Transcript tro2_ppt_lecture_04 - Louisiana Tech University

Lecture Presentation
Chapter 4
Chemical Quantities
and Aqueous
Reactions
Catherine MacGowan
Armstrong Atlantic State University
© 2013 Pearson Education, Inc.
Reaction Stoichiometry
• According to the Law of Conversion of Matter:
– Matter is neither created nor destroyed in a chemical reaction.
– A balanced chemical equation illustrates the law of
conversation of matter.
• In a balanced reaction:
– Total mass of reactants = Total mass of products
– A balanced reaction has the same type and quantity of
atoms on both sides of the reaction.
• Stoichiometry is based on the law of conversion of matter.
– Stoichiometry studies the quantitative aspects of chemical
reactions.
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Reaction Stoichiometry: What it means
4 Fe(s)
+
 2 Fe2O3(s)
3 O2(g)
This equation means:
4 atoms Fe +
4 moles Fe +
3 molecules O2

2 molecules
Fe2O3
3 moles O2

2 moles Fe2O3
4 moles Fe atoms
and 6 moles O atoms
atoms
23.4 g Fe + 96.0 g O2
415.4 g
of reactants
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=
Fe atoms
and 6 mole O
=
415.4 g Fe2O3
=
415.4 g
of products
Strategy Behind Solving STOICHIOMETRY Problems
1. Need a balanced reaction to determine the stoichiometric
relationship between:
a. Reactants and reactants or
b. Reactants and products or
c. Products and products
2. Go to the mole:
a. If mass is given, then divide by molecular mass:
mass (g) /mol. mass (g/mole) = mole
b. If volume and molarity (M) are given, then:
Ma x Va = moles A
3. Use the stoichiometric factor to convert from mole A to mole B to
solve problem.
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Using Stoichiometry to Predict
• Theoretical Yield and Limiting Reagent (Reactants)
• Percent Yield
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Problem:
The following unbalanced equation is the chemical reaction
associated with photosynthesis.
CO2(g) + H2O(l)  C6H6O6(s) + O2(g)
With adequate water, suppose a plant consumes 37.8 grams of CO2
during the week.
Determine how many grams of glucose (C6H6O6) would be produced
by the plant during this one-week period.
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Problem Strategy:
Determine how many grams of glucose (C6H6O6) would be produced
by the plant if 37.8 grams of CO2 is consumed.
1. Need a balanced equation:
6 CO2(g) + 6 H2O(l)  C6H6O6(s) + 6 O2(g)
2. Determine moles of CO2 consumed.
37.8 g CO2 x (1 mol CO2/44.0 g) = 0.859 mol CO2
3. Determine how many moles of C6H6O6 would be produced. Use the
stoichiometric relationship between CO2 and C6H6O6. In this reaction the
relationship is 6 mole CO2 to 1 mole C6H6O6.
0.859 mol CO2 x (1 mol C6H6O6/6 mol CO2) = 0.143 mol C6H6O6
4. Determine the grams of C6H6O6 produced.
0.143 mol C6H6O6 x (1 mol / 180.0 g/1 mol C6H6O6) = 25.8 g C6H6O6
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Theoretical and Percent Yield
• Theoretical Yield
– Predicting what could/should/would be produced
from a given amount of reactant(s)
• Limiting reagent (reactant)
– The determining reactant
– Using stoichiometric relationship(s) from balanced
reaction
• Percent Yield
– Actual amount produced
– % yield = (actual yield/theoretical yield) x 100
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Problem: Ammonium nitrate (NH4NO3) decomposes
to N2O(g) and H2O(g).
1. Predict how many grams of H2O will be produced
from the decomposition of 0.454 kg of NH4NO3.
2. If 131 grams of water is produced, what is the %
yield for this reaction?
Strategy:
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1. Write a balanced reaction.
2. Determine the theoretical yield.
3. Calculate the percent (%) yield.
Problem Solution: Ammonium nitrate (NH4NO3) decomposes
to N2O(g) and H2O(g).
Balance equation:
NH4NO3(s)  N2O(g) + 2 H2O(g)
1. Predict how many grams of H2O will be produced from the
decomposition of 0.454 kg of NH4NO3.
0.454 kg NH4NO3 x (1000 g/1 kg) x (1 mol NH4NO3/80.04 g)
= 5.68 mol NH4NO3
5.68 mol NH4NO3 x (2 mol H2O/1mol NH4NO3 ) = 11.4 mol H2O
11.4 mol H2O x (18.0 g/1 mol H2O) = 204 grams H2O
204 grams of H2O is the predicted or theoretical yield.
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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 (limiting reagent).
– The limiting reactant is 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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11
Predicting Theoretical Yield
Problem:
Determine the theoretical yield for the
following reaction:
Mg(s) + O2(g)  MgO(s)
when 42.5 g Mg(s) and 33.8 g O2(g) are
reacted.
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Strategy:
1. Check to see if reaction given is balanced.
2. Determine limiting reactant and theoretical yield.
Solution:
1. Balance reaction: 2 Mg(s) + O2(g)  2 MgO(s)
2. Determine limiting reactant * theoretical yield.
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Problem:
Ammonium nitrate (NH4NO3) decomposes
to N2O(g) and H2O(g).
Balance equation:
NH4NO3(s)  N2O(g) + 2 H2O(g)
If 131 grams of water is produced, what is the % yield for this
reaction?
% yield = (131 g/204 g) x 100 = 64.2%
62.4% is the percent yield for the reaction.
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Problem:
When 28.6 kg of carbon reacted with 88.2 kg
of TiO2, 42.8 kg of Ti was obtained.
Reaction: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Strategy: 1. Determine limiting reactant.
2. Calculate theoretical yield.
3. Calculate percent yield.
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Problem Solution:
Reaction:
When 28.6 kg of carbon reacted with 88.2 kg of
TiO2, 42.8 kg of Ti was obtained.
TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) It is balanced!
