Transcript Packet 1 - Kentucky Community and Technical College System

Packet 4
Chemical Quantities and Aqueous
Reactions
“Nothing in the world is as soft and yielding as water. Yet for
dissolving the hard and inflexible, nothing can surpass it.”
Tao Te Ching, ca. 500 B.C.
“Divide, and conquer.”
Phillip of Macedon
last modified: 7/20/2015
CHE 170 Packet 4 - 1
Concept Area I: Terminology
stoichiometry
stoichiometric
coefficients/factors
limiting
reactant/reagent
theoretical yield
actual/experimental
yield
percent yield
solute
solvent
solution
concentrated
dilute
Molarity, M
electrolyte (strong &
weak)
nonelectrolyte
acid (strong & weak)
base (strong & weak)
soluble
insoluble
slightly soluble
precipitate
precipitation reaction
molecular or
complete formula
equation
complete ionic
equation
net ionic equation
spectator ions
acid-base reaction
hydronium ion
salt
oxidation
reduction
oxidation-reduction,
redox
oxidation state or
numbers
oxidizing agent
reducing agent
oxidizer
reducer
CHE 170 Packet 4 - 2
The Mole Refresher
Remember, one mole is the amount of
substance that contains as many particles
(atoms, molecules, whatever) as there are in
12.0 g of 12C.
And remember that there is Avogadro’s
number of particles in a mole of any
substance.
What is Avogadro’s number?
CHE 140 Packet 8 - 3
Solution Concentration: Molarity
To indicate how much solute we have in a
solution, Molarity is commonly used. It is
abbreviated with a capital “M”. It is defined:
mols of solute
M
1 L of solution
“Molarity” is usually used as a noun. If an
adjective is needed to mean the same thing,
chemists use the term “Molar”.
I will usually always capitalize “Molarity” and
“Molar” since the symbol is a capital “M”.
CHE 170 Packet 4 - 4
A “road map” between grams, moles,
number of particles and Molarity!
mass of element or
compound
molar mass(es)
from periodic table
number of
particles
Avogadro’s number
6.022×1023
moles of element
or compound
Volume
Molarity
CHE 170 Packet 4 - 5
mols of solute
M
1 L of solution
This picture
shows very
clearly that a
one liter of
solution (in the
volumetric
flask) is quite
different than
one liter of
solvent (the
water in the
volumetric
flask and the
graduated
cylinder).
CHE 170 Packet 4 - 6
Let’s do an example for M!
What concentration is the solution we
made on the previous slide?
We took 25.0 g of CuSO4·5 H2O and
made a 1.00 liter solution.
Do the homework (problems 4.25—4.34) and make sure
comfortable with calculating Molarity because we will be adding it
CHE 170 Packet 4 - 7
to stoichiometry at the end of these notes!
Molarity Handout
Let’s Calculate Molarity!
1. Sodium bicarbonate is also known as baking
soda. If 38.9 g of it is dissolved in enough
water to make 500 mL of solution, what is
the Molarity of the solution?
2. If 25.3 g of sodium carbonate is dissolved in
enough water to make 250 mL of solution.
What is the Molar concentration of the
sodium ions?
3. How many grams of sodium phosphate do
we need to make 1.50 L of a 0.125 M
sodium phosphate solution?
CHE 170 Packet 4 - 8
Dilute vs.
Concentrated
Solutions can be
dilute or
concentrated.
dilute:
concentrated:
The concentration of a solution is dependent
only on how much solute is present in the
solvent (not whether the solute is strong or weak).
0.1 M NaCl is a dilute salt solution
5.0 M NaCl is a concentrated solution
CHE 170 Packet 4 - 9
Strong vs. Weak
Acids and bases can be classified as either strong or
weak. Salts are either strong or not, never weak.
Whether a solute is strong or weak has nothing to
do with the concentration (dilute or concentrated).
What determines if strong or weak?
CHE 170 Packet 4 - 10
Concept Area II: Stoichiometry
a. You should be able to interconvert between
grams and moles of reactants and products in
a balanced chemical equation.
b. You should be able to calculate theoretical
yield, actual yield, and percent yield.
CHE 170 Packet 4 - 11
Roadmap meets Stoichiometry worksheet
Stoichiometry!
Stoichiometry is the study of the
aspects of chemical reactions.
It rests on the principle of conservation of
matter.
Balanced chemical reactions are very
important in stoichiometry.
It is also still very important, or even more
important, to write down all units in any
stoichiometric calculation!
.
CHE 170 Packet 4 - 12
General Plan for Stoichiometric Calculations
Step 1) write a balanced chemical reaction.
Step 2) write down what we know
Step 3) write down what we want to know
Step 4) solve the problem using our new and
improved roadmap!
