Download Chapter 4 Chemical Quantities and Aqueous Reactions

Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 4
Chemical Quantities and
Aqueous Reactions
•4.2 Reaction Stoichiometry: How Much Carbon
Dioxide?
•4.3 Limiting Reactant, Theoretical Yield, and
Percent Yield
•4.4 Solution Concentration and Solution
Stoichiomentry
•4.5 Types of Aqueous Solutions and Solubility
•4.6 Precipitation Reactions
•4.7 Representing Aqueous Reactions:
Molecular, Ionic, and Complete Ionic Equations
•4.8 Acid-Base and Gas-Evolution Reactions
•4.9 Oxidation-Reduction Reactions
2008, Prentice Hall
Do You Understand Molecular Mass?
Prozac, C17H18F3NO, is an antidepressant
that inhibits the uptake of serotonin in the
brain. What is the molar mass of
C17H18F3NO.
309.326 g/mol
Burns 4/e Chap 4
Slide 2 of 75
Do You Understand Molecular Mass?
The artificial sweetener aspartame (Nutra
Sweet) C14H18N2O5 is used to sweeten
foods, coffee, and soft drinks. How many
moles of aspartame are present in 225
grams of aspartame?
294.34 g/mol
Burns 4/e Chap 4
Slide 3 of 75
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
Burns 4/e Chap 4
Slide 4 of 75
Reaction Stoichiometry
• the numerical relationships between chemical amounts
in a reaction is called 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
Burns 4/e Chap 4
Slide 5 of 75
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
16 mol CO2
22.0 moles C8H18 
 176 moles CO2
2 mol C8H18
Burns 4/e Chap 4
Slide 6 of 75
Example – Estimate the mass of CO2 produced in
2004 by the combustion of 3.4 x 1015 g gasoline
• 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 Concept Plan will
be:
g C8H18
mol C8H18
mol CO2
Burns 4/e Chap 4
g CO2
Slide 7 of 75
Example – Estimate the mass of CO2 produced in 2004
by the combustion of 3.4 x 1015 g gasoline
Given:
Find:
Concept Plan:
3.4 x 1015 g C8H18
g CO2
g C8H18
mol C8H18
1 mol
114.22 g
mol CO2
16 mol CO 2
2 mol C8 H18
g CO2
44.01 g
1 mol
Relationships: 1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18 = 16 mol CO2
Solution:
1 mol C8H18
16 mol CO2 44.01 g CO2
3.4 10 g C8H18 


