Types of Chemical Reactions and Solution Stoichiometry

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Chapter 4

• Types of Chemical Reactions and Solution
  Stoichiometry

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Chapter 4 Types of Chemical Reactions and
Solution Stoichiometry
4.1 Water, the Common Solvent 4.2 The Nature of
Aqueous Solutions Strong and Weak
Electrolytes 4.3 The Composition of Solutions 4.4
Types of Chemical Reactions 4.5 Precipitation
Reactions 4.6 Describing Reactions in
Solution 4.7 Selective Precipitation 4.8
Stoichiometry of Precipitation Reactions 4.9
Acid-Base Reactions 4.10 Oxidation-Reduction
Reactions 4.11 Balancing Oxidation-Reduction
Equations 4.12 Simple Oxidation-Reduction
Titrations
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Precipitation of silver chromate by adding
potassium chromate to a solution of silver
nitrate. K2CrO4 (aq) 2 AgNO3 (aq) Ag2CrO4
(s) 2 KNO3 (aq)
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Figure 4.1 A space-filling model of the water
molecule.
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Figure 4.2 Polar water molecules interact with
the positive and negative ions of a salt,
assisting with the dissolving process.
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Figure 4.3(a) The ethanol molecule contains a
polar O-H bond similar to those in the water
molecule. (b) The polar water molecule interacts
strongly with the polar O-H bond in ethanol.
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The Role of Water as a Solvent The solubility
of Ionic Compounds
Electrical conductivity - The flow of electrical
current in a solution is a
measure of the solubility of ionic
compounds or a
measurement of the presence of ions in
solution. Electrolyte - A substance that
conducts a current when dissolved in
water. Soluble ionic compounds
dissociate completely and
may conduct a large current, and are called
Strong Electrolytes.
NaCl(s) H2O(l)
Na(aq) Cl -(aq)
When Sodium Chloride dissolves into water the
ions become solvated, and are surrounded by water
molecules. These ions are called aqueous and
are free to move throughout the solution, and are
conducting electricity, or helping electrons to
move throughout the solution.
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Electrical Conductivity of Ionic Solutions
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Figure 4.4 Electrical Conductivity
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Figure 4.5 HCL (aq) is completely ionized.
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Figure 4.6 An aqueous solution of sodium
hydroxide.
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Figure 4.7 Acetic acid (HC2H3O2) exists in
water mostly as undissociated molecules.
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Figure 4.8 The reaction of NH3 in water.
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Carbohydrates
Molecules that contain carbon and water!
CxH2yOy
H
CH2OH
O
C
CH2OH
H
C
O
H
H
C
HO
H
OH
C
C
C
OH
H
C
OH
O
CH2OH
C
C
H
H
OH
Sucrose C12H22O11 , C12(H2O)11 a
disaccharide
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Molarity (Concentration of Solutions) M
Moles of Solute Moles Liters of
Solution L
M
solute material dissolved into the solvent In
air , Nitrogen is the solvent and oxygen, carbon
dioxide, etc.

are the solutes. In sea water , Water is the
solvent, and salt, magnesium chloride, etc.

are the solutes. In
brass , Copper is the solvent (90), and Zinc is
the solute(10)
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LIKE EXAMPLE 4.1 (P 93)
Calculate the Molarity of a solution prepared by
bubbling 3.68g of Gaseous ammonia into 75.7 ml
of solution.
Solution
Calculate the number of moles of ammonia
1 mol NH3 17.03g
3.68g NH3
X 0.216 mol NH3
Change the volume of the solution into liters
1 L 1000 mL
75.7 ml X
0.0757 L
Finally, we divide the number of moles of solute
by the volume of the solution
0.216 mol NH3 0.0757 L
Molarity
____________ M NH3
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Preparing a Solution - I

• Prepare a solution of Sodium Phosphate by
  dissolving 3.95g of Sodium Phosphate into water
  and diluting it to 300.0 ml or 0.300 l !
• What is the Molarity of the salt and each of the
  ions?
• Na3PO4 (s) H2O(solvent) 3 Na(aq)
  PO4-3(aq)

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Preparing a Solution - II

• Mol wt of Na3PO4 163.94 g / mol
• 3.95 g / 163.94 g/mol 0.0241 mol Na3PO4
• dissolve and dilute to 300.0 ml
• M 0.0241 mol Na3PO4 / 0.300 l 0.0803 M
• 
  Na3PO4
• for PO4-3 ions ______________ M
• for Na ions 3 x 0.0803 M ___________ M

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Like Example 4.3 (P 95)
An isotonic solution, one with the same ionic
content as blood is about 0.14 M NaCl. Calculate
the volume of blood that would contain 2.5 mg Of
NaCl?
Find the moles in 1.0 mg NaCl
1 g NaCl 1000 mg NaCl
1 mol NaCl 58.45g NaCl
2.5 mg NaCl x x
4.28 x 10-5 mol

NaCl
What volume of 0.14 M NaCl that would contain the
amount of NaCl (4.28 x 10-5 mol NaCl)
0.14 M NaCl L solution
V x 4.28 x 10-5 mol
NaCl
Solving for Volume gives
4.28 x 10-5 mol NaCl 0.14 mol NaCl L
solution
V
______________________ L
Or _________ ml of Blood!
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Figure 4.9 Steps involved in the preparation of
a standard solution.
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Like Example 4.4 (P 97)
A Chemist must prepare 1.00 L of a 0.375 M
solution of Ammonium Carbonate, what mass of
(NH4)2CO3 must be weighed out to prepare this
solution?
First, determine the moles of Ammonium Carbonate
required
0.375 M (NH4)2CO3 L solution
1.00 L x
0.375 M (NH4)2CO3
This amount can be converted to grams by using
the molar mass
94.07 g (NH4)2CO3 mol (NH4)2CO3
0.375 M (NH4)2CO3 x
35.276 g (NH4)2CO3
Or, to make 1.00L of solution, one must weigh out
35.3 g of (NH4)2CO3, put this into a 1.00 L
volumetric flask, and add water to the mark on
the flask.
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Make a Solution of Potassium Permanganate
Potassium Permanganate is KMnO4 and has a
molecular mass
of 158.04
g / mole
Problem Prepare a solution by dissolving 1.58
grams of KMnO4 into sufficient
water to make 250.00 ml of solution.
1 mole KMnO4 158.04 g KMnO4
1.58 g KMnO4 x
0.0100 moles KMnO4
0.0100 moles KMnO4 0.250 liters
Molarity
______________ M
Molarity of K ion K ion MnO4- ion
_____________ M
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Figure 4.10 (a) A measuring pipette (b) A
volumetric pipette.
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Figure 4.11 (a) A measuring pipette (b) Water
is added to the flask. (c) The resulting solution
is 1 M acetic acid.
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Dilution of Solutions

