Aqueous-solution Reactions
Classify a reaction by
Homogeneous chemical reactions:
gas phase
solutions
aqueous-solution (common occurrence)
non-aqueous-solution
Heterogeneous (more than one phase) chemical reactions:
gas / liquid
gas / solid
liquid / solid
Know the meaning of terms
aqueous solution reactions
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Nature of Aqueous Solutions
Nature of compounds
molecular substances (polar, non-polar, H-bonding)
non-electrolytes
ionic substances (acids, bases, salts)
strong electrolytes (completely ionized in solution)
week electrolytes (not completely ionized in solution)
Know your terms and species (in the solution)
aqueous solution reactions
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Dissolving a Strong Electrolyte
See them in your imaginationaqueous solution reactions
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aqueous solution reactions
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Strong Electrolytes
Strong acids: HNO3, H2SO4, HCl, HClO4
Strong bases: MOH (M = Na, K, Cs, Rb etc)
Salts: All salts dissolving in water are completely ionized. The ions
may react with water (to be discussed in Chem 123)
Stoichiometry & concentration relationship
NaCl (s)  Na+ (aq) + Cl– (aq)
Ca(OH)2 (s)  Ca+(aq) + 2 OH– (aq)
AlCl3 (s)  Al3+ (aq) + 3 Cl– (aq)
(NH4)2SO4 (s)  2 NH4 + (aq) + SO42– (aq)
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Concentration of Ions
A bottle labeled as 0.100 M Al2(SO4)3.
[Al3+] = _____ M (mol / L)
[SO42–] = _____ M
Assume sea water is 0.438 M NaCl, 0.0512 M MgCl2, and 0.001 M CaCl2
[Na+] = _____ M
[Mg2+] = _____ M
[Ca2+] = _____ M
[Cl–] = _____ M
Know how to calculate your quantities
aqueous solution reactions
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Precipitations
When ions form a solid that is not very soluble, a solid is formed. Such
a phenomenon is called precipitation.
The formation of a precipitation is also an equilibrium phenomenon (a
subject to be covered in Chem123)
Ag+ (aq) + Cl– (aq)  AgCl(s)
or
AgCl(s)  Ag+ (aq) + Cl– (aq)
Ksp = [Ag+][Cl–] is a constant
 the solubility product
aqueous solution reactions
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Precipitation Reactions
Heterogeneous Reactions
Spectator ions or bystander ions
Ag+ (aq) + NO3– (aq) + Cs+ (aq) + I– (aq)  AgI (s) + NO3– (aq) + Cs+ (aq)
Ag+ (aq) + I– (aq)  AgI (s) (net reaction)
or
Ag+ + I– AgI (s)
Mostly insoluble
Soluble ions
Alkali metals, NH4+
nitrates, ClO4-,
acetate
Mostly soluble ions
Halides, sulfates
aqueous solution reactions
Silver halides
Metal sulfides, hydroxides
carbonates, phosphates
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Acid-base Reactions
HCl (g)  H+ (aq) + Cl– (aq)
NaOH (s)  Na+ (aq) + OH– (aq)
neutralization reaction: H+ (aq) + OH– (aq)  H2O (l)
Explain these reactions
Mg(OH)2 (s) + 2 H+ Mg2+ (aq) + 2 H2O (l)
CaCO3 (s) + 2 H+ Ca2+ (aq) + H2O (l) + CO2 (g)
Mg(OH)2 (s) + 2 HC2H3O2  Mg2+ (aq) + 2 H2O (l) + 2 C2H3O2 – (aq)
acetic acid
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Oxidation-reduction reactions
Oxidation reaction must be accompanied by reductions
– redox reactions
Increasing oxidation state is oxidation (loss e–, LEO)
Decreasing oxidation state is reduction (gain e–, GER)
What elements are oxidized and reduced in each reaction?
Work out the oxidation state changes for them as well!
2 KClO3  2 KCl + 3 O2
Fe2O3 + 3 CO  2 Fe + 3 CO2
MnO2 + 4 H+ + 2 Cl–  Mn2+ + 2 H2O + Cl2
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Review Oxidation States
Oxidation states is an assigned number.
Formal charge concept may be used to assign oxidation states.