Mass
reactant
Mass
product
molar mass(es)
from periodic table
molar mass(es)
from periodic table
Moles
reactant
use stoichiometric
factors from balanced
chemical equation
Moles
product
CHE 170 Packet 4 - 13
Some Problems using:
2 NH4NO3 → 2 N2 + 4 H2O +O2
Mole ratio problem: If 37.2 mol NH4NO3
explode, how many mols of H2O are produced?
Mass problem: If we want to produce 12.0 g
oxygen, how many grams of NH4NO3 do we need?
CHE 170 Packet 4 - 14
Let’s try another one!
Using 5.00 g of H2O2, what mass of O2 and H2O can
be obtained when it completely decomposes?
Step 1: Write a balanced chemical equation!
Step 2: Write down what we know!
Step 3: Write down what we want to know.
Step 4: Solve problem using roadmap.
CHE 170 Packet 4 - 15
Using 5.00 g of H2O2, what mass of O2 and of H2O
that can be obtained when it completely decomposes?
So, first all roads lead to moles! Calculate moles of H2O2 from
grams given.
Then, use stoichiometric coefficients to calculate moles of O2 and
H2O.
Finally, calculate mass of O2 and H2O from moles calculated.
See notes page for step by step worked out.
CHE 170 Packet 4 - 16
We have been assuming that we get 100%
yields (reaction always goes to completion).
Do we really?
No! Just like most things (like
popcorn), reactions usually do not
produce as much as they
theoretically could.
So, we take our actual yield (our
experimentally measured
quantity) and compare it to our
theoretical yield (the amount we’d
get if everything reacted). To tell
us our percent yield.
CHE 170 Packet 4 - 17
Theoretical & Actual Yields
We started with 20 popcorn
kernels.
What is our theoretical yield
of popped popcorn?
What was our actual or
experimental yield of popped
popcorn?
CHE 170 Packet 4 - 18
Percent Yield Handout
So, to summarize and continue…
Theoretical yield is what we
get if all reacts.
Actual yield is what we
or
get.
To calculate percent yield:
actual yield
100
%
theoretical yield
So, what is our percent yield in our popcorn example
where only 16 of the 20 kernels popped?
Experimental yield should always be less than theoretical
yield. Why?
CHE 170 Packet 4 - 19
Concept Area III: Limiting Reagents
and Excess Reagents
a. You should be able to identify the limiting
reactant in a reaction mixture.
b. You should be able to calculate the amount
of product produced based on the limiting
reactant.
c. You should be able to calculate the amount
of excess reactant left over from a reaction
mixture.
CHE 170 Packet 4 - 20
Limiting Reactant
We all know:
1 hot dog + 1 bun → 1 hot dog “sandwich”
So, let’s go to the store
hot dog package has
hot dogs
hot dog bun package has
buns
Hmmmm… how many hot dog sandwiches can we
make if we only buy one package of each? How
many of what would be left over?
Determining the limiting reactant in a chemical
reaction works the same way!
CHE 170 Packet 4 - 21
Limiting Reactant
So, what will we look at to see which limits in a
reaction?
The number of
of the
!
The chemicals leftover are the excess reactants
or reagents.
For this reaction:
N2(g) + 3 H2(g) → 2 NH3(g)
If we start with the top
flask, which of the
bottom flasks shows
the reaction at
completion?
Tro page 122
CHE 170 Packet 4 - 22
Limiting Reactants
React solid Zn with 0.100 mol HCl(aq):
Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)
So, how many moles of Zn are needed
to completely react with the HCl?
1 mol Zn
0.100 mol HCl 
 0.0500 mol Zn
Let’s look at data for
2 mol HCl
three reactions:
Rxn 1
Rxn 2
Rxn 3
mass Zn (g)
7.00
mol Zn
0.107
mol HCl
0.100
mol HCl/mol Zn 0.93/1
Lim. Reactant
3.27
0.0500
0.100
2.00/1
1.31
0.0200
0.100
5.00/1
CHE 170 Packet 4 - 23
Let’s try this one!
SiO2
+
HF
→
SiF4↑
+ 2 H2O
Suppose we have 6.0 moles of silicon dioxide and 11 moles of
HF. Determine excess and limiting reagents, and how
much SiF4 gas and water are theoretically formed.
What steps should we take?
First, write a balanced chemical equation!
Now, what do we know and want to know?
Solve!
Correct mol:mol ratio? If no, which one do we have
too much of? too little?
Use limiting reagent to calculate theoretical amount of
products produced.
Let’s continue on next slide…
CHE 170 Packet 4 - 24
SiO2 + 4 HF → SiF4↑ + 2 H2O
Need 1 mol SiO2 for every mols HF
Have 6 mol SiO2, so would need
mols HF
We have 11 mols of HF, is that enough?
.
So, the limiting reagent is?
Thus, theoretical mol amounts that can be produced:
CHE 170 Packet 4 - 25
Of course, in a lab we can’t directly measure moles,
so let’s try a more realistic problem!