1. Determine limiting reactant.
28.6 kg carbon:
28.6 kg C x (1000 g/1 kg) = 2.86 x 104 g C
88.2 kg TiO2
88.2 kg TiO2 x (1000 g/1 kg) = 8.82 x 104 g TiO2
2.86 x 104 g C x (1 mol C/12.0 g) x (1 mol Ti/2 mol C)
= 1.19 x 103 mol Ti
THIS IS THE MAXIMUM MOLE OF Ti
THAT CAN BE PRODUCED IF ALL
CARBON IS USED.
8.82 x 104 g TiO2 x (1 mol TiO2/79.9 g) x (1 mol Ti/1 mol TiO2)
= 1.10 x 103 mol Ti
THIS IS THE MAXIMUM MOLE OF Ti
THAT CAN BE PRODUCED IF ALL TiO2
IS USED.
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Problem Solution:
Reaction:
When 28.6 kg of carbon reacted with 88.2 kg of
TiO2, 42.8 kg of Ti was obtained.
TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
2. Determining the theoretical yield from limiting reactant.
= 1.19 x 103 mol Ti
THIS IS THE MAXIMUM MOLE OF Ti
THAT CAN BE PRODUCED IF ALL
CARBON IS USED.
= 1.10 x 103 mol Ti
THIS IS THE MAXIMUM MOLE OF Ti
THAT CAN BE PRODUCED IF ALL TiO2
IS USED.
1.10 x 103 < 1.19 x 103; THEREFORE, the limiting reagent is TIO2 and
the theoretical yield in moles of Ti is 1.10 x 103.
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Problem Solution:
Reaction:
When 28.6 kg of carbon reacted with 88.2 kg of
TiO2, 42.8 kg of Ti was obtained.
TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
3. Determining percent yield:
The theoretical yield in moles of Ti is 1.10 x 103.
1.10 x 103 mol Ti x (47.9 g/1 mol Ti) = 5.29 x 104 g Ti
42.8 kg is actual yield of Ti.
42.8 kg x (1000 g/1 kg) = 4.28 x 104 g Ti was produced.
(4.28 x 104 g Ti/5.29 x 104 g Ti) x 100 = 80.9%
80.9% is the percent yield for this reaction.
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PROBLEM: Given the following chemical reaction:
Al(s) + O2(g)  Al2O3(s)
1. Determine how many grams of Al2O3 can form when
5.40 grams of Al and 8.10 grams of O2 are reacted.
2. If 4.50 grams of Al2O3 was produced, what is the
percent yield for this reaction?
3. Which reactant was in excess and how much (grams)
of this reactant remained after the reaction came to
completion?
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Step 1: Balance the reaction and calculate theoretical yield for each
reactant.
Balanced reaction:
4 Al(s) + 3 O2(g)  2 Al2O3(s)
5.40 g Al x (1 mol/27.0 g Al) = 0.200 mol Al
0.200 mol Al x 2 mol Al2O3 = 0.100 mol Al2O3
4 mol Al
stoichiometric factor
0.100 mol Al2O3 x 101.96 g = 10.2 g Al2O3
1 mol Al2O3
This means that if all 5.40 g Al were consumed, then only 10.2 grams of
Al2O3 COULD be produced.
NOW DETERMINE THE THEORITICAL YIELD IF ALL 8.10 grams of O2
WERE USED.
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Step 1: Balance the reaction and calculate theoretical yield for each
reactant.
Balanced reaction:
4 Al(s) + 3 O2(g)  2 Al2O3(s)
8.10 g O2 x ( 1 mol/32.00 g) = 0.253 mol O2
0.253 mol O2 x 2 mol Al2O3 = 0.169 mol Al2O3
3 mol O2
stoichiometric factor
0.169 mol Al2O3 x 101.96 g = 17.2 g Al2O3
1 mol Al2O3
This means that if all 8.10 grams of O2 were consumed, then 17.2 grams is
the MOST that COULD be produced.
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COMPARE THEORETICAL YIELDS TO DETERMINE THE LIMITING
REACTANT:
• If all 8.10 g O2 were used, then 17.2 g of Al2O3 would be produced.
• If all 5.40 g Al were used, then 10.2 g of Al2O3 would be produced.
10.2 g < 17.2 g
• The limiting reactant is Al.
• Theoretical yield is 10.2 g Al2O3.
To determine the percent yield of the reaction:
(4.50 g/10.2) x 100 = 44.1%
44.1% is the percent yield for this reaction.
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Which reactant will remain when the reaction is complete?
• Al was the limiting reactant.
– Therefore, O2 was in excess. But by how much?
– First find how much oxygen gas was required.
– Then find how much oxygen gas is in excess.
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How to Determine the Amount of Excess
Reagent Left Over
4 Al + 3 O2
0.200 mol = LR
products
0.253 mol
0.200 mol Al x (3 mol O2/4 mol Al) = 0.15 mol O2
0.150 mol of O2 is required to react with all 0.200 mol of Al.
O2 available – O2 required = excess O2
0.253 mol O2 - 0.150 mol O2 = 0.103 mol O2 left over
= 0.103 mol O2 in excess, or 3.30 grams O2
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PROBLEM:
Zinc reacts with acids to produce H2 gas. If you
have 10.0 g of Zn metal, what volume of 2.50 M HCl is
needed to covert the Zn metal completely to Zn2+ ions?
Step 1: Write the balanced equation.
Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)
Step 2: Calculate amount of Zn.
10.0 g Zn x (1 mol Zn/65.39 g) = 0.153 mol Zn
Step 3:
Use the stoichiometric factor.
0.153 mol Zn x (2 mol HCl/1 mol Zn) = 0.306 mol HCl
Step 4:
Calculate volume of HCl required.
0.306 mol HCl x (1.00 L/2.50 mol) = 0.122 L HCl
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Problem:
76.80 grams of apple juice (malic acid) requires 34.56 mL
of 0.664 M NaOH to reach the endpoint in a titration.
What is the (w/w)% of malic acid in this sample?
Reaction:
C4H6O5(aq) + 2 NaOH(aq) --> Na2C4H4O5(aq) + 2 H2O(l)
(malic acid)
Step 1:
Calculate amount of NaOH used.