Work out the oxidation states of all elements in these species:
NH3
Cl–
N2H4
Cl2
NH2OH N2
N2O
ClO–
ClO2– ClO2
CO
H2C2O4 C2O42– C2H6 CH4
PH3
P4
H3PO4 PO43–
H2S
HS–
S2–
S6
SO2
NO
ClO2–
NO2– NO2 NO3–
ClO3– ClO4–
CO2
CO32–
SO3–
SO3
S2O32– SO42–
H2O2
MnO2 KMnO4 MnO4–
Review stoichiometry
K2CrO4 CrO4–
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K2Cr2O7 Cr2O7–
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Half Reactions
These reactions explained during the lecture:
Zn = Zn2+ + 2 e–
Cu2+ + 2 e– = Cu
net (electron transfer)
Zn (s) + Cu2+ (aq) = Zn2+ (aq) + Cu (s)
For these half reactions,
Zn = Zn2+ + 2 e–
2 H+ + 2 e– = H2
get and explain the net reaction yourself
Explain how reactions proceed
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Balance Half Equations
1. Identify the key element that undergoes an oxidation state change.
2. Balance the number of atoms of the key element on both sides.
3. Add the appropriate number of electrons to compensate for the
change of oxidation state.
4. Add H+ (in acid medium), or OH- (in basic medium), to balance the
charge on both sides of the half-reactions; and H2O, if necessary, to
balance the equations.
Page 9 of handout
Balance
MnO4 – + ____ + _____  Mn2+ + __ H2O
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Balance these half equations
Zn (s)  Zn2+(aq)
I will illustrate the balance of these
half reaction equations and those
equations in other slides during the
lecture. If you are not at my lecture,
you should practice their balance to
acquire your skills.
Cu2+ (aq)  Cu (s)
H2(g)  H+ (aq)
I –(aq)  I2
Fe2+ (aq)  Fe3+ (aq)
SO32– (aq) SO42– (aq)
2 S2O32–(aq)  S4O62–(aq) + 2e–1
Cr2O72–(aq)  Cr3+ (aq)
The textbook gives a slight different
method to balance redox equations
and please find out the difference.
Both ways wok. You may use either
method.
aqueous solution reactions
Work on these from time to time
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Balance Redox Reaction equations
Add two half reaction equations so that you can cancel all electrons
to obtain a balanced redox reaction equation
H2O2 + I –  I2 + H2O
MnO4 – + H2O2  Mn2+ + O2
MnO4 – + SO3 2– Mn2+ + SO42–
MnO4 – + Fe2+  Mn2+ + Fe3+
Cr2O7 2– + UO2+  Cr3+ + UO22+
Work on these from
time to time to
refresh your skills.
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Disproportionation Reactions
Balance disproportionation reaction (the same substance is both
oxidized and reduced)
H2O2  2 H2O + O2
S2O32– SO42– + S (s)
S2O32– SO2 + S (s)
aqueous solution reactions
Work on these from time to time
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Analyze and Learn the Skills
Analyze this example and learn the skills to help you overcome any difficulty.
Balance the equation: S2O32– SO42– + S (s)
Task:
Identify the element oxidized and reduced: oxidation states of S in S2O32– SO42– and S are
+4, +6, and 0 respectively. S is both oxidized and reduced.
The oxidation half reaction:
The reduction half reaction:
S2O32– SO42– + 8 e– (2 S from +2  +6, (2×4=8))
S2O32– + 4 e – (2S from +2  0, (2×2=4)) S
Balance the charge with H+
S2O32–SO42– + 8 e– + 10 H+ (both sides have 2-)
S2O32– + 4 e – + 6 H+ S
(both sides
have 0)
Add water to balance: S2O32– + 5 H2O SO42– + 8 e– + 10 H+
S2O32– + 4 e – + 6 H+ S + 3 H2O
Make # of e the same: S2O32– + 5 H2O SO42– + 8 e– + 10 H+
2 S2O32– + 8 e – + 12 H+ 4 S + 6 H2O
The balanced equation: 3 S2O32– + 2 H+ SO42– + 4 S + 10 H+
solution
reactions here!!!
Make sure you fill in the details too aqueous
much to
be included
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Balance Redox Reaction in
Basic Solutions
Redox reactions may have different products depends on the acidity
(pH) of the solutions.
In basic solutions, there are more OH– ions than H+ ions. Thus, it is
sensible to have OH– appearing in the equations than to have H+ ions.
Balance these reactions in a basic solution:
MnO4 – + CN –  MnO2 + OCN –
Any one of several ways to assign oxidation states for CN works.
MnO4 – + SO32–  MnO2 + SO42–
aqueous
Practice balance these from time
tosolution
time reactions
to polish you skills!