Calculate the theoretical yield of zinc sulfide, in grams,
that can be made from 0.488 g zinc and 0.503 g of S8.
If the actual yield is 0.606 g of zinc sulfide, what is the
percent yield?
Those numbers are difficult to compare, so let’s get whole
numbers by dividing both by the smaller number. CHE 170 Packet 4 - 26
Calculate the theoretical yield of zinc sulfide, in grams, that
can be made from 0.488 g zinc and 0.503 g of S8. If the
actual yield is 0.606 g of zinc sulfide, what is the percent
yield?
Okay, we have a
8 Zn + S8 → 8 ZnS
1 mol Zn
–3
 7.4610– 3 mol Zn 1.9610  3.81  4
65.39g Zn
1 molS8
0.503g S8 
 1.9610– 3 molS8  1.9610– 3  1.00  1
256.5 g S8
0.488g Zn 
ratio of 4 mols
Zn for each mol
S8.
We need 8 mols
Zn for each mol
S8.
Therefore, since
4<8, Zn is
limiting!
This is the what?
CHE 170 Packet 4 - 27
Limiting Reactant Practice Handout
Calculate the theoretical yield of zinc sulfide, in grams, that
can be made from 0.488 g zinc and 0.503 g of S8. If the
actual yield is 0.606 g of zinc sulfide, what is the percent
yield?
8 Zn + S8 → 8 ZnS
7.46 10 – 3 mol Zn 
8 mol ZnS 97.46 g ZnS

 0.727 g ZnS
8 mol Zn
1 mol ZnS
How do we calculate percent yield again?
CHE 170 Packet 4 - 28
Concept Area IV: Introduction to Solutions
a.
You should understand the difference between a
solute, solvent and solution.
b. You should understand the term electrolyte and be
able to classify a compound as a strong electrolyte,
weak electrolyte or a nonelectrolyte.
c. You should know what ions are formed when an
electrolyte dissolves in water.
d. You should know the basic solubility rules and how
to read a solubility table for more advanced rules.
CHE 170 Packet 4 - 29
Water, water everywhere…
Most of the solutions we form are water based.
Why?
Well, ¾ of the Earth is covered by water.
And, 70-80% of us is water.
Because water is so abundant it is very useful to use as a
solvent. The fact that so many ionic and molecular
chemicals are soluble in it makes it even more useful.
Many chemists are even trying to do all their
experiments in water since it is so much more
environmentally friendly, green chemistry.
Since water is so common as a solvent, we have a
special name for solutions that use water as the
solvent, aqueous. Remember that it can be
abbreviated “aq” when labeling species in a chemical
CHE 170 Packet 4 - 30
reaction.
Solution Composition
solute:
; it is
normally the component
of a solution present in
the smaller amount
Tro page 127
solvent:
; it is normally
the component of a
solution present in the
greater amount
solution: a mixture of
substances that has a
uniform composition; a
homogeneous mixture
CHE 170 Packet 4 - 31
Different kinds of Solutions
A solution is a homogeneous mixture of two or
more substances in a single phase: solid,
liquid, gas.
Rubies are a solution of red chromium
compounds with transparent aluminum
oxide. Fillings are also a metal
solution as are all metal alloys!
When we blow up
a balloon, we are
filling it with a
gaseous solution
of O2, CO2 and
other gases.
Mixing sugar into
water makes a
sugar water
solution once the
sugar dissolves.
CHE 170 Packet 4 - 32
Solubility
Solubility is how well a solute dissolves in a
solvent to form a
.
The more soluble or miscible something is in
a solvent, the
it is to make a solution.
Sometimes something is so soluble we could
put large amounts in. Others are so insoluble
that maybe one molecule dissolves in a
swimming pool full of solvent.
What determines a compound’s solubility in a
particular solvent?
CHE 170 Packet 4 - 33
Aqueous Solutions of Ionic versus
Molecular Compounds
Although many solutes are soluble/miscible in water,
there are two different ways they can dissolve in it.
1) They can separate into ions.
What are some examples?
2) The solute can be attracted to the solvent.
(Preview of intermolecular forces we’ll learn later.)
What are some examples (no ionic compounds!)?
Once again we have special terms; compounds of type 1
are called electrolytes and type 2 are called
nonelectrolytes.
CHE 170 Packet 4 - 34
What ions do we get?
Remember how we learned to put together
ions using charge balance?
Pb2+ and S2– would form?
Ca2+ and F– would form?
Al3+ and O2– would form?
Let’s now reverse that and break them apart!
What ions would these soluble compounds
form in water?
NaOH
Pb(NO3)2
Al2(SO4)3
CHE 170 Packet 4 - 35
How does water dissolve something?
First, the shape of
water makes one
side of it partially
negative and the
other side partially
positive.
Ions are either
positive or negative,
right?