M x V = (0.664 M)(0.03456 L) = 0.0229 mol NaOH
Step 2:
Calculate amount of malic acid titrated.
0.0229 mol NaOH x (1 mol malic acid/2 mol NaOH)
= 0.0115 mol malic acid
Step 3:
Calculate the grams of malic acid in 0.0115 moles.
0.0115 mol malic acid x (134 g/1 mol malic acid) = 1.54 g
Step 4:
Calculate the weight (w/w)% malic acid in apple sample.
(1.54 g/76.80) x 100 = 2.01 (w/w)%
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Chemical Analysis Problem:
An impure sample of the mineral contains Na2SO4.
The mass of mineral sample = 0.123 g
All of the Na2SO4 in the sample is converted to insoluble
BaSO4.
The mass of BaSO4 is 0.177 g.
Determine the mass percent of Na2SO4 in the mineral.
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Answer:
Reaction equations:
Na2SO4(aq) + BaCl2(aq)  2 NaCl(aq) + BaSO4(s)
0.177 g BaSO4 x (1 mol/233.4 g) = 7.58 x 10-4 mol BaSO4
7.58 x 10-4 mol BaSO4 x (1 mol Na2SO4/1 mol BaSO4)
= 7.58 x 10-4 mol Na2SO4
7.58 x 10-4 mol Na2SO4 x (142.0 g/1 mol) = 0.108 g Na2SO4
(0.108 g Na2SO4/0.123 g sample)100% = 87.6% Na2SO4
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Solution Chemistry
• Solution Terminology
• Concentration
• Molarity
• Solution Preparation
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Terminology of Solution Chemistry
SOLVENT: the component whose physical state is
preserved when solution forms
SOLUTE: the other solution component; usually dissolved
in the solvent
The SOLVENT and SOLUTE MUST have similar polarities!
A SOLUTION is a homogeneous mixture composed of a
solute and a solvent.
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Concentration
• The amount of solute in a solution is given by its
concentration.
• The more concentrated a solution is, the more solute
is present in the solution.
• Concentration is dependent on the amount of solute
in solution.
– A dilute solution has the same number of moles
as a concentrated solution, BUT the moles
present are distributed over a greater area
(volume).
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Molarity
•
It is a concentration unit.
• moles of solute per 1 liter of solution
• mol/L = M
•
Molarity shows the relationship between the moles of
solute and liters of solution.
•
It is used because it describes how many molecules of
solute are in each liter of solution.
Molarity = amount of solute in moles
amount of solution in liters
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Molar Concentrations
Copper (II) chloride dissolves in water:
CuCl2(aq)  Cu2+(aq) + 2 Cl-(aq)
• For every one mole of CuCl2, one mole of Cu2+ and
two moles of Cl- ions are produced.
• 1.0 molar (M) CuCl2 yields 1.0 M Cu2+ ions and 2.0 M
Cl-.
So if [CuCl2] = 0.30 M, then
[Cu2+] = 0.30 M and [Cl-] = 2 x 0.30 M = 0.60 M
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Solution Preparation
A solution can be prepared in two ways:
– by dissolving a solid solute into a liquid solvent:
• Molarity x volume = moles
– where moles = (grams of solute/mol. mass of solute)
– by mixing a liquid solute with a liquid solvent:
• Dilution
– C1V1 = C2V2
– where
» C1 is the concentration of the original solution
» V1 is the amount C1 (solute)
» C2 is the new concentration
» V2 is the volume of the new concentration (C2)
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Preparing 1.00 L of a 1.00 M NaCl Solution
from a Solid Solute
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Problem:
Strategy:
Solution:
Determine the molarity of a solution that has
25.5 g KBr dissolved to a volume of 1.75 L
solution.
Convert grams KBr to moles of KBr.
Use the definition for molarity.
Divide moles of KBr by solution volume.
25.5 g KBr x (1 mol KBr/119.0 g )
= 0.214 mol KBr
0.214 mol KBr = M x V
0.214 mol KBr = M x 1.75 L
(0.214 mol KBr/1.75 L) = 0.122 M
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Problem:
Determine the mass of CaCl2 need to make 1.75 L of 1.50 M
solution.
Strategy:
Determine the moles of CaCl2.
From moles of CaCl2, determine the grams of CaCl2.
Solution:
M X V = moles
1.50 mol/L x 1.75 L = 3.28 moles CaCl2
mole = grams solute/mol. mass of solute
3.28 mol CaCl2 = x/(110.98 g CaCl2/1 mol CaCl2)
grams solute = 3.28 mol CaCl2 x (110.98 g/1 mol CaCl2)
x = 291 grams CaCl2
291 grams of CaCl2 is dissolved in a total of 1.75 L of water to
make a 1.50 M solution.
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Dilution: Making a Solution from a Solution
•
Often, solutions are stored as concentrated 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.
– NOTE: The moles of solute in solution 1 = the moles of solute in
solution 2.
– Only the volume changes.
• Amount of moles remain the same; they are just spread out over
a larger area.
•
The concentrations and volumes of the stock and new solutions are
inversely proportional.
C1∙V1 = C2∙V2
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Dilutions
FORMULA:
Cinitial x Vinitial = Cfinal x Vfinal
or
M1V1 = M2V2
IMPORTANT:
The moles of solute in the DILUTED (new) solution is
EQUAL to the moles of solute in the ORIGINAL (undiluted)
solution.
MOLES of SOLUTE DO NOT CHANGE when a solution is diluted.
The solution’s CONCENTRATION DOES CHANGE!!
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Diluting a Solution
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PROBLEM:
A 0.50 M NaOH is prepared from 50.0 mL of 3.0 M.
What is the volume of this diluted solution?
Solution: Cinitial x Vinitial = Cfinal x Vfinal
3.0 M x 50.0 mL = 0 .50 M x Vfinal
Volume of final solution = 3.0 x 102 mL
Conclusion:
Add 250 mL of water to 50.0 mL of 3.0 M NaOH to
make 300 mL of 0.50 M NaOH.
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In a dilution the moles are unchanged.
Why is this?