114.22 g C8H18 2 mol C8H18 1 mol CO2
15
 1.0 1016 g CO2
Check: since 8x moles of CO as C H , but the molar mass of C H is
2
8 18
8 18
3x CO2, the number makes sense
Practice
•
According to the following equation, how
many milliliters of water are made in the
combustion of 9.0 g of glucose?
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
1.
2.
3.
4.
convert 9.0 g of glucose into moles (MM 180)
convert moles of glucose into moles of water
convert moles of water into grams (MM 18.02)
convert grams of water into mL
a) How? what is the relationship between mass and
volume?
density of water = 1.00 g/mL
Burns 4/e Chap 4
Slide 9 of 75
Practice
According to the following equation, how many
milliliters of water are made in the combustion of
9.0 g of glucose?
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
9.0 g C6 H12O 6 x
1 mole C6 H12O 6
6 mole H 2O
18.0 g H 2O 1 mL H 2O
x
x
x
2
1 mole C6 H12O 6 1 mole H 2O 1.00 g H 2O
1.80 x 10 g
 5.4 mL H 2O
Burns 4/e Chap 4
Slide 10 of 75
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
Burns 4/e Chap 4
Slide 11 of 75
Things Don’t Always Go as Planned!
• many things can happen during the course of an
experiment that cause the loss of product
• the amount of product that is made in a reaction
is called the actual yield
– generally less than the theoretical yield, never more!
• the efficiency of product recovery is generally
given as the percent yield
actual yield
Percent Yield 
100%
theoretica l yield
Burns 4/e Chap 4
Slide 12 of 75
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 1 molecule of CH4 reacts with 2
molecules of O2
H
H
C
H
H
O
+
O
O
+
O
C
O
+
H
+
O
O
Burns 4/e Chap 4
H
O
H
H
Slide 13 of 75
Limiting and Excess Reactants in the
Combustion of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
• If we have 5 molecules of CH4 and 8 molecules
of O2, which is the limiting reactant?
H
C
H
H
H
H
H
H
H
C
H
H
H
C
C
H
H
H
H
H
H
H
C
+
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
?
H
H
Slide 14 of 75
Limiting and Excess Reactants in the Combustion
of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
H
C
H
H
H
H
H
H
H
C
H
H
H
C
C
H
H
H
H
H
H
H
C
+
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
H
H
since less CO2
can be made
from the O2 than
the CH4, the O2
is the limiting
reactant
1 molecules CO2
5 molecules CH 4 
 5 molecules CO2
1 molecules CH 4
1 molecules CO2
8 molecules O 2 
 4 molecules CO2
2 molecules O 2
Slide 15 of 75
Example 4.4
Finding Limiting Reactant,
Theoretical Yield, and
Percent Yield
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 CO(g)
Burns 4/e Chap 4
Slide 17 of 75
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
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
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent Yield.
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
• Write a Concept Plan:
kg
g
C
C
1000 g
1 kg
kg
TiO2
smallest
mol Ti
1000 g
1 kg
47.87 g
1 mol Ti
g
TiO2
g Ti
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
1 mol C
12.01 g C
1 mol TiO 2
79.87 g TiO 2
1 kg
1 000 g
mol
C
mol
TiO2
kg Ti
T.Y.
1 mol Ti
2 mol C
1 mol Ti
1 mol TiO 2
% yield 
mol
Ti
mol
Ti
act. yield
theor. yield
}
smallest
amount is
from
limiting
reactant
% Yield
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)
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 Concept Plan:
1000 g 1 mole C 1 mol Ti
28.6 kg C 


 1.1907 103 mol Ti
1 kg 12.01 g C 2 mol C
1000 g 1 mole TiO 2
1 mol Ti
88.2 kg TiO 2 


 1.1043 103 mol Ti
1 kg 79.87 g TiO 2 1 mol TiO 2
Limiting Reactant
smallest moles of Ti
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 Concept Plan:
47.87 g Ti 1 kg
1.1043 10 mol Ti 

 52.9 kg Ti
1 mol
1000 g
3
Theoretical Yield
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 Concept Plan:
Actual Yield
100%  Percent Yield
Theoretica l Yield
42.8 kg Ti
100%  80.9%
52.9 kg Ti
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%
Since Ti has lower molar mass than TiO2, the T.Y. makes sense
The Percent Yield makes sense as it is less than 100%.
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)
Burns 4/e Chap 4
Slide 26 of 75
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)
Given:
Find:
9.05 g NH3, 45.2 g CuO
g N2
Concept Plan: g NH
3
mol NH3
1 mol N 2
2 mol NH3
1 mol
17.03 g
g CuO
mol CuO
1 mol
79.55 g
smallest moles N2
Relationships:
mol N2
mol N2
1 mol N 2
3 mol CuO
1 mol
28.02 g
g N2
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
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)
Solution:
1 mol NH3
1 mol N 2
9.05 g NH3 