So, the “negative”
side of water
surrounds cations,
and the “positive”
side of water
surrounds the
anions as we see in
this bottom picture.
Tro page 133
CHE 170 Packet 4 - 36
Ionic Compounds
Let’s look at an example of
an ionic compound,
potassium permanganate,
dissolving/dissociating in
water:
What’s the formula?
So, the ions formed in the
solution would be?
Ionic compounds are also
electrolytes, let’s examine
what this means…
CHE 170 Packet 4 - 37
Electrolytes & Nonelectrolytes
Electrolytes when put into an aqueous solution
form what?
Do nonelectrolytes form them?
Pure water does not conduct electricity. Water
with only nonelectrolytes in it does not
conduct electricity. Water with electrolytes
present does conduct electricity.
Why?
Hint: what do electrolytes do that pure
water and nonelectrolytes don’t?
CHE 170 Packet 4 - 38
Here’s a picture that
represents what
happens to the ions
when electricity is put
through a solution of
ions.
The movement of the ions
allows for the electrons
to flow.
If there weren’t ions,
electrons would be
unable to move through
the solution.
CHE 170 Packet 4 - 39
Notice each picture below represents what is happening at the
submicroscopic level. We can see the water molecules interacting
with the ions or molecules present.
2+
2+
2+
CHE 170 Packet 4 - 40
These pictures
are also trying
to represent
what is
happening at
the
submicroscopic
level, but the
water
molecules have
been omitted
for clarity.
CHE 170 Packet 4 - 41
Electrolytes, Weak Electrolytes and Nonelectrolytes
Okay, so we now know the difference between electrolytes
and nonelectrolytes. But, what makes one electrolyte
strong and another weak?!?!
Strong electrolytes dissociate (break apart) or ionize
, and thus, are
.
conductors of electricity. Experimental proof:
.
Weak electrolytes dissociate or ionize
and thus, are
.conductors of electricity.
Experimental proof:
Nonelectrolytes dissociate or ionize
and thus, are
.conductors of electricity.
Experimental proof:
,
.
,
.
Remember! Strong and weak do not refer to concentration
in this context!
CHE 170 Packet 4 - 42
Electrolytes Handout
Strong Electrolytes – ions in solution
1) All soluble salts (even if only slightly soluble).
2) Strong bases – Group 1A and 2A hydroxides
3) Strong acids – HCl, HBr, HI, HNO3, HClO4, H2SO4
Weak Electrolytes – ions & molecules in solution
1) None (salts are either soluble or they’re not)
2) Weak bases – NH3, etc.
3) Weak acids – CH3CO2H, etc.
Nonelectrolytes – no ions, just molecules (or nothing) in sol’n
1) Insoluble salts
4) All other compounds not already covered.
CHE 170 Packet 4 - 43
Solubility Rules
Tro page 136
CHE 170 Packet 4 - 44
Solubility Rules handout
Solubility Rules
Learn how to use solubility tables! A table similar
to the one from our book (shown on previous
slide) will be provided on exams (a copy of it is
on the notes page).
However it will be
missing the…
Key rule – memorize!
Salts of alkali metals,
ammonium, nitrate, and
acetate are almost always
soluble (no common
exceptions).
SOLUBLE COMPOUNDS
Almost all salts of Na+, K+,
NH4+
Salts of nitrate, NO3–
chlorate, ClO3–
perchlorate, ClO4–
acetate, CH3CO2–
EXCEPTIONS
Almost all salts of Cl–, Br–, I–
Halides of Ag+, Hg22+, Pb2+
Compounds containg F–
Fluorides of Mg2+, Ca2+, Sr2+, Ba2+, Pb2+
Salts of sulfate, SO42–
Sulfates of Mg2+, Ca2+, Sr2+, Ba2+, Pb2+
INSOLUBLE COMPOUNDS
EXCEPTIONS
All salts of carbonate, CO32–
phosphate, PO43–
oxalate, C2O42–
Salts of NH4+ and the alkali metal cations
chromate, CrO42–
Most metal sulfides, S2–
Most metal hydroxides and
oxides
See notes page for the solubility table provided during exams.
Ba(OH)2 is soluble
CHE 170 Packet 4 - 45
Concept Area V: Writing Different Types
of Chemical Reactions
a. You should be able to write a complete
balanced equation for the following types:
molecular/complete formula, complete ionic,
and net ionic.
b. You should be able to recognize a
precipitation forming reaction and predict
the identity of the precipitate that forms.
c. You should be able to complete a
precipitation forming reaction.
CHE 170 Packet 4 - 46
AB + CD → AD + CB
This type of reaction can be called an exchange
reaction, double displacement or metathesis.
There are three basic types: a.) precipitation,
b.) acid-base, and c.) gas-forming.