• Take 25.00 ml of the 0.0400 M KMnO4
• Dilute the 25.00 ml to 1.000 l - What is the
  resulting Molarity of the diluted solution?
• moles Vol x M
• 0.0250 l x 0.0400 M 0.00100 Moles
• 0.00100 Mol / 1.00 l _______________ M

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Figure 4.13 Reactant solutions (a) Ba(NO3)3(aq)
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Figure 4.13 Reactant solutions (b) K2CrO4(aq).
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Figure 4.12 When yellow aqueous potassium
chromate is added to a colorless barium nitrate
solution, yellow barium chromate precipitates.
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Figure 4.14 Reaction of K2CrO4 (aq) and
Ba(NO3)2 (aq).
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Figure 4.15 Precipitation of silver chloride by
mixing solutions of silver nitrate and potassium
chloride.
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Figure 4.16 Photos and molecular-level
representations illustrating the reaction of
KCL(aq) with AgNO3(aq) to form AgCl(s).
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Table 4.1 (P102) Simple Rules for Solubility
of Salts in
Water

• Most nitrate (NO3-) salts are soluble.
• Most salts of Na, K, and NH4 are soluble.
• Most chloride salts are soluble. Notable
  exceptions are AgCl,
• PbCl2, and Hg2Cl2.
• Most sulfate salts are soluble. Notable
  exceptions are BaSO4,
• PbSO4, and CaSO4.
• Most hydroxide salts are only slightly soluble.
  The important
• soluble hydroxides are NaOH, KOH, and
  Ca(OH)2
• (marginally soluble).
• Most sulfide (S2-), carbonate (CO32-), and
  phosphate (PO43-)
• salts are only slightly soluble.