 0.266 mol N 2
17.03 g NH3 2 mol NH3
1 mol CuO
1 mol N 2
45.2 g CuO 

 0.189 mol N 2
79.55 g CuO 3 mol CuO
28.02 g N 2
0.189 mol N 2 
 5.30 g N 2
1 mol N 2
Check:
units are correct, and since there are fewer moles
of N2 than CuO in the reaction and N2 has a
smaller mass, the number makes sense
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
Burns 4/e Chap 4
Slide 29 of 75
Describing Solutions
• since solutions are mixtures, the composition can
vary from one sample to another
– pure substances have constant composition
– salt water 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 identical
Burns 4/e Chap 4
Slide 30 of 75
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
• quantitatively, the relative
amount of solute in the solution
is called the concentration
Burns 4/e Chap 4
Slide 31 of 75
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
amount of solute (in moles)
molarity, M 
amount of solution (in L)
Burns 4/e Chap 4
Slide 32 of 75
Preparing 1 L of a 1.00 M NaCl Solution
Burns 4/e Chap 4
Slide 33 of 75
Example 4.5 – Find the molarity of a solution that
has 25.5 g KBr dissolved in 1.75 L of solution
•
Sort
Information
•
Strategize
Given:
Find:
Concept Plan:
25.5 g KBr, 1.75 L solution
Molarity, M
g KBr
mol KBr
1 mol
119.00 g
Relationships:
•
Follow the
Concept Plan
to Solve the
problem
•
Check
M
mol
L
M
L sol’n
1 mol KBr = 119.00 g,
M = moles/L
Solution:
1 mol KBr
 0.21429 mol KBr
119.00 g KBr
moles KBr 0.21429 mol KBr
molarity, M 

 0.122 M
L solution
1.75 L
25.5 g KBr 
Check: since most solutions are between 0 and
18 M, the answer makes sense
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
1 L solution
2 mol sugar
2 mol sugar
1 L solution
Burns 4/e Chap 4
Slide 35 of 75
Example 4.6 – How many liters of 0.125 M NaOH
contains 0.255 mol NaOH?
•
Sort
Information
•
Strategize
Given: 0.125 M NaOH, 0.255 mol NaOH
Find:
liters, L
Concept Plan:
mol NaOH
L sol’n
1 L solution
0.125 mol NaOH
Relationships:
•
•
Follow the
Concept Plan
to Solve the
problem
Check
0.125 mol NaOH = 1 L solution
Solution:
0.255 mol NaOH 
1 L solution
 2.04 L solution
0.125 mol NaOH
Check: since each L has only 0.125 mol NaOH,
it makes sense that 0.255 mol should
require a little more than 2 L
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
Burns 4/e Chap 4
Slide 37 of 75
Example 4.7 – To what volume should you dilute
0.200 L of 15.0 M NaOH to make 3.00 M NaOH?
•
Sort
Information
•
Strategize
Given: V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M
V2, L
Find:
Concept Plan:
V1, M1, M2
V2
M1  V1
 V2
M2
Relationships:
•
Follow the
Concept Plan
to Solve the
problem
•
Check
Solution:
M1V1 = M2V2
mol 

15.0
  0.200 L 
L 

 1.00 L
mol 

3.00


L


Check: since the solution is diluted by a factor
of 5, the volume should increase by a
factor of 5, and it does
Solution Stoichiometry
• since 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
Burns 4/e Chap 4
Slide 39 of 75
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)
•
•
Sort
Information
Strategize
Given:
0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2
Find:
L KCl
Concept Plan:
L Pb(NO3)2
mol Pb(NO3)2
0.175 mol
1 L Pb(NO3 ) 2
Relationships:
•
•
Follow the
Concept
Plan to
Solve the
problem
Check
mol KCl
2 mol KCl
1 mol Pb(NO3 ) 2
L KCl
1 L KCl
0.150 mol
1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol,
1 mol Pb(NO3)2 = 2 mol KCl
Solution:
0.150 L Pb(NO3 ) 2 
0.175 mol
2 mol KCl
1 L KCl