All involve the exchange of ions to form a new
. As my high school chemistry
instructor liked to say,
“Switch partners and dance! ”
CHE 170 Packet 4 - 47
a.) Precipitation Reactions
Example:
2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2↓
A precipitate is an insoluble compound formed by a
reaction in solution, usually aqueous. The
precipitate forms a
, that then falls to the
bottom due to gravity.
We use solubility rules to predict what will happen
when two soluble salts are combined.
The formation of a precipitate will drive a reaction
to completion. Why?
Well, why would the following not really happen.
Think submicroscopically!
NH4OH(aq) + NaCl(aq) → NaOH(aq) + NH4Cl(aq)
Tro page 137
CHE 170 Packet 4 - 48
a.) Precipitation Reactions
2 AgNO3(aq) + CaCl2(aq) → Ca(NO3)2(aq) + 2 AgCl ↓
CHE 170 Packet 4 - 49
How to predict precipitates when solutions
of two ionic compounds are mixed
 Step 1 Write the reactants as they actually
exist before any reaction occurs (the complete
ionic equation). Remember that when a salt
dissolves, its ions completely separate.
 Step 2 Consider the various solids that could
form. To do this, simply exchange the anions
(or the cations) of the added salts.
 Step 3 Use the solubility rules to decide
whether a solid forms and, if so, to predict
the identity of the solid.
CHE 170 Packet 4 - 50
Writing Chemical Equations, Part II
Molecular or complete formula – previously
used when we learned how to write chemical
equations. Note that ionic reactions should use the term
complete formula, but most people just call it molecular.
Complete Ionic – just like above, only all
soluble ionic compounds are written as separate
ions as they actually exist, not together
Net Ionic –just like complete ionic only ions
that don’t change (spectators) aren’t written
Beginning on the next slide, let’s work an example
CHE 170 Packet 4 - 51
to help us learn how to use!
Molecular or Complete Formula Equations
AgNO3(aq) + CaCl2(aq) →
This is a complete formula equation; it’s not
technically a molecular equation because we have
ions and not molecules. But, remember that
most times people will go ahead and call it a
molecular equation anyway.
Note that we have written everything together as
formulas. We’ve even shown what state the
chemicals are in to make it complete.
CHE 170 Packet 4 - 52
2 AgNO3(aq) + CaCl2(aq) → 2 AgCl(s)+ Ca(NO3)2(aq)
Complete Ionic Equations
Note that here we wrote everything how it actually exists
in the solution. The soluble salts ionize and are
written as ions. The insoluble salts don’t ionize and
are written with the ions together. The states may
still be put in; we didn’t have room here.
So in a complete ionic reaction, separate all the ionic
compounds that are “aq” and leave together all the
“s”, “ℓ” and “g”!
Could we write one of these for a molecular equation?
CHE 170 Packet 4 - 53
2 Ag+ + 2 NO3– + Ca2+ + 2 Cl– → 2 AgCl + Ca2+ + 2 NO3–
Net Ionic Equations
Note here we only wrote those species that
were different in products and reactants.
The rest did not participate, they watched.
So, they are called spectator ions.
CHE 170 Packet 4 - 54
Your Turn!
Write all three kinds of equations for the following:
1. Al(NO3)3(aq) + 3 NaOH(aq) → Al(OH)3(s) + 3 NaNO3(aq)
complete ionic
net ionic
2. When aqueous solutions of copper(II) nitrate and potassium
carbonate are mixed, a precipitate forms.
complete formula
complete ionic
net ionic
CHE 170 Packet 4 - 55
Helpful Tips
How do we know what state something is when
the problem doesn’t say and we’re expected to
know?
Remember, that many periodic tables color-code to
indicate what state the elements are typically found in.
Most pure covalent compounds are gases or liquids.
Most pure ionic compounds are solids.
If the reaction is done in water (may say aqueous or
solution to indicate), most of the compounds will be
aqueous except…
insoluble salts which will be a solid precipitate.
covalent gases which will bubble away as a gas.
CHE 170 Packet 4 - 56
Concept Area V: Acids & Bases
a.
You should be able to recognize acids and bases
and the reactions involving them.
b. You should be able to write a molecular, complete
ionic and net ionic reaction for acid-base reactions.
c. You should be able to complete an acid-base
reaction.
d. You should know the difference between
strong/weak acids and bases and give common
examples of each.
e. You should be able to recognize a gas-forming
reaction.
CHE 170 Packet 4 - 57
Arrhenius Acid & Base Definitions
base:
acid:
CHE 170 Packet 4 - 58
Acids and Bases Handout
Some Common Acids & Bases
Acids
HCl
HNO3
H2SO4
H2CO3
H3PO4
CH3CO2H
hydrochloric acid
nitric acid
sulfuric acid
carbonic acid
phosphoric acid
acetic acid
Bases
NaOH
KOH
Ca(OH)2
Mg(OH)2
Ba(OH)2
NH3
sodium hydroxide
potassium hydroxide
calcium hydroxide
magnesium hydroxide
barium hydroxide
ammonia
Which ones are strong?