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The Solubility of Ionic Compounds in Water
The solubility of Ionic Compounds in water
depends upon the relative strengths of the
electrostatic forces between ions in the ionic
compound and the attractive forces between the
ions and water molecules in the solvent. There
is a tremendous range in the solubility of ionic
compounds in water! The solubility of so called
insoluble compounds may be several orders of
magnitude less than ones that are
called soluble in water, for example
Solubility of NaCl in water at 20oC 365
g/L Solubility of MgCl2 in water at 20oC 542.5
g/L Solubility of AlCl3 in water at 20oC 699
g/L Solubility of PbCl2 in water at 20oC 9.9
g/L Solubility of AgCl in water at 20oC 0.009
g/L Solubility of CuCl in water at 20oC 0.0062
g/L
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The Solubility of Covalent Compounds in Water
The covalent compounds that are very soluble in
water are the ones with -OH group in them and
are called Polar and can have strong polar
(electrostatic)interactions with water. Examples
are compound such as table sugar, sucrose
(C12H22O11) beverage alcohol, ethanol
(C2H5-OH) and ethylene glycol (C2H6O2) in
antifreeze.
H
Methanol Methyl Alcohol
C
H
O H
H
Other covalent compounds that do not contain a
polar center, or the -OH group are considered
Non-Polar , and have little or no interactions
with water molecules. Examples are the
hydrocarbons in Gasoline and Oil. This leads to
the obvious problems in Oil spills, where the oil
will not mix with the water and forms a layer on
the surface!
Octane C8H18 and / or Benzene C6H6
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When a solution of Na2SO4 (aq) is added to a
solution of Pb(NO3)2, the white solid PbSO4(s)
forms.
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Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - I
Problem How many moles of each ion are in each
of the following a) 4.0 moles of sodium
carbonate dissolved in water b) 46.5 g of
rubidium fluoride dissolved in water c) 5.14 x
1021 formula units of iron (III) chloride
dissolved in water d) 75.0 ml of 0.56M
scandium bromide dissolved in water e) 7.8
moles of ammonium sulfate dissolved in
water a) Na2CO3 (s)
2 Na(aq) CO3-2(aq)
moles of Na 4.0 moles Na2CO3 x
8.0 moles Na and 4.0 moles of
CO3-2 are present
H2O
2 mol Na 1 mol Na2CO3
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Determining Moles of Ions in Aqueous Solutions
of Ionic Compounds - II
H2O
b) RbF(s)
Rb(aq) F -(aq)
1 mol RbF 104.47 g RbF
moles of RbF 46.5 g RbF x
0.445 moles RbF
thus, 0.445 mol Rb and 0.445 mol F - are present
H2O
c) FeCl3 (s)
Fe3(aq) 3 Cl -(aq)
moles of FeCl3 9.32 x 1021 formula units
1 mol FeCl3 6.022 x 1023 formula
units FeCl3
x
0.0155 mol FeCl3
3 mol Cl - 1 mol FeCl3
moles of Cl - 0.0155 mol FeCl3 x
_________ mol Cl -
and ____________ mol Fe3 are also present.
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Determining Moles of Ions in Aqueous Solutions
of Ionic Compounds - III
H2O
d) ScBr3 (s)
Sc3(aq) 3 Br -(aq)
Converting from volume to moles
1 L 103 ml
0.56 mol ScBr3 1 L
Moles of ScBr3 75.0 ml x x
0.042 mol ScBr3
3 mol Br - 1 mol ScBr3
Moles of Br - 0.042 mol ScBr3 x
0.126 mol Br -
0.042 mol Sc3 are also present
H2O
e) (NH4)2SO4 (s)
2 NH4(aq) SO4- 2(aq)
2 mol NH4 1 mol(NH4)2SO4
Moles of NH4 7.8 moles (NH4)2SO4 x
____ mol NH4
and ______ mol SO4- 2 are also present.
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Solid Fe(OH)3 forms when aqueous KOH and Fe(NO3)3
are mixed.
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Precipitation Reactions Will a Precipitate form?
If we add a solution containing Potassium
Chloride to a solution containing Ammonium
Nitrate, will we get a precipitate?
KCl(aq) NH4NO3 (aq) K(aq)
Cl-(aq) NH4(aq) NO3-(aq)
By exchanging cations and anions we see that we
could have Potassium Chloride and Ammonium
Nitrate, or Potassium Nitrate and Ammonium
Chloride. In looking at the solubility table it
shows all possible products as soluble, so there
is no net reaction!
KCl(aq) NH4NO3 (aq) No Reaction!
If we mix a solution of Sodium sulfate with a
solution of Barium Nitrate, will we get a
precipitate? From the solubility table it shows
that Barium Sulfate is insoluble, therefore we
will get a precipitate!
Na2SO4 (aq) Ba(NO3)2 (aq)
BaSO4 (s) 2 NaNO3 (aq)
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Precipitation Reactions A solid product is formed
When ever two aqueous solutions are mixed, there
is the possibility of forming an insoluble
compound. Let us look at some examples to see
how we can predict the result of adding two
different solutions together.
Pb(NO3)2 (aq) NaI(aq) Pb2(aq) 2
NO3-(aq) Na(aq) I-(aq)
When we add These two solutions together, the
ions can combine in the way they came into the
solution, or they can exchange partners. In this
case we could have Lead Nitrate and Sodium
Iodide, or Lead Iodide and Sodium Nitrate formed,
to determine which will happen we must look
at the solubility table(P 141) to determine what
could form. The table indicates that Lead Iodide
will be insoluble, so a precipitate will form!
Pb(NO3)2 (aq) 2 NaI(aq)
PbI2 (s) 2 NaNO3 (aq)
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Predicting Whether a Precipitation Reaction
Occurs Writing Equations
a) Calcium Nitrate and Sodium Sulfate solutions
are added together.
Molecular Equation
Ca(NO3)2 (aq) Na2SO4 (aq)
CaSO4 (s) 2 NaNO3 (aq)
Total Ionic Equation
Ca2(aq)2 NO3-(aq) 2 Na(aq) SO4-2(aq)
CaSO4 (s) 2 Na(aq) 2 NO3-(aq)
Net Ionic Equation
Ca2(aq) SO4-2(aq)
CaSO4 (s)
Spectator Ions are Na and NO3-
b) Ammonium Sulfate and Magnesium Chloride are
added together.
In exchanging ions, no precipitates will be
formed, so there will be no Chemical reactions
occurring! All ions are spectator ions!
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Figure 4.17 Selective precipitation of Ag,
Ba2, and Fe3 ions.
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Species present, Balanced net ionic equation.
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Like Example 4.7 (P 108)
Calculate the mass of solid sodium iodide that
must be added to 2.50 L of a 0.125 M lead nitrate
solution to precipitate all of the lead as PbI2
(s)!
The chemical equation for the reaction is
Pb(NO3)2 (aq) 2 NaI(aq) PbI2
(s) 2 NaNO3 (aq)
Two times as much sodium iodide is needed to
precipitate the Lead ions. The number of moles
of sodium iodide needed is
0.125 Mol Pb2 1.0 L soln.
2 mol I- 1 mol Pb2
2.50 L x x
0.625 mol I-
The mass of sodium iodide is
149.9 g NaI 1 mol NaI
1 mol NaI 1 mol I-
0.625 mol I- x x
__________ g NaI
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Like Example 4.8 (P 108)
When aqueous solutions of silver nitrate and
sodium chloride are mixed, silver chloride is
precipitated. What mass of silver chloride would
be formed by the addition of 75.00 ml to 3.17 M
NaCl and 128 ml of 2.44 M silver nitrate?
The stoichiometric relationship comes from the
chemical equation
AgNO3 (aq) NaCl(aq) AgCl(s)
NaNO3 (aq)
There is a one to one relationship, therefore the
number of moles are the same, but which is in the
lowest quantity?
VAgNO3 x MAgNO3 0.128 L x 2.44 M 0.312 mol Ag
VNaCl x MNaCl 0.07500 L x 3.17 M 0.238 mol Cl-
Since the Chloride ion is smaller, it is
limiting, and we use it to calculate the mass of
AgCl, since we can only obtain 0.238 mol of AgCl
Mass AgCl 0.238 mol x 143.35 g AgCl/ mol
_________ g
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Like Example 4.9 (P 109)
What mass of Pb2 could by precipitated from a
solution by the addition of 0.785 L of 0.0015 M
Sodium Iodide solution?
Find the stoichiometric relationship from the
chemical equation
Pb2(aq) 2 I-(aq) PbI2 (s)
It will take twice the iodide ion to precipitate
the Lead ions
Moles I - VNaI x MNaI 0.785 L x 0.0015 Moles
0.00118 mol I-
Liter
0.00118 mol I- 2 mol I-/ mol Pb2
Moles Lead ion
0.000590 mol Pb2
Mass of Lead 207.2g Pb x 0.000590 moles
____________ g Pb
mol Pb
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Acids - A group of Covalent molecules which lose
Hydrogen ions to water molecules in
solution
When gaseous hydrogen Iodide dissolves in water,
the attraction of the oxygen atom of the water
molecule for the hydrogen atom in HI is greater
that the attraction of the of the Iodide ion for
the hydrogen atom, and it is lost to the water
molecule to form an Hydronium ion and an Iodide
ion in solution. We can write the Hydrogen atom
in solution as either H(aq) or as H3O(aq) they
mean the same thing in solution. The presence of
a Hydrogen atom that is easily lost in solution
is an Acid and is called an acidic solution.
The water (H2O) could also be written above the
arrow indicating that the solvent was water in
which the HI was dissolved.
HI(g) H2O(L) H(aq) I
-(aq)
HI(g) H2O(L) H3O(aq) I
-(aq)
H2O
HI(g) H(aq) I
-(aq)
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Strong Acids and the Molarity of H Ions in
Aqueous Solutions of Acids
Problem In aqueous solutions, each molecule of
sulfuric acid will lose two protons to yield two
Hydronium ions, and one sulfate ion. What is the
molarity of the sulfate and Hydronium ions in a
solution prepared by dissolving 155g of
concentrate sulfuric acid into sufficient water
to produce 2.30 Liters of acid solution? Plan
Determine the number of moles of sulfuric acid,
divide the moles by the volume to get the
molarity of the acid and the sulfate ion. The
hydronium ions concentration will be twice the
acid molarity. Solution Two moles of H are
released for every mole of acid
H2SO4 (l) 2 H2O(l)
2 H3O(aq) SO4- 2(aq)
1 mole H2SO4 98.09 g H2SO4
Moles H2SO4
1.58 moles H2SO4
155 g H2SO4
x
1.58 mol SO4-2 2.30 l solution
Molarity of SO4- 2
0.687 Molar in SO4- 2
Molarity of H 2 x 0.687 mol H 1.37 Molar
in H
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The gravimetric procedure.(P109-110)
.
1 mol CaC2O4 H2O 146.12g CaC2O4 H2O
.
0.2920 g CaC2O4 H2O X
1.998 x 10-3 mol