From the previous problem:
The number of moles of NaOH in original solution was 0.15 moles NaOH:
(3.0 mol/L)(0.050 L) = 0.15 moles NaOH
The number of moles of NaOH in diluted solution also equals 0.15 moles
NaOH.
The concentration [M] for the diluted solution is 0.50 M, which contains
0.15 moles of NaOH in 0.30 liters.
(0.50 mol/L)(0.300 L) = 0.15 moles NaOH
The original solution concentration was 3.0 M, which contains 0.15 moles of
NaOH in 0.050 L, and the dilute solution also contains 0.15 moles of NaOH, but
its concentration is 0.50 M NaOH in 0.330 L.
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Dilution: One More Point of View
PROBLEM:
Make a 1:3 dilution of 100.0 mL of 0.15 M HCl solution.
What does this mean?
A 1:3 dilution of 0.15 M HCl means that the new
concentration will be 0.050 M HCl.
Or
1/3 x 0.15 M HCl = 0.050 M HCl
NOTE: Its new volume is 3x that of the original volume, or 300.0 mL.
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Problem:
To what volume should you dilute 0.200 L of 15.0 M NaOH to
make 3.00 M NaOH?
Strategy:
Use the dilution formula
Solution:
C1
is 15.0 M NaOH, the concentration of the original
(concentrated) solution
V1
is 0.200 L of the 15.0 M that is to be diluted
C2
is 3.00 M NaOH or the concentration of the diluted
solution
V2
is the unknown quantity – it is the amount (mL) of the
15.0 M NaOH that will be diluted to make 3.00 M NaOH
C1∙V1 = C2∙V2
15.0 M NaOH x 0.200 L = 3.00 M NaOH x V2
V2 = (15.0 M NaOH x 0.200 L)/3.00 M NaOH
V2 = 1.00L
0.200 L of 15.0 M solution will be diluted to 1.00 L to make a 3.00 M NaOH
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Problem: What is the concentration of a solution prepared by diluting
45.0 mL of 8.25 M HNO3 to 135.0 mL?
C1∙V1 = C2∙V2
Strategy:
Use dilution formula
Solution:
C1 is 8.25 M HNO3, the concentration of the original (concentrated)
solution
V1 is 45.0 mL of the 8.25 M HNO3 that is to be diluted
C2 is the unknown quantity – it is the new concentration for HNO3
V2 is 135.0 mL, the new volume.
8.25 M HNO3 x 45.0 mL = C2 x 135.0 mL
C2 = (8.25 M HNO3 x 45.0 mL)/135.0 mL
C2 = 2.75 M HNO3
When 45.0 mL of 8.25 M HNO3 is diluted to 135.0 mL, the
concentration of the solution is 2.75 M HNO3.
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Solution Stoichiometry
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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PROBLEM:
Determining the concentration of unknown base by
titration. In a titration, 1.065 g of H2C2O4 (oxalic acid) requires
35.62 mL of NaOH to reach the equivalence point of the titration.
What is the molarity (M) of the NaOH?
Strategy:
• Need a balanced chemical equation to know stoichiometric
relationships
• H2C2O4(aq) + 2 NaOH(aq)  Na2C2O4(aq) + 2 H2O(l)
• Conversions
• oxalic acid  moles of oxalic acid  moles of NaOH
• Moles of NaOH  molarity (M) NaOH
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Answer:
H2C2O4(aq) + 2 NaOH(aq)  Na2C2O4(aq) + 2 H2O(l)
Step 1: Calculate amount of H2C2O4.
1.065 g x (1 mol/90.4 g H2C2O4) = 0.0118 mol H2C2O4
Step 2: Calculate amount of NaOH required.
0.0118 mol H2C2O4 x (2 mol NaOH/1 mol H2C2O4)
= 0.0237 mol NaOH
Step 3: Calculate concentration of NaOH.
M x V = moles
M = 0.0237 mol NaOH/0.03562 L
M = 0.6654
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Problem:
Standardization (actual concentration of base with
KHP (potassium hydrogen phthalate, KHC8H4O4)) of
KHP required 26.75 mL of NaOH to reach the
endpoint.
Step 1:
Balance reaction:
KHP(aq) + NaOH  NaKP(aq) + H2O(l)
Step 2:
Calculate moles of KHP titrated to reach endpoint.
0.520 g KHP x (1 mol KHP/204.3 g)
= 2.55 x 10-3 mol KHP
Step 3:
Use the stoichiometric relationship to determine
moles of NaOH titrated.
2.55 x 10-3 moles KHP x (1 mol NaOH/1 mol KHP)
= 2.55 x 10-3 mol NaOH
Step 4:
Determine the molarity (M or mol/L) of NaOH.
2.55 x 10-3 NaOH/0.02675 = 0.0953 M NaOH
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Problem:
Determination of an unknown acid’s molarity by titration
of a standardized base:
It took 32.15 mL of 0.0953 M NaOH to reach the equivalence point
when 25.00 mL of unknown concentration of HCl was titrated.
Find HCl molarity.
Step 1:
HCl(aq) + NaOH  NaCl(aq) + H2O(l)
Step 2:
Calculate moles of NaOH titrated to reach endpoint.
0.03215 L x 0.0953 M NaOH = 3.07 x 10-3 moles
Step 3:
Use the stoichiometric relationship to determine moles of
HCl titrated.
3.07 x 10-3 moles NaOH x (1 mol HCl/1 mol NaOH)
= 3.07 x 10-3 mol HCl
Step 4:
Determine the molarity (M or mol/L) of HCl.
3.07 x 10-3 HCl/0.02500 L = 0.123 M HCl
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PROBLEM:
Determining the concentration of an unknown acid:
It takes 27.85 mL of 0.664 M NaOH to reach the endpoint of a reaction
when 15.0 mL aliquot of H2SO4 is titrated. What is the molarity of the
unknown sample of H2SO4?
Strategy:
•
Need a balance chemical equation to know stoichiometric relationships
• H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
•
Conversions
• MNaOH x volumeNaOH = moles of NaOH
• moles NaOH  moles of H2SO4
• Moles of H2SO4  molarity (M) H2SO4
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Answer:
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
Step 1: Calculate moles of NaOH titrated to reach endpoint.