1 L Pb(NO3 ) 2 1 mol Pb(NO3 ) 2 0.150 mol
 0.350 L KCl
Check:
since 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 volume Pb(NO3)2
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
Slide 41 of 75
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
Burns 4/e Chap 4
Slide 42 of 75
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
Burns 4/e Chap 4
Slide 43 of 75
Molecular View of
Electrolytes and Nonelectrolytes
• in order 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 all 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
Burns 4/e Chap 4
Slide 44 of 75
Salt vs. Sugar Dissolved in Water
ionic compounds dissociate
into ions when they dissolve
molecular compounds do not
dissociate when they dissolve
Burns 4/e Chap 4
Slide 45 of 75
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 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)
Burns 4/e Chap 4
Slide 46 of 75
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
Na2SO4(aq)  2 Na+(aq) + SO42-(aq)
CH3COOH(aq)  H+(aq) + CH3COO-(aq)
Burns 4/e Chap 4
Slide 47 of 75
Classes of Dissolved Materials
Burns 4/e Chap 4
Slide 48 of 75
Solubility of Ionic Compounds
• some ionic compounds, like NaCl, dissolve very well in
water at room temperature
• other ionic compounds, like AgCl, dissolve hardly at all
in water at room temperature
• compounds that dissolve in a solvent are said to be
soluble, while 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
Burns 4/e Chap 4
Slide 49 of 75
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
Burns 4/e Chap 4
Slide 50 of 75
Solubility Rules
Compounds that Are Generally Soluble in Water
Compounds Containing the Exceptions
Following Ions are Generally (when combined with ions on the
Soluble
left the compound is insoluble)
Li+, Na+, K+, NH4+
none
NO3–, CH3COO–
none
Cl–, Br–, I–
Ag+, Hg22+, Pb2+
SO42–
Ag+, Ca2+, Sr2+, Ba2+, Pb2+
Burns 4/e Chap 4
Slide 51 of 75
Solubility Rules
Compounds that Are Generally Insoluble
Exceptions
Compounds Containing the (when combined with ions on the
Following Ions are Generally left the compound is soluble or
Insoluble
slightly soluble)
OH–
S2–
CO32–, PO43–
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+
Burns 4/e Chap 4
Slide 52 of 75
Precipitation Reactions
• reactions between aqueous solutions of ionic
compounds that produce an ionic compound
that is insoluble in water are called
precipitation reactions and the insoluble
product is called a precipitate
Burns 4/e Chap 4
Slide 53 of 75
2 KI(aq) + Pb(NO3)2(aq)  PbI2(s) + 2
KNO3(aq)
Slide 54 of 75
No Precipitate Formation =
No Reaction
KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq)
all ions still present,  no reaction
Burns 4/e Chap 4
Slide 55 of 75
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
Burns 4/e Chap 4
Slide 56 of 75
Process for Predicting the Products of
a Precipitation Reaction
5. If either product is insoluble, write the formulas
for the products after the arrow – writing (s)
after the product that is insoluble and will
precipitate, and (aq) after products that are
soluble and will not precipitate
6. Balance the equation
Burns 4/e Chap 4
Slide 57 of 75
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
•
cross charges and reduce
K2CO3(aq) + NiCl2(aq)  KCl + NiCO3
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 soluble, write no reaction
does not apply since NiCO3 is insoluble
Burns 4/e Chap 4
Slide 59 of 75
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)
Burns 4/e Chap 4
Slide 60 of 75
Ionic Equations
• equations which 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 which describe the actual dissolved species are
called complete ionic equations
– aqueous strong electrolytes are written as ions
• soluble salts, strong acids, strong bases
– insoluble substances, weak electrolytes, and nonelectrolytes