CaC2O4 H2O 1.998
x 10-3 mol Ca2 X
8.009 x 10-2g Ca2 Mass Ca is
x 100 ______________
.
.
40.08 g Ca2 1 mol Ca2
8.009 x 10-2 0.4367 g
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Species present, Balanced net ionic equation.
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Acid - Base Reactions Neutralization Rxns.
An Acid is a substance that produces H (H3O)
ions when dissolved in water, and is a proton
donor. A Base is a substance that produces OH -
ions when dissolved in water. the OH- ions react
with the H ions to produce water, H2O, and are
therefore proton acceptors. Acids and Bases are
electrolytes, and their strength is categorized
in terms of their degree of dissociation in
water to make hydronium or hydroxide ions. Strong
acids and bases dissociate completely, and are
strong electrolytes. Weak acids and bases
dissociate weakly and are weak electrolytes.
The generalized reaction between an Acid and a
Base is
HX(aq) MOH(aq)
MX(aq) H2O(L)
Acid Base
Salt Water
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Selected Acids and Bases
Acids
Bases
Strong
Strong Hydrochloric, HCl
Sodium hydroxide, NaOH Hydrobromic,
HBr Potassium
hydroxide, KOH Hydroiodoic, HI
Calcium hydroxide, Ca(OH)2
Nitric acid, HNO3
Strontium hydroxide, Sr(OH)2 Sulfuric acid,
H2SO4 Barium
hydroxide, Ba(OH)2 Perchloric acid,
HClO4 Weak
Weak Hydrofluoric, HF
Ammonia, NH3 Phosphoric acid,
H3PO4 Acetic acid, CH3COOH (or
HC2H3O2)
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Writing Balanced Equations for
Neutralization Reactions - I
Problem Write balanced chemical reactions
(molecular, total ionic, and
net ionic) for the following Chemical reactions
a) Calcium Hydroxide(aq) and Hydroiodoic
acid(aq) b) Lithium Hydroxide(aq) and
Nitric acid(aq) c) Barium Hydroxide(aq)
and Sulfuric acid(aq) Plan These are all strong
acids and bases, therefore they will make
water and the corresponding salts. Solution
a) Ca(OH)2 (aq) 2HI(aq)
CaI2 (aq) 2H2O(l) Ca2(aq)
2 OH -(aq) 2 H(aq) 2 I -(aq)


Ca2(aq) 2 I -(aq) 2 H2O(l)
2 OH -(aq) 2 H(aq)
2 H2O(l)
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Writing Balanced Equations for Neutralization
Reactions - II
b) LiOH(aq) HNO3 (aq)
LiNO3 (aq) H2O(l) Li(aq) OH
-(aq) H(aq) NO3-(aq)

Li(aq) NO3-(aq) H2O(l)
OH -(aq) H(aq)
H2O(l)
c) Ba(OH)2 (aq) H2SO4 (aq)
BaSO4 (s) 2 H2O(l) Ba2(aq) 2
OH -(aq) 2 H(aq) SO42-(aq) BaSO4
(s) 2 H2O(l)