0.664 M NaOH x 0.02785 L NaOH = 1.86 x 10-1 moles NaOH
Step 2: Use the stoichiometric relationship to determine moles of
H2SO4 titrated.
1.86 x 10-1 moles NaOH x (1 mol H2SO4/2 mol NaOH)
= 0.0930 mol H2SO4
Step 3: Determine the molarity (M) of H2SO4 sample.
0.093 mol H2SO4 / 0.0150 L = 0.620 M H2SO4
© 2013 Pearson Education, Inc.
Aqueous Chemistry
• Aqueous Solutions
• Electrolytes
• Solubility
• Reaction Types
• Equation Writing
• Molecular, Complete, and Net Ionic
• Solution Stoichiometry
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Ionization and Miscible Terminology
Dissociation and Ionization
•
When ionic compounds dissolve in
water, the anions and cations are
separated from each other. This is
called dissociation.
Dissolving and Miscible
•
Na2S(aq)  2 Na+(aq) + S2–(aq)
K2SO4(aq)  2 K+(aq) + SO42−(aq)
H2SO4(aq)  2 H+(aq) + SO42–(aq)
•
•
When compounds (solutes) dissolve
in water (water) and form ions, the
process is referred to as ionization.
© 2013 Pearson Education, Inc.
When compounds (solutes)
dissolve in water as molecules,
we say that that the compound
is miscible.
–
Sugar(s) + water  sugar
water(l)
Ions are NOT formed.
Electrolytes
•
Electrolytes can be described as:
– Strong:
• Chemical substances that IONIZE COMPLETELY into their ions
– Examples: Soluble salts and strong acids or bases
– HCl (aq)  H+ + Cl- or CuCl2  Cu2+ + 2 Cl• Can conduct electrical current
– Weak:
• Chemical substances that IONIZE PARTIALLY into their ions
– Examples: weak acids or weak bases
» CH3COOH(aq)  CH3COO- (aq) + H+(aq)
• Can conduct electrical current
– Nonelectrolytes:
• Chemical substances that DISSOLVE in water but NOT AS IONS
• They do not conduct electricity.
– Example: polar substances such as sugar or alcohol
» C6H12O6(s)  C6H12O6(aq)
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Electrolytes and Nonelectrolytes
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Salt (Ionic compound) vs.
Sugar Dissolved in Water
Salts (ionic compound) ionize when
dissolved in water to form ions
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Molecular compounds do not
dissociate when they dissolve;
they DO NOT form ions.
Solutions are formed when solutes dissolve
in solvents.
•
For a solution to form, the solute and solvent must have similar
polarities.
– “Likes dissolve in likes.”
– There are attractive forces between the solute particles holding them
together.
•
When a solution forms:
– The solute-solute interactions and solvent-solvent interactions must
be overcome to form NEW solute-solvent interactions.
• If the attractions between solute and solvent are strong enough,
the solute will dissolve and a solution is formed.
© 2013 Pearson Education, Inc.
Acid vs. Base
Acids
Bases
•
Molecular compounds that
ionize when dissolved in water
– When acids ionize, they
form the cation H+ and an
anion.
•
Molecular compounds that
ionize when dissolved in water
– When bases ionize, they
form the anion OH- and a
cation.
•
The degree of ionization varies.
– Acids that ionize virtually
100% are called strong
acids.
•
The degree of ionization varies.
– Bases that ionize virtually
100% are called strong
bases.
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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NaOH(aq)  OH-(aq) + Na+(aq)
– Acids that only ionize a
small percentage are called
weak bases.
NH3(aq) + H2O  OH-(aq) + NH4+(aq)
Solubility of Ionic Compounds
•
Compounds that dissolve in a solvent are said to be soluble, while those
that do not are said to be insoluble.
– Example:
• NaCl is soluble in water.
• AgCl is insoluble in water.
•
Solubility of a compound is affected by:
– Temperature
– Volume
– Pressure for gas solutes
•
Even insoluble compounds dissolve, just not enough to be meaningful.
•
Predicting whether a compound will dissolve in water is 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)
Compounds Containing the
Following Ions Are Generally
Soluble:
Exceptions
(when combined with ions on the
left, the compound is insoluble)
Li+, Na+, K+, NH4+
none
NO3–, C2H3O2–
none
Cl–, Br–, I–
Ag+, Hg22+, Pb2+
SO42–
Ag+, Ca2+, Sr2+, Ba2+, Pb2+
© 2013 Pearson Education, Inc.
Solubility Rules
(Compounds That Are Generally Insoluble in Water)
Exceptions
Compounds Containing the
Following Ions Are Generally
Insoluble:
OH–
S2–
CO32–, PO43–
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(when combined with ions on the left,
the compound is soluble or slightly
soluble)
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+
Which of the following salts are soluble in
water?
KOH
AgBr
CaCl2
Pb(NO3)2
PbSO4
© 2013 Pearson Education, Inc.
Which of the following salts are 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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Reaction Types
• Double Displacement:
• Metathesis
• Precipitation reactions
• Gas Forming
• Metal + Acid
• Acid-Base Neutralization
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Criteria for Chemical Reactions to Occur
Kinetics and Thermodynamics
• With the kinetics and thermodynamics conditions met, then
the driving forces behind a chemical reaction are:
– Making a gas
– Making a liquid
– Making a precipitate (solid)
– Exchange of electrons (redox reaction)
• At least one of the four driving forces must be observed for
a chemical reaction to have occurred.
© 2013 Pearson Education, Inc.
Chemical Reactions: Types
Two main categories:
• Decomposition
– A reactant decomposes to two or more products
• Synthesis
– Two or more reactants combine to make product(s)
– Types of synthesis reactions
• Double Displacement
– Precipitation reactions
• Single Displacements
– Gas-forming reactions
• Reduction/Oxidation
– Combustion reactions
– Corrosion
• Neutralization reactions
– Acid and base
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Double Displacement Reactions
•
There is an “exchange of ions” or a “switching of partners” between
reactants to form products.