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Oxidizing and Reducing Agents
Oxidizing agent or oxidant such as O2 or F2 is a substance that is able
to oxidize other substances. It is reduced in the process (gains
electron or decreases oxidation state).
Describe reducing agent or reductant
OxIdant
NH3 N2H4
Species
Cannot be
reduced
further
NH2OH N2
N2O
NO
Reductant
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NO2 – NO2
NO3 –
Species
Cannot be
oxidized
further
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Titrations
Titration is a method used in volumetric analysis. The addition of a
solution is carefully controlled so that stoichiometric amounts can be
read from a burette.
Titration can be carried out for instantaneous or rapid reactions such
as neutralization and oxidation reactions.
Explain these terms:
neutralization,
titration,
quivalence point,
half equivalence point,
indicators
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Conductance in Titration
Conductance measurement of a HCl
solution titrated by NaOH is shown:
measured (total) conductance
area due to the ions labeled
conductance
Do all ions have the same
conductance?
Why or why not?
Why does the total conductance
vary?
Is conductance of an ion depends on
the concentration?
H+
OH–
Na+
Cl–
V of NaOH added
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Explain physical properties of chemicals
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Stoichiometry in Solution Chemistry
For titration calculations, the amount of reactant m is evaluated from
the concentration C and volume V by
m=C*V
For example, when m1 amount of acid is neutralized by m2 amount of
base, m1 = m1. For redox reactions, similar relationship can also be
used, but the stoichiometric relationship should be kept in mind.
Amount in mmol = C in M * V in mL
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Volumetric Analysis
A 5.00-mL sample of vinegar requires 38.08 mL of 0.1000 M NaOH
solution to reach the equivalence point. What is the concentration of
acetic acid in the vinegar by weight percent if its density is 1.01 g / mL?
Solution:
Net reaction:
38.08 mL OH–
5 mL vinegar
OH– + HC2H3O2  H2O + C2H3O2 – (aq)
0.1000 mol
1000 mL
= 0.0453 HC2H3O2 in vinegar
= 4.53 % HC2H3O2 in vinegar
1 mol HC2H3O2
1 mol OH–
If you find this solution difficult
to understand, use your own
method to solve it
60.05 g HC2H3O2
1 mol HC2H3O2
1 mL vinegar
1.01 g vinegar
The vinegar has 4.53 % of acetic acid by mass.
What’s the concentration of aqueous
a 7% solution
vinegar?
reactions
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Another way of thinking
A 5.00-mL sample of vinegar requires 38.08 mL of 0.1000 M NaOH
solution to reach the equivalence point. What is the concentration of
acetic acid in the vinegar by weight percent if its density is 1.01 g / mL?
Solution: a slight variation and hope you find it easier to follow
Net reaction:
38.08 mL OH–
OH– + HC2H3O2  H2O + C2H3O2 – (aq)
0.1000 mol
1000 mL
1 mol HC2H3O2
1 mol OH–
60.05 g HC2H3O2
1 mol HC2H3O2
1.01 g vinegar
1 mL vinegar
4.53 % ofHC2H3O2 in vinegar =
= 0.0453 HC2H3O2 in vinegar
5 mL vinegar
Mass of acetic acid
Mass of sample
= 4.53 % HC2H3O2 in vinegar
aqueous solution reactions
What the concentration of a 7% vinegar?
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Chemical Analysis Application
How much 0.1000 M KMnO4 solution is required to reach the equivalence
point in a titration of 1.00 g oxidize oxalic acid (oa = H2C2O4.2H2O)?
Solution:
2 MnO4–
Redox reaction:
+ 5 H2C2O4 + 6 H+ = 2 Mn2+ + 8 H2O + 10 CO2
1.00 g oa
1 mol oa
2 mol MnO4
1000 mL
126 g oa
5 mol oa
0.1000 mol
Molar mass of
H2C2O4.2H2O
= 126 g mol-1
Do problem 5.97
(balanced? Chk pls)
= 31.75 mL MnO4–
Note mole
ratio in the
balanced
equation.
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Back Titration
A 1.00-g sample containing MnO2 dissolved in solution is treated with 2.00 g
of oxalic acid (oa = H2C2O4.2H2O). Then 50.00 mL 0.1000 M KMnO4 is
required for the titration of the excess oa. What is the % MnO2 by mass?