Ba2(aq) 2 OH -(aq) 2 H(aq) SO42-(aq)
BaSO4 (s) 2 H2O(l)
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Figure 4.18 The titration of an acid with a
base.
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Like Example 4.10 (P 113)
What volume of 0.468 M H2SO4 is needed to
neutralize 215.00 ml of a 0.125 M LiOH solution?
Calculate the number of moles of base
Vbase x Mbase 0.21500 L x 0.125 M 0.0269
mol LiOH
From the balanced equation find the moles of acid
needed
2 LiOH(aq) H2SO4 (aq) 2 H2O(l)
Li2SO4 (aq)
Since there are two protons per molecule, we will
need half as much sulfuric acid as we have
lithium hydroxide or 0.0134 mol H2SO4
Volume of acid
Moles acid Macid
0.0134 moles 0.468 Mol L
Vacid
0.0286 L H2SO4
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A firefighter in a protective suit with an oxygen
tank neutralizes an acid spill.
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Finding the Concentration of Base from an
Acid - Base Titration - I
Problem A titration is performed between Sodium
Hydroxide and Potassium Hydrogenphthalate (KHP)
to standardize the base solution, by placing
50.00 mg of solid Potassium Hydrogenphthalate in
a flask with a few drops of an indicator. A
buret is filled with the base, and the initial
buret reading is 0.55 ml at the end of the
titration the buret reading is 33.87 ml. What is
the concentration of the base? Plan Use the
molar mass of KHP (204.2 g/mol) to calculate the
number of moles of the acid, from the balanced
chemical equation, the reaction is equal molar,
so we know the moles of base, and from the
difference in the buret readings, we can
calculate the molarity of the base. Solution
HKC8H4O4 (aq) OH -(aq)
KC8H4O4-(aq) H2O(aq)
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Potassium Hydrogenphthalate KHC8H4O4
O
C
K
O
O
C
H
O
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Finding the Concentration of Base from an
Acid - Base Titration - II
50.00 mg KHP 204.2 g KHP 1 mol KHP
1.00 g 1000 mg
moles KHP x
0.00024486 mol KHP
Volume of base Final buret reading - Initial
buret reading 33.87
ml - 0.55 ml 33.32 ml of base
one mole of acid one mole of base therefore
0.00024486 moles of acid will yield 0.00024486
moles of base in a volume of 33.32 ml.
0.00024486 moles 0.03332 L
molarity of base
__________ moles per liter
molarity of base ___________________ M
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Like Example 4.12 (P114-115)
A powered residue contains some ascorbic
acid(Vitamin C, mol wt 176g/mol) and the rest
is a non acidic compound. If 10.0g of the powder
is neutralized by 20.00 ml of 1.5 M sodium
hydroxide, a strong base, and the remaining base
titrated with hydrochloric acid using 10.50 ml
of 1.80 M. What is the percentage of ascorbic
acid?
Mol acid 0.01050 L x
1.80 Mol/L 0.0189 mol HCl
Mol base 0.0200 L x 1.50 Mol/L 0.0300 mol
NaOH The difference between the base and acid
will be the moles of ascorbic acid!
Reacted base 0.0300 0.0189 0.0111 mol
Ascorbic acid Mass ascorbic acid 0.0111 mol x
176g/mol 1.95 g Ascorbic acid
1.95g 10.00g
ascorbic acid x 100
____________
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Figure 4.19 Reaction of solid sodium and
gaseous chlorine to form solid sodium chloride.
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Oxidation of copper metal by nitric acid.
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Highest and Lowest oxidation numbers of
Chemically reactive main-group Elements
1A
2A
3A
4A
5A
6A
7A
Period
4
4
5
6
7
1
2
3
-3
-2
-1
F
O
N
C
B
2
Li
Be
Cl
3
S
P
Si
Al
Na
Mg
Br
4
Se
As
Ge
Ga
K
Ca
non-metals
I
5
Te
Sb
Sn
In
Rb
Sr
metalloids
At
6
Po
Bi
Pb
Tl
Cs
Ba
metals
7
Ra
Fr
84
Period
IA
VIIIA
Main Group Elements
He
H
1
IIA
IIIA
IVA
VA
VIA
VIIA
1 -1
Ne
F
O
N
C
B
Be
Li
2
4,2 -1,-4
all from
1
2
3
-1
-1,-2
5 -3
Ar
Na
Mg
Al
Si
P
S
Cl
-1
3
7,5 3,1
6,4 2,-2
5,3 -3
1
2
3
4,-4
Kr
K
Ca
Ga
Ge
As
Se
Br
-1
4
4,2 -4
7,5 3,1
2
6,4 -2
5,3 -3
1
2
3, 2
Xe
Rb
Sr
In
Sn
Sb
Te
I
-1
5
3,2 1
4,2, -4
7,5 3,1
6,4 2
6,4 -2
5,3 -3
1
2
Rn
Cs
Ba
Tl
Pb
Bi
Po
At
-1
6
7,5 3,1
6,4 2,-2
1
4,2
2
2
3,1
3
85
Transition Metals
Possible Oxidation States
VIIIB
IIIB
IVB
VB
VIB
VIIB
IB
IIB
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
2
4,3 2
7,6 4,3
6,3 2
5,4 32
3
2
2,1
2
3,2
3,2
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
5,4 2
8,5 4,3
7,5 4
6,5 4,3
3
4,3
2
1
4,2
4,3
Hg
Au
Pt
Ir
Os
Re
W
Ta
Hf
La
2
5,4 3
6,5 4
7,5 4
8,6 4,3
4,3 1
3
4,3
2,1
3,1
4,2
86
Determining the Oxidation Number of an
Element in a Compound
Problem Determine the oxidation number (Ox. No.)
of each element in the following
compounds. a) Iron III Chloride b)
Nitrogen Dioxide c) Sulfuric acid Plan We
apply the rules in Table 4.3, always making sure
that the Ox. No. values in a compound
add up to zero, and in a polyatomic
ion, to the ions charge. Solution a) FeCl3
This compound is composed of monoatomic ions. The
Ox. No. of Cl- is -1, for a total of -3.
Therefore the Fe is 3. b) NO2 The Ox. No. of
oxygen is -2 for a total of -4. Since the
Ox. No. in a compound must add up to zero, the
Ox. No. of N is 4. c) H2SO4 The Ox. No. of H
is 1, so the SO42- group must sum to -2.
The Ox. No. of each O is -2 for a total of -8. So
the Sulfur atom is 6.
87
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88
Method to remove chlorinated organic molecules
from Ground water
Fe(s) RCl(aq) H(aq) Fe2(aq)
RH(aq) Cl-(aq)
89
Figure 4.20 A summary of an oxidation-reduction
process, in which M is oxidized and X is reduced.
90
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91
Recognizing Oxidizing and Reducing Agents - I
Problem Identify the oxidizing and reducing
agent in each of the Rx a) Zn(s) 2 HCl(aq)
ZnCl2 (aq) H2 (g) b) S8 (s)
12 O2 (g) 8 SO3 (g) c) NiO(s)
CO(g) Ni(s) CO2 (g) Plan First
we assign an oxidation number (O.N.) to each atom
(or ion) based on the rules in Table 4.3. The
reactant is the reducing agent if it contains an
atom that is oxidized (O.N. increased in the
reaction). The reactant is the oxidizing agent if
it contains an atom that is reduced ( O.N.
decreased). Solution a) Assigning oxidation