AX + BY
AY + BX
1. Determine formulas of possible products.
– (+) ion from one reactant with (–) ion from other
– Balance charges of combined ions to get formula of each product.
2. Determine solubility of each product in water.
– Use the solubility rules.
– If product is insoluble or slightly soluble, it will precipitate and an
equation can be written.
– If neither product will precipitate, there is no reaction and an
equation CAN NOT be written.
© 2013 Pearson Education, Inc.
Precipitation Reactions Are a Type of
Double Displacement Reaction
• 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.
• The “driving force” is the formation of an insoluble compound,
a precipitate (solid).
• Solubility rules are used to determine whether or not a product
forms a solid (insoluble).
– BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2 NaCl(aq)
© 2013 Pearson Education, Inc.
© 2013 Pearson Education, Inc.
Problem: Write the equation for the precipitation reaction between an
aqueous solution of potassium carbonate and an aqueous solution of
nickel(II) chloride.
Strategy for writing an equation for a double displacement reaction:
1.
2.
3.
4.
5.
6.
Write the formulas of the reactants.
Determine the possible products.
a) Determine the ions present.
b) Exchange the ions.
c) Write the formulas of the products.
Determine the solubility of each product.
If both products soluble, write no reaction.
Write (aq) next to soluble products and (s) next to insoluble products.
Balance the equation.
© 2013 Pearson Education, Inc.
Answer:
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.
K2CO3(aq) + NiCl2(aq)  KCl + NiCO3
3. Determine the solubility of each product.
KCl is soluble.
NiCO3 is insoluble.
© 2013 Pearson Education, Inc.
Answer continued:
4. If both products soluble, write no reaction.
Does not apply since NiCO3 is insoluble.
5. For soluble products write (aq), 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)
© 2013 Pearson Education, Inc.
Practice—Predict the products and
Problem:the
Write
the equation for the following
balance
equation.
precipitation reactions:
a) KCl(aq) + AgNO3(aq)  ?
b) Na2S(aq) + CaCl2(aq)  ?
© 2013 Pearson Education, Inc.
Problem: Write the equation for the following
precipitation reactions:
a) KCl(aq) + AgNO3(aq)  ?
Answer:
KCl(aq) + AgNO3(aq)  AgCl(s) + KNO3(aq)
b) Na2S(aq) + CaCl2(aq)  ?
Answer:
No reaction because Na2S(aq) + CaCl2(aq)
yields soluble products CaS(aq) + NaCl(aq).
© 2013 Pearson Education, Inc.
Net Ionic Equations
Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2 KNO3(aq)
The above chemical reaction is written as a “complete equation.”
– Because Pb(NO3)2 and K2CrO4 are strong electrolytes we can
write
Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + CrO42-(aq)
 PbCrO4(s) + 2 K+(aq) + 2 NO3-(aq)
This is written now as a complete “ionic equation.”
Question: What about K+ and NO3- ions?
Answer: They are “spectator ions.” These ions DO NOT
participate in the reaction.
© 2013 Pearson Education, Inc.
Net Ionic Equations
Complete ionic equation:
Pb2+(aq) + 2 NO3- (aq) + 2 K+(aq) + CrO42-(aq)
 PbCrO4(s) + 2 K+(aq) + 2 NO3-(aq)
• Spectator ions are left out when writing net ionic
equations.
Pb2+(aq) + CrO42-(aq) PbCrO4(s)
is the NET IONIC EQUATION for this reaction.
© 2013 Pearson Education, Inc.
Problem: Write the molecular, complete ionic, and
net ionic equation for each of the following
reactions.
a) K2SO4(aq) + 2 AgNO3(aq) ?
b) Na2CO3(aq) + 2 HCl(aq)  ?
c) Ni(NO3)2(aq) + Mg(OH)2(aq)  ?
d) Li3PO4(aq) + CaCl2(aq)  ?
© 2013 Pearson Education, Inc.
a) K2SO4(aq) + 2 AgNO3(aq)  ?
Molecular:
K2SO4(aq) + 2 AgNO3(aq)  Ag2SO4(s) + 2 KNO3(aq)
Complete ionic:
2 K+(aq)+ SO42- (aq) + 2 Ag+(aq) + 2 NO3- (aq) 
Ag2SO4(s) + 2 K+(aq) + 2NO3-(aq)
Net ionic:
SO42-(aq) + 2 Ag(aq)  Ag2SO4(s)
b) Na2CO3(aq) + 2 HCl(aq)  ?
Molecular:
Na2CO3(aq) + 2 HCl(aq)  CO2(g) + 2 NaCl(aq) + H2O(l)
Complete ionic:
2 Na+(aq) + CO32-(aq) + 2 H+(aq) + 2 Cl-(aq) 
CO2(g) + 2 Na+(aq) + H2O(l) + 2 ClNet ionic:
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CO32- (aq) + 2 H+(aq)  CO2(g) + H2O(l)
c) Ni(NO3)2(aq) + Mg(OH)2 (aq)  ?
Molecular:
Ni(NO3)2(aq) + Mg(OH)2(aq)  Ni(OH)2(s) + Mg(NO3)2(aq)
Complete ionic:
2 Ni2+(aq) + 2 NO3- (aq) + Mg2+(aq) + 2 OH-(aq) 
Ni(OH)2(s) + 2 Mg2+(aq) + 2 NO3- (aq)
Net ionic:
2 Ni2+(aq) + 2 OH-(aq)  Ni(OH)2(s)
d) Li3PO4(aq) + CaCl2(aq)  ?
Molecular:
2 Li3PO4(aq) + 3 CaCl2(aq)  Ca3(PO4)2 (s) + 6 LiCl(aq)
Complete ionic:
6 Li+ + 2 PO43-(aq) + 3 Ca2+ + 6 Cl- (aq) 
Ca3(PO4)2(s) 6 Li+(aq) + 6 ClNet ionic:
© 2013 Pearson Education, Inc.
2 PO43- (aq) + 2 Ca2+(aq)  Ca3(PO4)2 (s)
Acid-Base Reactions
•
The “driving force” for many strong acid- strong base reactions is the formation of
water.