Which ones are weak?
CHE 170 Packet 4 - 59
Diagram that shows relative strengths of a
few acids and bases
CHE 170 Packet 4 - 60
b.) Acid-Base Reactions
Example:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(ℓ)
ACID + BASE YIELDS SALT & “WATER”!
Since the final product is neither acidic or basic
(if perfect mol:mol ratio), this reaction can also be
called a
reaction.
The driving force of this reaction is the formation
of a covalent compound, water – a liquid.
Acid-base reactions are common, and once we
recognize are much easier to solve!
CHE 170 Packet 4 - 61
2 H2O
H3O+ + OH–
Water is in constant
equilibrium between
hydronium ions and
hydroxide ions.
A proton, H+, never exists alone in an aqueous
solution, it is always a hydronium ion, H3O+.
However, chemists will write simply H+ in an
aqueous solution as a shorthand.
We may also use this shorthand when presence of
the hydronium ion itself is not important. Just
remember when writing H+, it’s really H3O+!
–
CHE 170 Packet 4 - 62
Acids & Bases
We’ve mentioned strong acids/bases and
weak acids/bases. What determines which
type is which again?
Complete the chemical equations for the
following:
HCl + H2O
b) CH3CO2H + H2O
c) NaOH + H2O
d) NH3 + H2O
a)
CHE 170 Packet 4 - 63
Completing Acid-Base Reactions
Complete the following acid-base reactions.
Remember, acid + base yields salt + “water”!
1. HCl(aq) + NaOH(aq) →
2. NH3(aq) + HBr(aq) →
3. CH3CO2H(aq) + KOH(aq) →
CHE 170 Packet 4 - 64
Strong
vs.
Weak
versus
Concentrated
vs.
Dilute
Science uses these terms differently than non-science uses. Do
you know how to use scientifically?
Label the following strong/weak and concentrated/dilute
appropriately.
1.
2.
3.
4.
5.
6.
7.
0.001 M NaCl (aq)
12.0 M HCl (aq)
6.0 M CH3COOH (aq)
0.05 M NaOH (aq)
8.0 M RbOH (aq)
0.0001 M CaSO4(aq)
0.10 M NH3 (aq)
CHE 170 Packet 4 - 65
Practice!
Complete the equations below. Then, for each
equation, label strong, weak, or non
electrolyte, and draw a picture of the solution.
The first is done as an example.
H2O
K3PO4(s) → 3 K+(aq) +PO43–(aq)
soluble salt → strong electrolyte
H2 O
HNO3(ℓ) →
H+(aq) +NO3–(aq)
H2O
C2H3OH(ℓ) → C2H3OH (aq)
CHE 170 Packet 4 - 66
And, more practice!
1. Which is the best electrical conductor? Why?
a.
b.
c.
d.
0.10 M NaCl
0.10 M CH3CH2OH
0.10 M CH3COOH
0.10 M C5H12O6
2. Which has the highest H+ concentration? Why?
a.
b.
c.
d.
0.10 M HCl
0.10 M H2SO4
0.10 M CH3COOH
0.10 M NH3
CHE 170 Packet 4 - 67
c.) Gas-Forming Reactions
Example:
CuCO3(s) + 2 HNO3(aq) → Cu(NO3)2(aq) + H2CO3(aq)
H2CO3(aq)
CO2(g) + H2O(ℓ)
Notice that one of our products (H2CO3) was able to
continue reacting to form a gas.
The driving force for this reaction is the gas being
evolved in the second step.
What would the net ionic equation be?
a) CO32– + 2 H+
CO2 + H2O
b) CuCO3 + 2 H+
Cu2+ + CO2 + H2O
CHE 170 Packet 4 - 68
Examples of Gas-Forming Reactions
Tro page 144
CHE 170 Packet 4 - 69
Concept Area VI: Redox
a.
You should be able to recognize an oxidationreduction, redox, reaction
b. You should be able to calculate oxidation numbers
for each atom in a chemical species.
c. You should be able to identify:
what’s oxidized
what’s reduced
what gains electrons
what loses electrons
the oxidizing agent
the reducing agent
CHE 170 Packet 4 - 70
Oxidation-Reduction Reactions, Redox
Remember how some elements, especially
transition metals, can form differently charged
ions? Well, sometimes they change because of
a chemical reaction!
Those charges are called oxidation states.
Why? Well an atom/ion can change its
oxidation state and does not have to ever be an
ion by itself. Plus, it doesn’t make sense to refer
to the charge on an ion in a molecule!
When an atom/ion changes oxidation states,
what is happening to it?
CHE 170 Packet 4 - 71
Oxidation-Reduction Reactions, Redox
So, when atoms/ions change their oxidation number by
gaining or losing electrons they become reduced or
oxidized, respectively.