numbers
-1
-1
1
0
0
2
Zn(s) 2 HCl(aq) ZnCl2
(aq) H2 (g)
HCl is the oxidizing agent, and Zn is the
reducing agent!
92
Recognizing Oxidizing and Reducing Agents - II
b) Assigning oxidation numbers
S 0 S6
-2
S is Oxidized
6
0
0
O0 O-2
S8 (s) 12 O2 (g) 8
SO3 (g)
O is Reduced
S8 is the reducing agent and O2 is the oxidizing
agent
c) Assigning oxidation numbers
C2 C4 C is oxidized
Ni2 Ni0 Ni is Reduced
-2
-2
-2
0
2
4
2
NiO(s) CO(g)
Ni(s) CO2 (g)
CO is the reducing agent and NiO is the oxidizing
agent
93
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94
Activity Series of the Metals
Strongly reducing Weakly reducing
Li Li e- K K e-
These elements react
rapidly with aqueous H ions Ba Ba2
2 e- (acid) or
with liquid H2O to release H2 gas. Ca
Ca2 2 e- Na Na e- Mg
Mg2 2e- Al Al3 3e-
These elements react with
aqueous H ions or with Mn Mn2 2e-
steam to release H2
gas. Zn Zn2 2e- Cr
Cr3 3e- Fe Fe2 2e- Co
Co2 2e- Ni Ni2 2e-
These elements react with
aqueous H ions to Sn Sn2 2e-
release H2 gas. H2
2 H 2e- Cu Cu2
2e- Ag Ag e-
These elements do not react with aqueous
H ions Hg Hg2 2e-
to release H2 gas. Pt
Pt2 2e- Au Au3 3e-
95
Examples of Activity Series Problems
Cu(s) 2 Ag(aq) Cu2(aq) 2
Ag(s) 2 Fe(s) 3 Cu2(aq) 2
Fe3(aq) 3 Cu(s) Mg(s) Zn2(aq)
Mg2(aq) Zn(s) 2 K(s) Sn2(aq)
2 K(aq) Sn(s) Pt(s) Ni2(aq)
N. R. 2 Al(s) 6 H(aq)
2 Al3(aq) 3 H2 (g) Au(s)
H(aq) N. R.
96
Problem Calculate the mass of metallic Iron that
must be added to 500.0 liters
of a solution containing
0.00040M of Pt2(aq) ions in solution to reclaim
all of the Platinum.
Solution V x M moles
500.0L x 0.00040 Mol/L 0.20 Mol
Pt2 assume that the Iron goes to Fe3
therefore we will need only 2 moles of
Iron for every 3 moles of Platinum 0.20
mol Pt2 x 0.133 mol Fe
0.133 mol Fe x
____________ g Fe
2 mol Fe 3 mol Pt2
55.85 g Fe mol Fe
97
Balancing REDOX Equations
The oxidation number method
Step 1) Assign oxidation numbers to all
elements in the equation. Step 2) From the
changes in oxidation numbers, identify the
oxidized and reduced species.
Step 3) Compute the number of
electrons lost in the oxidation and
gained in the reduction from the oxidation
number changes. Draw tie-lines
between these atoms to show electron
changes. Step 4) Multiply one or both of these
numbers by appropriate factors to
make the electrons lost equal the electrons
gained, and use the factors as
balancing coefficients. Step 5) Complete the
balancing by inspection, adding states of matter.
98
REDOX Balancing using Ox. No. Method - I
2 e-
0
-2
___ H2 (g) ___ O2 (g)
___ H2O(g)
2
2
1
- 1 e-
0
electrons lost must electrons gained therefore
multiply Hydrogen reaction by 2! and we are
balanced!
99
REDOX Balancing Using Ox. No. Method - II
-1e-
2
3
Fe2(aq) MnO4-(aq) H3O(aq)
Fe3(aq) Mn2(aq) H2O(aq)
2
5 e-
7
Multiply Fe2 Fe3 by five to correct for the
electrons gained by the Manganese.
5 Fe2(aq) MnO4-(aq) H3O(aq)
5 Fe3(aq) Mn2(aq) H2O(aq)
Make four water molecules from protons from the
acid, and the oxygen from the MnO4-, this will
require 8 protons, or Hydronium ions. This will
give a total of 12 water molecules formed.
5 Fe2(aq) MnO4-(aq) 8 H3O(aq) 5
Fe3(aq) Mn2(aq) 12 H2O(aq)
100
Balancing Oxidation-Reduction Equations Occurring
in Acidic Solution by the Half-Reaction Method.
101
Balancing Oxidation-Reduction Equations Occurring
in Basic Solution by the Half-Reaction Method.
102
Balancing Redox Equations in Aqueous Acid
and Base Solutions
ACID You may add either H ( H3O ), or water
( H2O ) to either side of the
chemical equation. BASE You may add either
OH -, or water to either side of the
chemical equation.
H OH - H2O
H OH - H2O
H H2O H3O
103
REDOX Balancing by Half-Reaction Method-I
Fe2(aq) MnO4-(aq)
Fe3(aq) Mn2(aq) acid solution
Identify Oxidation and Reduction Half Reactions
Fe2(aq) Fe3(aq) e-
oxidation half-reaction
MnO4-(aq) Mn2(aq)
add H to the reactants and that will give water
as a product!
MnO4-(aq) 8H3O(aq) 5e- Mn2(aq)
12H2O(l)
reduction
half-reaction
Sum the two half-reactions
Fe2(aq) Fe3(aq) e- x5
MnO4-(aq) 8H3O(aq) 5e-
Mn2(aq) 12H2O(l)
MnO4-(aq) 8H3O(aq)5e- 5Fe2(aq)
5Fe3(aq)5e- Mn2(aq) 12H2O(l)
104
REDOX Balancing by Half-Reaction Method -II
MnO4-(aq) SO32-(aq)
MnO2 (s) SO42-(aq) basic solution
Oxidation SO32-
SO42-(aq) 2e - Add OH- to the reactant
side, and water to the product side to get
oxygen to balance since we have one more oxygen
on sulfate than on sulfite.
SO32-(aq) 2 OH-(aq)
SO42-(aq) H2O(l) 2e -
Reduction MnO4-(aq)
3e - MnO2 (s) Add water
to the reactant side and OH- to the product side
to take up the oxygen lost when MnO4- goes to
MnO2 and loses two oxygen atoms.
MnO4-(aq) 2 H2O(l) 3e -
MnO2 (s) 4 OH-(aq)
Multiply the oxidation equation by 3 to make the
electrons 6. Multiply the reduction equation by 2
to make the electrons 6, and add the two.
3 SO3-2(aq) 2 MnO4-(aq) H2O(l) 3
SO4-2(aq) 2 MnO2 (s) 2 OH-(aq)
105
REDOX Balancing by Half-Reaction Method-III
MnO4-(aq) SO32-(aq)
MnO2 (s) SO42-(aq) acidic solution
Oxidation SO32-(aq)
SO42-(aq) 2 e - Add water to the reactant
side to supply an oxygen and add two protons to
the product side that will remain plus the two
electrons. SO32-(aq)
H2O(l) SO42-(aq) 2 H(aq) 2 e -
Reduction MnO4-(aq) 3 e-
MnO2 (s) Add water to the product side
to take up the extra oxygen from Mn cpds, and add
Hydrogen to the reactant side .
MnO4-(aq) 3 e- 4H MnO2
(s) 2 H2O(l)
Multiply the oxidation equation by 3, and the
reduction equation by 2, and add them canceling
out the electrons, protons and water molecules.
3SO32-(aq) 2MnO4-(aq) 2H(aq) 3
SO42-(aq) 2MnO2 (s) H2O(l)
106
REDOX Balancing using Ox. No. Method - III
7
4
3 e -
( Acidic Solution )
MnO4-(aq) SO32-(aq) MnO2 (s)
SO42-(aq)
4
6
- 2 e -
To balance the electrons, we must multiply the
sulfite by 3, and the permanganate by 2. We then
have to account for the oxygen imbalance by
adding acid to the reactant side, and water to
the product side.
2 MnO4-(aq) 3 SO32-(aq) H3O(aq) 2
MnO2 (s) 3 SO42-(aq) H2O(aq)
For the final balance it is necessary to realize
that protons needed to bind up the oxygen atoms
must be balanced, and since we have called Hion