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
– Net ionic equation:
OH- (aq) + H+ (aq)  2 H2O(l)
•
A common product of many acid-base reactions is water and a SALT, MX.
HX + MOH  MX + H2O
Mn+ comes from base and Xn- comes from acid.
•
Acid-base reactions are referred to as NEUTRALIZATION reactions.
– NEUTRALIZATIONS
• When the number of moles of acid in solution = the moles of base in
solution
• DOES NOT mean that the pH of the solution is necessarily 7
© 2013 Pearson Education, Inc.
© 2013 Pearson Education, Inc.
Gas-Evolution 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)
© 2013 Pearson Education, Inc.
Gas-Forming Reactions
Metal carbonates + acid  CO2(g) + salt + water
Na2CO3(aq) + 2HCl(aq)  H2O(l) + CO2(g) + 2 NaCl(aq)
Net ionic:
CO32- + 2 H+  H2O(l) + CO2(g)
CaCO3(s) + 2 HCl(aq)  CO2(g) + H2O(l) + CaCl2(aq)
Net ionic: CaCO3(s) + 2 H+  CO2(g) + H2O(l) + Ca2+
Metal (s) + acid  Gas + salt
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2(aq)
Net ionic:
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Mg(s) + 2 H+  H2(g) + Mg2+
Compounds That Undergo Gas-Evolution
Reactions
Reactant
Type
Reacting
with
Ion
Exchange
Product
Decom
-pose?
Gas
Formed
Example
metalnS,
metal HS
acid
H2S
no
H2S
K2S(aq) + 2 HCl(aq) 
2 KCl(aq) + H2S(g)
metalnCO3,
metal HCO3
acid
H2CO3
yes
CO2
K2CO3(aq) + 2 HCl(aq) 
2 KCl(aq) + CO2(g) + H2O(l)
metalnSO3
metal HSO3
acid
H2SO3
yes
SO2
K2SO3(aq) + 2 HCl(aq) 
2 KCl(aq) + SO2(g) + H2O(l)
(NH4)nanion
base
NH4OH
yes
NH3
KOH(aq) + NH4Cl(aq) 
KCl(aq) + NH3(g) + H2O(l)
© 2013 Pearson Education, Inc.
© 2013 Pearson Education, Inc.
Other Patterns in Reactions
• The precipitation, acid/base, and gas-evolving
reactions are all involved in exchanging the ions in the
solution.
• Other kinds of reactions involve transferring electrons
from one atom to another; these are called oxidation–
reduction reactions.
– 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 and Reduction Reactions
Oxidation:
•
•
•
Reduction:
The process that occurs when
– the oxidation number of an
element increases
– an element loses electrons
– a compound gains oxygen
– a compound loses hydrogen
– a half-reaction has electrons as
products
It occurs at the anode in an
electrochemical cell.
•
OXIDIZING AGENT is an electron
acceptor; it causes another species
to be OXIDIZED but it (agent) is
being reduced.
•
© 2013 Pearson Education, Inc.
•
The process that occurs when
– the oxidation number of an
element decreases
– an element gains electrons
– a compound loses oxygen
– a compound gains hydrogen
– a half-reaction has electrons as
reactants
It occurs at the cathode in an
electrochemical cell.
REDUCING AGENT is an electron
donor; it causes another species to
be REDUCED but it (agent) is being
oxidized.
Redox Reaction:
Transfer of Electron Reactions
2 Al(s) + 3 Cu2+(aq)  2 Al3+(aq) + 3 Cu(s)
Oxidation
Al(s)  Al3+(aq) + 3 e» Electrons are LOST by a substance.
» Ox. no. of Al increases as e- are donated by the metal.
» Therefore, Al is OXIDIZED.
» Al is the REDUCING AGENT in this balanced half-reaction.
Reduction:
Cu2+(aq) + 2 e-  Cu(s)
» Electrons are GAINED by a substance.
» Ox. no. of Cu decreases as e- are accepted by the ion.
» Therefore, Cu is REDUCED.
» Cu is the OXIDIZING AGENT in this balanced half-reaction.
If something is being reduced, something else is being oxidized.
© 2013 Pearson Education, Inc.
Oxidation and Reduction
• Oxidation and reduction MUST occur simultaneously.
• 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
oxidation
reduction
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+1  −2
© 2013 Pearson Education, Inc.
© 2013 Pearson Education, Inc.
Rules for Assigning Oxidation States
1.
Free elements in their atomic state have an oxidation state = 0.
Example:
Na = 0 and Cl2 = 0
2.
Alkali and alkaline earth metals always have an oxidation state
of +1 and +2, respectively.
3.
Fluoride ion has an oxidation state of -1.
4.
Hydrogen’s oxidation state is mostly +1 (H+), but when
combined with a metal it behaves as the hydride anion -1 (H-).
© 2013 Pearson Education, Inc.
Rules for Assigning Oxidation States
5. Oxygen most of the time has an oxidation state of -2 (O2- oxide
ion); however, it can have oxidations states of -1 (O22- or Operoxide ion) or -½ ( O½- super oxide ion) .
6.
The sum of the oxidation states of all the atoms in a compound
is 0.
Example:
Na = +1 and Cl = −1
NaCl: (+1) + (−1) = 0
7. 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
NO3–, (+5) + 3(-2) = −1
© 2013 Pearson Education, Inc.
Problem: Assign oxidation states, determine the elements
oxidized and reduced, and determine the oxidizing agent
and reducing agent in the following reaction.
The reaction:
Sn4+ + Ca → Sn2+ + Ca2+
Answer:
Sn4+ + Ca → Sn2+ + Ca2+
+4
0
+2
+2
Ca → Ca2+ going from 0 to +2: lost two electrons:
Ca is oxidized
Sn4+ → Sn2+ going from +4 to +2: gained two electrons
Sn4+ is reduced
Ca is the reducing agent; Sn4+ is the oxidizing agent.
© 2013 Pearson Education, Inc.
Redox Half-Reactions
•
Redox reactions MUST be charge and mass balanced!