If one of the reactants loses electrons, the other reactant
must gain them, right?
We’ll be sorting out all this vocabulary in a bit….. But,
Can oxidation occur without reduction?
Can something lose electrons when nothing gains
electrons?
Can reduction occur without oxidation?
Can something gain electrons when nothing loses
electrons?
CHE 170 Packet 4 - 72
Common Oxidizing and Reducing Agents
CHE 170 Packet 4 - 73
Assigning Oxidation Numbers
So far, we’ve always treated polyatomic ions as
a unit.
However, in a redox reaction that unit can be
broken apart.
So, we have to be able to determine the
oxidation numbers/states on the atoms within
the polyatomic ions.
While we’re at it how do we assign these
numbers to anything?
CHE 170 Packet 4 - 74
Assigning Oxidation Numbers Handout
Rules for Assigning Oxidation Numbers
1) The oxidation number of any free, uncombined element is
zero. Thus the atoms in Cl2, O3, S8, C and Xe all have the
oxidation number of zero.
2) The oxidation number of monatomic ions is the same as
the charge on that ion. So, you already know the oxidation
numbers for alkali and alkaline earth metals, halogens and
others you memorized.
3) Hydrogen’s oxidation number is almost always +1. The only
exceptions are when hydrogen is in its elemental state, and
when hydrogen combines with a metal to form a hydride
(where it would be –1).
4) Oxygen’s oxidation number is almost always –2. If you
have a peroxide the charge would be –1, and when it’s in its
CHE 170 Packet 4 - 75
elemental state the charge would be zero.
Rules for Assigning Oxidation Numbers
continued
5) Fluorine always has an oxidation number of –1 in
compounds.
6) The sum of all the oxidation numbers in a species must
equal the charge on that species. So, in an uncharged
species, the sum of all the oxidation numbers must be
zero; in an ion, the sum of all the oxidation numbers
must equal the charge on that ion.
7) The most electronegative element in a compound has a
negative oxidation number.
8) Use “charge balance” to obtain oxidation numbers of all
other atoms in the species.
CHE 170 Packet 4 - 76
An example of a redox reaction…
molecular (or complete formula) equation
CuSO4(aq) + Mg(s) → MgSO4(aq) + Cu(s)
complete ionic equation
Cu2+ + SO42– + Mg → Mg2+ + SO42– + Cu
Oxidation
net ionic equation
number of …
Cu2+ + Mg → Mg2+ + Cu
Now, let’s complete the table below!
oxidation
number as
reactant
copper
sulfur
oxygen
magnesium
oxidation
number as
product
CHE 170 Packet 4 - 77
Cu2+ + Mg → Mg2+ + Cu
Vocabulary time!
Copper’s oxidation number went from +2 to
0, so it gained electrons.
Therefore, it was reduced. Which makes sense
since copper’s oxidation number got smaller
(was reduced in size).
Since copper(II) ion was reduced, it underwent
the process of reduction.
Copper(II) ion had to get those electrons from
somewhere; so, it caused something else to be
oxidized or lose electrons. Thus, it is also
known as the oxidizing agent or the oxidizer!
Fun, eh? Now you try with Magnesium!
CHE 170 Packet 4 - 78
Cu2+ + Mg → Mg2+ + Cu
Your turn with the vocabulary:
Magnesium’s oxidation number went from
, so it
.
Therefore, it was
.
Since magnesium was
, it underwent
the process of
.
Magnesium electrons didn’t just disappear, they
went to another species; so, it caused something
else to
and get
.
Thus, magnesium is also the
or the
.
!
Get it?
CHE 170 Packet 4 - 79
Useful mnemonic device!
LEO the lion goes GER
CHE 170 Packet 4 - 80
Or we can use this mnemonic device!
OIL RIG
CHE 170 Packet 4 - 81
Redox Reactions Handout
Answer the following:
1. SnO2 + 2 C → Sn + 2 CO
oxidized?
reduced?
gained electrons?
lost electrons?
oxidizing agent?
reducing agent?
2. Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag
oxidized?
reduced?
gained electrons?
lost electrons?
oxidizing agent?
reducing agent?
Proper practice makes perfect for these!
CHE 170 Packet 4 - 82
Summary of Reaction Types
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Combustion
Burning something is also redox:
CH4 + 2 O2 → CO2 + 2 H2O + energy
Normally, burning requires oxygen and we
produce carbon dioxide and water.
Carbon monoxide, CO, is formed when
combustion is incomplete – there is not
enough oxygen present.
CHE 170 Packet 4 - 84
Summary of Common reaction Types with
their Driving Force!