- hydronium ions,therefore water will be formed!
2 MnO4-(aq) 3 SO32-(aq)2 H3O(aq) 2 MnO2
(s) 3 SO42-(aq) 3 H2O(aq)
107
REDOX Balancing by Half-Reaction Method-IV
MnO4-(aq) SO32-(aq)
MnO2(s) SO42-(aq) basic solution
balance the equation as if it were in acid, and
then convert it to base
2MnO4-(aq) 3SO32-(aq) 2H(aq)
2MnO2(s) 3SO42-(aq) H2O(l)
2MnO4-(aq) 3SO32-(aq) 2H(aq)
2MnO2(s) 3SO42-(aq) H2O(l)
To convert to base, add two OH- to each side of
the equation
2MnO4-(aq) 3SO32-(aq) 2H(aq) 2OH-(aq)
2MnO2(s) 3SO42-(aq) H2O(l)2OH-(aq)
On the reactant side, the H and the OH- cancel
to give water.
2MnO4-(aq) 3SO32-(aq)2H2O(l) 2MnO2(s)
3SO42-(aq) H2O(l)2OH-(aq)
Cancel out the water on each side of the
equation, and you are done!
2MnO4-(aq) 3SO32-(aq) H2O(l)
2MnO2(s) 3SO42-(aq) 2OH-(aq)
108
REDOX Balancing Using Ox. No. Method-IV
Zinc metal is dissolved in Nitric Acid to give
Zn2 and the ammonium ion from the reduced
Nitric acid, write the balanced chemical equation!
Zn(s) H(aq) NO3-(aq)
Zn2(aq) NH4(aq)
Oxidation method
- 2 e-
Zn(s) H(aq) NO3-(aq)
Zn2(aq) NH4(aq)
5
-3
8 e-
Multiply Zinc and Zn2 by 4, and ammonia by
unity. Since we have no oxygen on the product
side, add 3 water molecules to the product side,
requiring 10 H on the reactant side.
4 Zn(s) 10 H(aq) NO3-(aq)
4 Zn2(aq) NH4(aq) 3 H2O(l)
109
REDOX Balancing by Half-Reaction Method-V
Zn(s) H3O(aq) NO3-(aq)
Zn2(aq) NH4(aq)
Given
Oxidation Zn(s) Zn2
2 e-
Reduction H3O(aq) NO3-(aq) 8 e -
NH4(aq) H2O(l)
We will need three waters to pick up the oxygen
from the nitrate ion, and
for the hydrogen, we will need to have
10 hydrogen ions. Because the
Hydrogen ions came as
hydronium ions, we will need 10 more water
molecules. 10 H3O(aq) NO3-(aq) 8 e
- NH4(aq) 13 H2O(l)
Finally, if we are to add the two equations, we
must multiply the Ox. one by 4 to be able to
cancel out the electrons, so the final balanced
equation is
10 H3O(aq) NO3-(aq) 4 Zn(s) 4
Zn2(aq) NH4(aq) 13 H2O(l)
110
REDOX Balancing by Half-Reaction Method
-VI - A
In acid Potassium dichromate reacts with
ethanol(C2H5OH) to yield the blue-green solution
of Cr3, the reaction used in breathalyzers.
H3O(aq) Cr2O72-(aq) C2H5OH(l)
Cr3(aq) CO2 (g) H2O(l)
Oxidation C2H5OH(l)
CO2 (g) We need to balance
oxygen by adding water to the reactant side, and
balance Hydrogen by adding protons to the
product side.
C2H5OH(l) 3 H2O(l) 2 CO2
(g) 12 H(aq)
Since we wish to consider H as the Hydronium ion
- H3O , we must add 12 water molecules to the
reactant side, and make the H into H3O.
C2H5OH(l) 15 H2O(l) 2 CO2
(g) 12 H3O(aq) 12 e -
111
REDOX Balancing by Half-Reaction Method
- VI - B
Reduction Cr2O72-(aq)
Cr3(aq) Dichromate has two chromium
atoms, therefore the products need to have two
Cr3, and 3 electrons per atom. The oxygen atoms
from the dichromate need to be taken up as water
on the product side by adding protons to the
reactant side.
14H(aq) Cr2O72-(aq) Cr3(aq)
7 H2O(l)
Each Chromium atom changes oxidation from a 6 to
a 3 thereby accepting 6 electrons, so we add 6
electrons to the reactant side.
6e - 14 H3O(aq) Cr2O72-(aq)
2 Cr3(aq) 21 H2O(l)
Adding the two equations will give the final
equation
Ox C2H5OH(l) 15 H2O(l) 2
CO2 (g) 12 H3O(aq) 12 e -
Rd
6e - 14 H3O(aq) Cr2O72-(aq)
2 Cr3(aq) 21 H2O(l) x 2
C2H5OH(l) 16 H3O(aq) 2 Cr2O72-(aq) 2 CO2
(g) 4 Cr3(aq) 27 H2O(l)
112
REDOX Balancing by Half-Reaction Method
-VII - A
Silver is reclaimed from ores by extraction using
basic Cyanide ion.
OH-
Ag(s) CN-(aq) O2 (g)
Ag(CN)2-(aq)
Oxidation CN-(aq) Ag(s)
Ag(CN)2-(aq)
Since we need two cyanide ions to form the
complex, add two to the reactant side of the
equation. Silver is also oxidized, so it loses
an electron, so we add one electron to the
product side.
2 CN-(aq) Ag(s)
Ag(CN)2-(aq) e -
Reduction O2 (g) H2O(aq)
OH-(l)
Since oxygen is to form oxide ions, 4 electrons
need to be added to the reactant side, and 2
water molecules are needed to supply the
hydrogen to make hydroxide ions, yielding 4 OH-
ions.
4 e - O2 (g) 2 H2O(aq)
4 OH-(l)
113
REDOX Balancing by Half-Reaction Method
- VII - B
Adding the Reduction equation to the Oxidation
equation will require the Oxidation one to be
multiplied by 4 to eliminate the electrons.
Ox (x4) 8CN-(aq) 4 Ag(s)
4 Ag(CN)2-(aq) 4 e -
Rd 4 e - O2 (g) 2 H2O(l)
4 OH -(aq)
8 CN -(aq) 4 Ag(s) O2 (g) 2 H2O(l)
4 Ag(CN)2-(aq) 4 OH -(aq)
114
REDOX Balancing Using Ox. No. Method -V
0
1
-1 e -
Ag(s) CN -(aq) O2 (g)
Ag(CN)2-(aq) OH -(aq)
0
- 2
2 e -
To balance electrons we must put a 4 in front of
the Ag, since each oxygen loses two electrons,
and they come two at a time! That requires us to
put a 4 in front of the silver complex, yielding
8 cyanide ions.
4 Ag(s) 8 CN -(aq) O2 (g)
4 Ag(CN)2-(aq) OH -(aq)
We have no hydrogen on the reactant side
therefore we must add water as a reactant, and
since we also add oxygen, we must add two water
molecules, that well give us 4 hydroxide anions,
giving us a balanced chemical equation.
4 Ag(s) 8 CN -(aq) O2 (g) 2 H2O(l)
4 Ag(CN)2-(aq) 4 OH -(aq)
115
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116
Redox Titration- Calculation outline - I
Problem Calcium Oxalate was precipitated from
blood by the addition of Sodium Oxalate so
that calcium ion could be determined in the
blood sample. The sulfuric acid solution that
the precipitate was dissolved in required 2.05
ml of 4.88 x 10-4 M KMnO4 to reach
the endpoint. a) calculate the amount (mol) of
Ca2. b) calculate the Ca2 ion conc. Plan a)
Calculate the molarity of Ca2 in the H2SO4
solution. b) Convert the Ca2 concentration into
units of mg Ca2/ 100 ml blood.
Volume (L) of KMnO4 Solution
a)
M (mol/L)
Moles of KMnO4
b)
Molar ratio
Moles of CaC2O4
Chemical Formulas
c)
Moles of Ca2
117
Figure 4.21 Permanganate being introduced into
a flask of reducing agent.
118
Redox Titration - Calculation - I
Equation 2 KMnO4 (aq) 5 CaC2O4 (aq) 8 H2SO4
(aq) 2 MnSO4 (aq)
K2SO4 (aq) 5 CaSO4 (aq) 10 CO2 (g) 8 H2O(L)
a) Moles of KMnO4
Mol Vol x Molarity Mol 0.00205 L x 4.88 x 10-
4mol/L Mol 1.00 x 10 - 6mol KMnO4
b) Moles of CaC2O4
5 mol CaC2O4
Mol CaC2O4 1.00 x 10-6 mol KMnO4 x