•
Balancing redox reactions:
– Note the environment that the reaction is occurring in (neutral, acidic,
basic).
– Split the redox reaction into two separate half-reactions.
• Determine which species is being oxidized and which is being
reduced using oxidation states.
» The oxidation half-reaction has electrons as products.
» The reduction half-reaction has electrons as reactants.
– Mass and charge balance each half-reaction.
• Mass balancing
• Charge balancing
© 2013 Pearson Education, Inc.
Redox Half-Reactions
•
Mass and charge balance each half-reaction.
•
Mass balancing
• Balance your metal then nonmetal elements.
• Balance your H next.
• Balance you O last.
– Environment concerns, specifically acidic and basic environments
• Acidic
– Balance H as H+.
– Balance O as water, H2O,
• Basic
– Balance H as H+.
– Balance O as OH-.
– Counterbalance H+ with OH- to make H2O.
•
– Charge balancing
• Use oxidation states to determine number of electrons transferred.
» The oxidation half-reaction has electrons as products.
» The reduction half-reaction has electrons as reactants.
Add the two half-reactions and simplify.
© 2013 Pearson Education, Inc.
Problem:
Balance the following reaction:
Cl2 + I− + H2O → Cl− + IO3− + H+
Break into two half-reactions to mass and charge balance.
oxidation:
I− → IO3− + 6 e−
3 H2O+ I− → IO3− + 6 e− + 6 H+
reduction:
Cl2 + 2 e− → 2 Cl−
Final reactions can NOT have electrons in the final version.
3 H2O + I−
3 (Cl2 + 2 e−
→
→
IO3− + 6 e− + 6H+
2 Cl−)
3 Cl2 + I− + 3 H2O → 6 Cl− + IO3− + 6 H+
© 2013 Pearson Education, Inc.
Balancing Redox Reaction in Acidic Solution
Fe2+ + MnO4– → Fe3+ + Mn2+
Break into two half-reactions to mass and charge balance.
oxidation:
Fe2+ → Fe3+ + 1 e-
reduction:
MnO4− + 5 e− → Mn2+
8 H+ + MnO4− + 5 e− → Mn2+ + 4 H2O
Final reactions can NOT have electrons in the final version.
5 (Fe2+ →
8H+ + MnO4− + 5 e− →
Fe+3 + 1 e- )
Mn2+ + 4H2O
5 Fe2+ + MnO4– + 8 H+ → Mn2+ + 4 H2O + 5 Fe3
© 2013 Pearson Education, Inc.
Balancing Redox Reaction in Basic Solution
I- + MnO4– → I2 + MnO2
Break into two half-reactions to mass and charge balance.
oxidation:
I- → I2 + e−
2 I- → I2 + 2 e−
reduction:
MnO4− + 3 e− → MnO2
4 H+ + MnO4− + 3 e− → MnO2 + 2 H2O
Can NOT have H+ present in a basic solution so add OH- to H+.
4 OH- + 4 H+ + MnO4− + 3 e− → MnO2 + 2 H2O + 4 OH4 H2O + MnO4− + 3 e− → MnO2 + 2 H2O + 4 OH2 H2O + MnO4− + 3 e− → MnO2 + 4 OHFinal reactions can NOT have electrons in the final version.
3 (2 I- →
I2 + 2 e−)
2 (2 H2O + MnO4− + 3 e− → MnO2 + 4OH- )
6 I(aq) + 2 MnO4(aq) + 4 H2O(l)  3 I2(aq) + 2 MnO2(s) + 8 OH(aq)
© 2013 Pearson Education, Inc.
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)
© 2013 Pearson Education, Inc.
Problem:
Assign an oxidation state to the following:
• Br2
• K+
• LiF
• CO2
• SO42−
© 2013 Pearson Education, Inc.
Problem: Assign an oxidation state to the following:
• Br2
Br = 0;
• Br2 is in its elemental state.
• K+
K = +1
• LiF
Li = +1 and F = −1;
• LiF is a compound and compounds have
an overall charge of zero.
• CO2
O = −2 and C = +4;
• CO2 is a compound and compounds have
an overall charge of zero.
• SO42−
O = −2 and S = +6;
• SO4-2 ion has an overall charge of -2.
© 2013 Pearson Education, Inc.
Combustion Reactions
2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
• Reactions in which O2(g) is a reactant are called
combustion reactions.
• Combustion reactions release lots of energy.
• Combustion reactions are a subclass of
oxidation–reduction reactions.
© 2013 Pearson Education, Inc.
Combustion Products Predictions
To predict the products of a combustion reaction,
combine each element in the other reactant with
oxygen.
Reactant Contains:
Combustion Product
C
CO2(g)
H
H2O(g)
S
SO2(g)
N
NO(g) or NO2(g)
Metal (M)
M2O(s)
© 2013 Pearson Education, Inc.
Problem: Write the equation for the complete
combustion of CH3OH(l).
1.
Write the reactants.
– Combustion is reacting with O2(g).
CH3OH(l) + O2(g) 
2.
Combine O with each element to make a product.
– C + O make CO2(g); H + O make H2O(g)
CH3OH(l) + O2(g)  CO2(g) + H2O(g)
3.
Balance the equation.
2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(g)
© 2013 Pearson Education, Inc.
Problem: Complete and balance the following
reactions.
1. Combustion of acetic acid, HC2H3O2(l)
2. Combustion of isopropyl alcohol, C3H7OH(l)
© 2013 Pearson Education, Inc.
Problem: Complete and balance the following
reactions.
1. Combustion of acetic acid, HC2H3O2(l)
HC2H3O2(l) + O2(g) 
HC2H3O2(l) + O2(g)  CO2(g) + H2O(g)
HC2H3O2(l) + 2 O2(g)  2 CO2(g) + 2 H2O(g)
© 2013 Pearson Education, Inc.
Problem: Complete and balance the following
reactions
2. Combustion of isopropyl alcohol, C3H7OH(l)
C3H7OH(l) + O2(g) 
C3H7OH(l) + O2(g)  CO2(g) + H2O(g)
C3H7OH(l) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
© 2013 Pearson Education, Inc.