Solution:
2 MnO4–
Redox reaction:
+ 5 H2C2O4 + 6 H+ = 2 Mn2+ + 8 H2O + 10 CO2
H2C2O4 + MnO2 + 2 H+ = Mn2+ + 2 H2O + 2 CO2
50.00 mL MnO4
(2.000 - 1.575) g oa
(balanced? Chk pls)
0.1000 mol MnO4
5 mol oa
126 g oa
1000 mL
2 mol MnO4
1 mol oa
86.9 g MnO2
100 %
126 g oa
1.0 g Sample
Cool head 4 complicated problem, eh! aqueous solution reactions
= 1.575 g oa
= 29.0 % MnO2 by mass
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Review 1
Sufficient amount of AgNO3 is placed in a 10.00 mL of tab water, and
the AgCl solid is filtered and dried. The solid weighs 0.123 g. What is
the concentration of chloride ion?
Solution:
access and limiting reagent, and ppt
0.123 g AgCl
10.00 mL
1 mol AgCl
143.4 g AgCl
1 mol Cl1 mol AgCl
1000 mL
1L
= 0.0858 mol / L of Cl–
aqueous solution reactions
Express [Cl-] in mol and mass %.
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Review 2
A 0.1234-g sample (S) NaCl and sugar mixture contains 40% NaCl.
When dissolved in water, the solution is treated with AgNO3. How
much dry AgCl is the theoretical yield?
Solution:
percentage analysis
0.1234 g S
40 g NaCl
100 g S
= 0.1323 g AgCl
1 mol AgCl
58.5 g NaCl
143.4 g AgCl
1 mol AgCl
 theoretical yield
aqueous solution reactions
Use the factors to help you think!
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Review 3
A 0.2345-g sample (S) NaCl and CaCl2 mixture contains 40% NaCl.
When dissolved in water, the solution is treated with AgNO3. How
much dry AgCl is the theoretical yield?
Solution:
0.2345*0.40 g NaCl
0.2345*0.60 g CaCl2
143.4 g AgCl
58.5 g NaCl
= 0.2299 g AgCl
2*143.4 g AgCl
110 g CaCl2
= 0.3668 g AgCl
(0.2299 + 0.3668) g AgCl = 0.5967 g AgCl
Solve the same for a mixture of NaCl and KCl.
aqueous solution reactions
Find the percentage of NaCl if 0.5000 g AgCl is collected.
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Review 4
When 0.2345-g NaCl and CaCl2 mixture is dissolved in water, the
solution is treated with AgNO3. The mass of dry AgCl collected is
0.5967 g. What is the percentage of NaCl in the mixture?
Solution: Assume the sample contains x fraction of NaCl
0.2345 (x) g NaCl 143.4 g AgCl
58.5 g NaCl
+ 0.2345 (1 – x) g CaCl2
2*143.4 g AgCl
110 g CaCl2
= 0.5967 g AgCl
Solve for x in the equation: x = 0.40 = 40%
Solve the same problem for 0.6000 instead of 0.5967 g AgCl.
aqueous solution reactions
In this problem, what are the min. and max. masses of AgCl? (0.5748 – 0.6114 g)
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Review 5 - skills
obtain empirical formula from composition and combustion experiment
find molecular formula from empirical formula and molar mass
evaluate theoretical yield in a reaction, identify the limiting reagent
Identify limiting reagent and evaluation percent yield in a reaction
predict amounts of products or reagents required in reactions
find percent yield of a reaction but pay attention to limiting reagent
give concentration of common ions in a solution of several salts
calculate masses of reagents or products (AgCl) involving common ions
find percentage of a compound in a mixture by titration or precipitation
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Review 6 – skill 2
Identify oxidation states and recognize reagent oxidized
balance redox reaction equations
evaluate quantities of reagents or products involving redox reactions
stoichiometry calculation in titration experiments paying attention to molar
relationship such as H3PO4, Ca(OH)2, as well as redox reactions
evaluate quantities involving isotope composition of elements
calculate one of density, mass and volume in solid, liquid, and gas
name inorganic compounds
write proper net ionic equation of reaction (know your spectator ions)
aqueous solution reactions
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TA Hours & Rooms – fall 03
The Tutors handle the tutorial periods for CHEM 120/121 but they can
also provide assistance to individuals or small groups on a drop-in
basis according to the following schedule. (i.e. One of the Tutors will be
in each room for the specified period. You can drop in at any time
during that period to get some help.)
Day
Monday
Monday
Monday
Wednesday
Wednesday
Time
12:30-1:20
1:30-2:20
2:30-3:20
12:30-3:20
1:30-2:20
Room*
PHY 313
MC 4062
PHY 150
PHY 313
MC 4060
* These rooms are for one-on-one tutorial, not tutors office.
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