
1. free elements have an oxidation state = 0
 Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)
2. monatomic ions have an oxidation state equal
to their charge
 Na = +1 and Cl = −1 in NaCl
3. (a) the sum of the oxidation states of all the
atoms in a compound is 0
 Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0
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Rules for Assigning Oxidation States
3. (b) the sum of the oxidation states of all the
atoms in a polyatomic ion equals the charge
on the ion
 N = +5 and O = −2 in NO3–, (+5) + 3(−2) = −1
4. (a) Group I metals have an oxidation state of
+1 in all their compounds
 Na = +1 in NaCl
4. (b) Group II metals have an oxidation state of
+2 in all their compounds
 Mg = +2 in MgCl2
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Rules for Assigning Oxidation States
5. in their compounds, nonmetals have oxidation
states according to the table below
 nonmetals higher on the table take priority
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Example: Determine the oxidation states of all
the atoms in a propanoate ion, C3H5O2–
• There are no free elements or free ions in
•
propanoate, so the first rule that applies is Rule 3b
(C3) + (H5) + (O2) = −1
Because all the atoms are nonmetals, the next
rule we use is Rule 5, following the elements in
order:
 H = +1
 O = −2
(C3) + 5(+1) + 2(−2)Note:
= −1unlike charges,
(C3) = −2 oxidation states can
be fractions!
C = −⅔
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Practice – Assign an oxidation state
to each element in the following
• Br2
Br = 0, (Rule 1)
• K+
K = +1, (Rule 2)
• LiF
Li = +1, (Rule 4a) & F = −1, (Rule 5)
• CO2
O = −2, (Rule 5) & C = +4, (Rule 3a)
• SO42−
O = −2, (Rule 5) & S = +6, (Rule 3b)
• Na2O2 Na = +1, (Rule 4a) & O = −1 , (Rule 3a)
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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
0
+4 –2
+1 −2
oxidation
reduction
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Oxidation–Reduction
• Oxidation and reduction must occur simultaneously
 if an atom loses electrons another atom must take them
• The reactant that reduces an element in another
reactant is called the reducing agent
 the reducing agent contains the element that is oxidized
• The reactant that oxidizes an element in another
reactant is called the oxidizing agent
 the oxidizing agent contains the element that is reduced
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2 is the oxidizing agent
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Example: Assign oxidation states,
determine the element oxidized and reduced,
and determine the oxidizing agent and
reducing agent in the following reactions:
Reducing
agent
Oxidizing
agent
Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O
0
+7
−2
+1
+3
+4 −2
+1 −2
Reduction
Oxidation
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Practice – 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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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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Practice — Solution stoichiometry
• 43.8 mL of 0.107 M HCl is needed to
neutralize 37.6 mL of Ba(OH)2 solution.
What is the molarity of the base?
2 HCl + Ba(OH)2  BaCl2 + 2 H2O
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Practice – 43.8 mL of 0.107 M HCl is needed to neutralize
37.6 mL of Ba(OH)2 solution. What is the molarity of the
base? 2 HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + 2 H2O(aq)
Given: 0.0438
43.8 mL
L of 0.107 M HCl, 0.0376
37.6 mLLBa(OH)
Ba(OH)22
MBa(OH)
Ba(OH)22
Find: M
Conceptual
Plan:
L HCl
mol HCl
mol Ba(OH)2
M Ba(OH)2
L Ba(OH)2
Relationships: 1 mL= 0.001 L, 1 L HCl = 0.107 mol, 1 mol Ba(OH)2 : 2 mol HCl
Solution:
Check:
the units are correct, the number makes sense because
there are fewer moles than liters
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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
 a long glass tube with precise volume markings that
allows small additions of solution
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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
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Titration
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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.
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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
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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
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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
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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
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mol
NaOH
131
mol
HCl
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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
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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
• Apply the conceptual plan
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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.
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Practice − What is the concentration of
NaOH solution that requires 27.5 mL to
titrate 50.0 mL of 0.1015 M H2SO4?
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
50.0 mL 27.5 mL
0.1015 M ? M
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Practice — What is the concentration of NaOH solution that
requires 27.5 mL to titrate 50.0 mL of 0.1015 M H2SO4?
2 NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2 H2O(aq)
Given:
Find:
50.0 mLLof
0.0500
of0.1015
0.1015M
MHH22SO
SO44,,27.5
0.0275
mLLNaOH
NaOH
M NaOH
Conceptual L H SO
2
4
Plan:
mol H2SO4
mol NaOH
M NaOH
L NaOH
Relationships: 1 mL= 0.001L, 1 LH2SO4 = 0.1015mol, 2mol NaOH : 1mol H2SO4
Solution:
Check:
the units are correct, the number makes because the volume of
NaOH is about ½ the H2SO4, but the stoichiometry says you need
twice the moles of NaOH as H2SO4
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