Reaction Type
Driving Force
Precipitation
Formation of an insoluble
compound (see section 4.6)
Acid-base /
neutralization
Formation of water and/or proton
transfer (see section 4.8)
Gas-forming
Evolution of a water-insoluble gas
such as CO2 (see section 4.8)
Oxidation-reduction Electron transfer (see section 4.9)
CHE 170 Packet 4 - 85
Concept Area VIII: Solution Stoichiometry
a. You should be able to interconvert between
grams, moles, Molarity and volume of
solution of reactants or products in a
balanced chemical equation.
b. You should be able to calculate dilutions.
c. You should be able to use titration data to
calculate the Molarity of an unknown
solution.
d. You should be able to use and understand
the pH scale (this includes calculating pH
from H+ and H+ from pH).
CHE 170 Packet 4 - 86
Expanded General Plan for Stoichiometric Calculations
Step 1) write a balanced chemical reaction.
Step 2) write down what we know
Step 3) write down what we want to know
Step 4) solve the problem using our new and
improved roadmap!
Mass
reactant
Mass
product
molar mass(es)
from periodic table
molar mass(es)
from periodic table
Moles
reactant
Volume
Molarity
reactant
use stoichiometric
factor from balance
chemical equation
Moles
product
Volume
Molarity
product
CHE 170 Packet 4 - 87
Let’s try our newest roadmap out!
Metallic zinc reacts with hydrochloric acid to
produce hydrogen and ZnCl2. What volume
of 2.50 M HCl (in mL) is required to
completely consume 11.8 g of zinc?
Which of these solutions is correct?
1
2 1.00
11.8 
 
 0.144 mL
65.39 1 2.50
1 mol 1 mol 1.00 L 1000mL
11.8 g 



 36.1 mL
65.39 g 2 mol 2.50 mol
1L
11.8 g Zn
1 mol Zn 2 mol HCl 1.00 L HCl sol'n 1000mL HCl



 144 mL
65.39 g Zn 1 mol Zn
2.50 mol HCl
1 L HCl
CHE 170 Packet 4 - 88
Dilutions: M1V1=M2V2
How do we make 500 mL of a 0.00100 M solution of
K2Cr2O7 if we only have a 0.100 M K2Cr2O7 and
water to make it from?
M1V1  M 2 V2
(0.100 M)(V1 )  (0.00100M)(500mL)
(0.00100M)(500mL)
V1 
(0.100 M)
V1  5.00 mL
Ah, so we used 5.00 mL of the 0.100 M solution and diluted it to
500 mL (with approx. 495 mL of water) to make our solution!
Note: This formula is only to be used with dilutions! If
mixing an acid and a base (a reaction), this formula
CHE 170 Packet 4 - 89
cannot be used!
How to do a dilution in the lab:
CHE 170 Packet 4 - 90
How do we determine the concentration of an acid
using a base of known concentration or vice versa?
Known volume of a
base of known
concentration slowly
added to the acid.
Known volume of
an acid of
unknown
concentration.
By doing a titration!
Titrations are used to determine the
concentration of an unknown compound
(usually an acid or base) using a compound of
known concentration.
Usually, a titration is where we take a known
volume of an unknown concentration of acid
and slowly react it with a known volume and
concentration of a base (or vice versa).
In lab, we use a buret to deliver the base into an
Erlenmeyer flask or beaker holding the acid.
CHE 170 Packet 4 - 91
A titration in a lab…
CHE 170 Packet 4 - 92
A titration in a lab explained.
1. Add solution from the buret, usually base.
2. The reagent (base) reacts with compound
(acid) in solution in the flask.
3. Indicator shows (by changing colors) when the
exact stoichiometric reaction has occurred if
correct indicator chosen.
4. Net ionic equation: H+ + OH– → H2O.
5. At the equivalence point we have a perfect
mol:mol ratio. So, moles H+ = moles OH –.
CHE 170 Packet 4 - 93
Titration Problem:
So, if 38.55 mL of HCl is used to titrate 2.150 g of
Na2CO3 according to the following equation, what is
the Molarity of the HCl?
Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + CO2(g) + H2O(ℓ)
CHE 170 Packet 4 - 94
Some Problems for Practice!
26 g
1. If we react 1.0 L of 0.44 M HCl with 30 g NaOH
how many grams of NaCl can we make?
2. What is the Molarity of sulfuric acid if it takes 20.03
0.267 M mL of 0.100 M NaOH to neutralize 15.00 mL of the
acid?
3. If we react 3.25 g of zinc with excess 5.00 M
0.100 g
hydrochloric acid at STP, how much hydrogen gas
(1.11 L)
will be produced in grams (and liters)? Note, the
other product formed is ZnCl2.
4. What is the Molarity of a solution formed when 15.0
1.50 M
mL of 5.00 M HCl is diluted to 50.00 mL?
See notes page for these problems worked out.
CHE 170 Packet 4 - 95
The End of Packet 4
Any Questions?
CHE 170 Packet 4 - 96