2 mol KMnO4
Mol CaC2O4 2.50 x 10 -6 mol CaC2O4
c) Moles of Ca2
1 mol Ca2
Mol Ca2 2.50 x 10 -6 mol CaC2O4 x

1 mol CaC2O4
Mol Ca2 2.50 x 10 -6 mol Ca2
119
Redox Titration - Calculation Outline - II
Moles of Ca2/ 1 ml of blood
multiply by 100 a) Calc of mol Ca2 per
100 ml
Moles of Ca2/ 100 ml blood
M (g/mol) b) Calc of mass of Ca2 per
100 ml
Mass (g) of Ca2/ 100 ml blood
1g 1000mg c) convert g to mg!
Mass (mg) of Ca2 / 100 ml blood
120
Redox Titration - Calculation - II
a) Mol Ca2 per 100 ml Blood
Mol Ca2
Mol Ca2
x 100 ml Blood
100 ml Blood
1.00 ml Blood
Mol Ca2
2.50 x 10 -6 mol Ca2
x 100 ml
Blood
100 ml Blood
1.00 ml Blood
Mol Ca2
2.50 x 10 -4 mol Ca2
100 ml Blood
b) mass (g) of Ca2
Mass Ca2 Mol Ca2 x Mol Mass Ca/ mol Mass
Ca2 2.50 x 10 -4mol Ca2 x 40.08g Ca/mol
0.0100 g Ca2
c) mass (mg) of Ca2
Mass Ca2 0.0100g Ca2 x 1000mg Ca2/g Ca2
10.0 mg Ca2
100 ml Blood
121
Types of Chemical Reactions - I
I) Combination Reactions that are Redox
Reactions a) A Metal and a Non-Metal form
an Ionic compound b) Two Non-Metals form a
Covalent compound c) Combination of an
Element and a Compound II) Combination Reactions
that are not Redox Reactions a) A Metal
oxide and a Non-Metal form an ionic compound
with a polyatomic anion b) Metal
Oxides and water form Bases c) Non-Metal
Oxides and water form Acids III) Decomposition
Reactions a) Thermal Decomposition
i) Many ionic compounds with oxoanions form
a metal oxide and a gaseous
non-metal ii) Many Metal oxides,
Chlorates, and Perchlorates release
Oxygen b) Electrolytic Decomposition
122
Types of Chemical Reactions - II
IV) Displacement Reactions a) Single
-Displacement Reactions - Activity Series of the
Metals i) A metal displaces
hydrogen from water or an acid ii)
A metal displaces another metal ion from
solution iii) A halogen displaces a
halide ion from solution b) Double
-Displacement Reactions - i) In
Precipitation Reactions- A precipitate forms
ii) In acid-Base Reactions - Acid
Base form a salt water V) Combustion Reactions
- All are Redox Processes a) Combustion
of an element with oxygen to form oxides
b) Combustion of Hydrocarbons to yield Water
Carbon Dioxide
Reactants
Products