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)
Burns 4/e Chap 4
Slide 61 of 75
Ionic Equations
• ions that are both reactants and products are called
spectator ions
2K+(aq) + 2OH-(aq) + Mg+(aq) + 2NO3-(aq)  2K+(aq) + 2NO3-(aq) + Mg(OH)2(s)
• an ionic equation in which the spectator ions are
removed is called a net ionic equation
2OH-(aq) + Mg2+(aq)  Mg(OH)2(s)
Burns 4/e Chap 4
Slide 62 of 75
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
Burns 4/e Chap 4
Slide 63 of 75
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, like 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
Burns 4/e Chap 4
Slide 64 of 75
Common Acids
Chemical Name
Formula
Uses
Strength
Common Bases
Burns 4/e Chap 4
Slide 66 of 75
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Burns 4/e Chap 4
Slide 67 of 75
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
(H+ + NO3-) + (Ca2+ + OH-) 
b) Exchange the ions, H+1 combines with OH-1 to make H2O(l)
(H+ + NO3-) + (Ca2+ + OH-)  (Ca2+ + NO3-) + H2O(l)
c) Write the formula of the salt
 cross the charges
(H+ + NO3-) + (Ca2+ + OH-)  Ca(NO3)2 + H2O(l)
Burns 4/e Chap 4
Slide 68 of 75
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
a (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)
Burns 4/e Chap 4
Slide 69 of 75
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)
Burns 4/e Chap 4
Slide 70 of 75
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
Burns 4/e Chap 4
Slide 71 of 75
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
– aka the equivalence point
Burns 4/e Chap 4
Slide 72 of 75
Titration
Burns 4/e Chap 4
Slide 73 of 75
Titration
The base solution is the
titrant in the burette.
As the base is added to
the acid, 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.
Burns 4/e Chap 4
Slide 74 of 75
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
Burns 4/e Chap 4
Slide 75 of 75
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
Burns 4/e Chap 4
Slide 76 of 75
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
Burns 4/e Chap 4
Slide 77 of 75
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
CF:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
• Write a Concept Plan:
mL
NaOH
L
NaOH
mL
HCl
L
HCl
mol
NaOH
Burns 4/e Chap 4
mol
HCl
Slide 78 of 75
Example:
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
CF:
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 Solution Map:
= 1.25 x 10-3 mol HCl
Burns 4/e Chap 4
Slide 79 of 75
Example:
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
CF:
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 Concept Plan:
Burns 4/e Chap 4
Slide 80 of 75
Example:
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
CF:
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 since
the neutralization takes less HCl solution than
NaOH solution, so the HCl should be more concentrated.
Burns 4/e Chap 4
Slide 81 of 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)
Burns 4/e Chap 4
Slide 82 of 75
NaHCO3(aq) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
Burns 4/e Chap 4
Slide 83 of 75
Compounds that Undergo
Gas Evolving Reactions
Reactant
Type
Reacting
With
Ion
Exchange
Product
Decompose?
Gas
Formed
Example
metalnS,
metal HS
acid
H2S
no
H2S
K2S(aq) + 2HCl(aq) 
2KCl(aq) + H2S(g)
metalnCO3,
metal HCO3
acid
H2CO3
yes
CO2
K2CO3(aq) + 2HCl(aq) 
2KCl(aq) + CO2(g) + H2O(l)
metalnSO3
metal HSO3
acid
H2SO3
yes
SO2
K2SO3(aq) + 2HCl(aq) 
2KCl(aq) + SO2(g) + H2O(l)
(NH4)nanion
base
NH4OH
yes
NH3
KOH(aq) + NH4Cl(aq) 
KCl(aq) + NH3(g) + H2O(l)
Burns 4/e Chap 4
Slide 84 of 75
Example 4.15 - 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
(Na+1 + CO3-2) + (H+1 + NO3-1) 
b) Exchange the anions
(Na+1 + CO3-2) + (H+1 + NO3-1)  (Na+1 + NO3-1) + (H+1 + CO3-2)
c) Write the formula of compounds
 cross the charges
Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3
Burns 4/e Chap 4
Slide 85 of 75
Example 4.15 - When an aqueous solution of
sodium carbonate is added to an aqueous
solution of nitric acid, a gas evolves
3. Check to see either product 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)
Burns 4/e Chap 4
Slide 86 of 75
Example 4.15 - 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 a
(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 + CO2(g) + H2O(l)
Burns 4/e Chap 4
Slide 87 of 75
Other Patterns in Reactions
• the precipitation, acid-base, and gas evolving
reactions all involved 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)
Burns 4/e Chap 4
Slide 88 of 75
Combustion as Redox
2 H2(g) + O2(g)  2 H2O(g)
Burns 4/e Chap 4
Slide 89 of 75
Redox without Combustion
2 Na(s) + Cl2(g)  2 NaCl(s)
2 Na  2 Na+ + 2 e
Cl2 + 2 e  2 Cl
Burns 4/e Chap 4
Slide 90 of 75
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 reaction involves 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+2O– (s)
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Burns 4/e Chap 4
Slide 91 of 75
Oxidation and Reduction
• in order 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
Burns 4/e Chap 4
Leo
Slide 92 of 75
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
Burns 4/e Chap 4
Slide 93 of 75
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
Burns 4/e Chap 4
Slide 94 of 75
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
Burns 4/e Chap 4
Slide 95 of 75
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
Nonmetal
Oxidation State
Example
F
-1
CF4
H
+1
CH4
O
-2
CO2
Group 7A
-1
CCl4
Group 6A
-2
CS2
Group 5A
-3
NH3
Burns 4/e Chap 4
Slide 96 of 75
Practice – Assign an Oxidation State to
Each Element in the following
• Br2
• K+
• LiF
• CO2
• SO42• Na2O2
Burns 4/e Chap 4
Slide 97 of 75
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)
Burns 4/e Chap 4
Slide 98 of 75
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
Burns 4/e Chap 4
Slide 99 of 75
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
Burns 4/e Chap 4
Slide 100 of 75
Identify the Oxidizing and Reducing
Agents in Each of the Following
3 H2S + 2 NO3– + 2 H+  3 S + 2 NO + 4
H2O
MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O
Burns 4/e Chap 4
Slide 101 of 75
Identify the Oxidizing and Reducing
Agents in Each of the Following
3 H2S + 2 NO3– + 2 H+  3 S + 2 NO + 4
red ag
ox ag
H2O
+1 -2
+5 -2
+1
0
+2 -2
+1 -2
oxidation
reduction
ox ag
red ag
+4 -2
+1 -1
MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O
+2 -1
0
+1 -2
oxidation
reduction
Burns 4/e Chap 4
Slide 102 of 75
Combustion Reactions
• Reactions in which O2(g) is a
reactant are called
combustion reactions
• Combustion reactions release
lots of energy
• Combustion reactions are a
subclass of oxidationreduction reactions
2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
Burns 4/e Chap 4
Slide 103 of 75
Combustion Products
• to predict the products of a combustion
reaction, combine each element in the other
reactant with oxygen
Reactant
Combustion Product
contains C
CO2(g)
contains H
H2O(g)
contains S
SO2(g)
contains N
NO(g) or NO2(g)
contains metal
M2On(s)
Burns 4/e Chap 4
Slide 104 of 75
Practice – Complete the Reactions
• combustion of C3H7OH(l)
• combustion of CH3NH2(g)
Burns 4/e Chap 4
Slide 105 of 75
Practice – Complete the Reactions
C3H7OH(l) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
CH3NH2(g) + 3 O2(g)  CO2(g) + 2 H2O(g) + NO2(g)
Burns 4/e Chap 4
Slide 106 of 75
Important Terms
• p174
Burns 4/e Chap 4
Slide 107 of 75
Homework
• You should examine and be able to answer all of
the ‘Problems’…some of them (or similar) may
be on the test
• To be handed in for grading: 4.25, 4.32, 4.38,
4.45, 4.46, 4.51, 4.56, 4.64, 4.70, 4.72, 4.78,
4.82
• Bonus: 4.100
• For fun: 4.103, 4.107
Burns 4/e Chap 4
Slide 108 of 75