1. Types of Chemical Reactions and Solution Stoichiometry
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry

2. 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

3. Precipitation of
silver chromate by
adding potassium
chromate to a
solution of silver
nitrate.
K2CrO4 (aq) + 2 AgNO3 (aq)
Ag2CrO4 (s) + 2 KNO3 (aq)

4. Figure 4.1: A space-filling model of the water molecule.

6. Figure 4.2: Polar water molecules interact with the positive and negative ions of a salt, assisting with the dissolving process.

8. 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.

9. 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.

10. Electrical Conductivity of Ionic Solutions

11. Figure 4.4: Electrical Conductivity

12. Figure 4.5: HCL (aq) is completely ionized.

17. Figure 4.6: An aqueous solution of sodium hydroxide.

18. Figure 4.7: Acetic acid (HC2H3O2) exists in water mostly as undissociated molecules.

19. Figure 4.8: The reaction of NH3 in water.

20. Carbohydrates Molecules that contain carbon and water! CxH2yOy H CH2OH
Sucrose C12H22O11 , C12(H2O)11 a disaccharide

21. 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.
In brass , Copper is the solvent (90%), and Zinc is the solute(10%)

22. 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 = mol NH3
Change the volume of the solution into liters:
1 L
1000 mL
75.7 ml X = L
Finally, we divide the number of moles of solute by the volume
of the solution:
0.216 mol NH3 L
Molarity = = ____________ M NH3

23. Preparing a Solution - I
Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to ml or l !
What is the Molarity of the salt and each of the ions?
Na3PO4 (s) + H2O(solvent) = 3 Na+(aq) + PO4-3(aq)

24. Preparing a Solution - II
Mol wt of Na3PO4 = g / mol
3.95 g / g/mol = mol Na3PO4
dissolve and dilute to ml
M = mol Na3PO4 / l = M
Na3PO4
for PO4-3 ions = ______________ M
for Na+ ions = 3 x M = ___________ M

25. 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 = 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!

26. Figure 4.9: Steps involved in the preparation of a standard solution.

27. Like Example 4.4 (P 97)
A Chemist must prepare 1.00 L of a 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 = 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 = 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.

28. Make a Solution of Potassium Permanganate
Potassium Permanganate is KMnO4 and has a molecular mass
of g / mole
Problem: Prepare a solution by dissolving 1.58 grams of KMnO4
into sufficient water to make ml of solution.
1 mole KMnO4
g KMnO4
1.58 g KMnO4 x = moles KMnO4
moles KMnO4
0.250 liters
Molarity = = ______________ M
Molarity of K+ ion = [K+] ion = [MnO4-] ion = _____________ M

29. Figure 4.10: (a) A measuring pipette (b) A volumetric pipette.

30. Figure 4. 11: (a) A measuring pipette (b) Water is added to the flask
Figure 4.11: (a) A measuring pipette (b) Water is added to the flask. (c) The resulting solution is 1 M acetic acid.

31. Dilution of Solutions Take 25.00 ml of the 0.0400 M KMnO4
Dilute the ml to l - What is the resulting Molarity of the diluted solution?
# moles = Vol x M
l x M = Moles
Mol / 1.00 l = _______________ M

34. Figure 4.13: Reactant solutions: (a) Ba(NO3)3(aq)

35. Figure 4.13: Reactant solutions: (b) K2CrO4(aq).

36. Figure 4.12: When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates.

37. Figure 4.14: Reaction of K2CrO4 (aq) and Ba(NO3)2 (aq).

38. Figure 4.15: Precipitation of silver chloride by mixing solutions of silver nitrate and potassium chloride.

39. Figure 4.16: Photos and molecular-level representations illustrating the reaction of KCL(aq) with AgNO3(aq) to form AgCl(s).

40. 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.

41. 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 = 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 = g/L
Solubility of CuCl in water at 20oC = g/L

42. 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 H C
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 = C8H and / or Benzene = C6H6

43. When a solution of Na2SO4 (aq) is added to a solution of Pb(NO3)2, the white solid PbSO4(s) forms.

45. 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) g of rubidium fluoride dissolved in water
c) x 1021 formula units of iron (III) chloride dissolved in water
d) ml of 0.56M scandium bromide dissolved in water
e) moles of ammonium sulfate dissolved in water
a) Na2CO3 (s) 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

46. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II
H2O
b) RbF(s) Rb+(aq) + F -(aq)
1 mol RbF
g RbF
moles of RbF = 46.5 g RbF x = moles RbF
thus, mol Rb+ and mol F - are present
H2O
c) FeCl3 (s) Fe+3(aq) + 3 Cl -(aq)
moles of FeCl3 = 9.32 x 1021 formula units
1 mol FeCl3
6.022 x 1023 formula units FeCl3 x
= mol FeCl3
3 mol Cl -
1 mol FeCl3
moles of Cl - = mol FeCl3 x = _________ mol Cl -
and ____________ mol Fe+3 are also present.

47. Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - III
H2O
d) ScBr3 (s) Sc+3(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 = mol ScBr3
3 mol Br -
1 mol ScBr3
Moles of Br - = mol ScBr3 x = mol Br -
0.042 mol Sc+3 are also present
H2O
e) (NH4)2SO4 (s) 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.

48. Solid Fe(OH)3 forms when aqueous KOH and Fe(NO3)3 are mixed.

49. 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)

50. 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) Pb+2(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)

53. 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!

54. Figure 4.17: Selective precipitation of Ag+, Ba2+, and Fe3+ ions.

56. Species present, Balanced net ionic equation.

57. Like Example 4.7 (P 108)
Calculate the mass of solid sodium iodide that must be added to
2.50 L of a 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 = 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

58. 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 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 = L x 2.44 M = mol Ag+
VNaCl x MNaCl = L x 3.17 M = 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 mol of
AgCl:
Mass AgCl = mol x g AgCl/ mol = _________ g

59. Like Example 4.9 (P 109)
What mass of Pb2+ could by precipitated from a solution by the
addition of L of 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 = L x Moles = mol I-
Liter
mol I-
2 mol I-/ mol Pb2+
Moles Lead ion = = mol Pb2+
Mass of Lead = 207.2g Pb x moles = ____________ g Pb
mol Pb

61. 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)

62. 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) 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 = = Molar in SO4- 2
Molarity of H+ = 2 x mol H+ = 1.37 Molar in H+

63. The gravimetric procedure.(P109-110)
1 mol CaC2O4 H2O
146.12g CaC2O4 H2O .
g CaC2O4 H2O X = x 10-3 mol
CaC2O4 H2O
1.998 x 10-3 mol Ca2+ X = x 10-2g Ca2+
Mass % Ca is: x 100% = ______________% . .
40.08 g Ca2+
1 mol Ca2+
8.009 x 10-2 g

64. Species present, Balanced net ionic equation.

65. 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

66. 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, HNO Strontium hydroxide, Sr(OH)2
Sulfuric acid, H2SO Barium hydroxide, Ba(OH)2
Perchloric acid, HClO4
Weak Weak
Hydrofluoric, HF Ammonia, NH3
Phosphoric acid, H3PO4
Acetic acid, CH3COOH
(or HC2H3O2)

68. 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) H2O(l)

69. 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)

70. Figure 4.18: The titration of an acid with a base.

72. Like Example 4.10 (P 113)
What volume of M H2SO4 is needed to neutralize ml
of a M LiOH solution?
Calculate the number of moles of base:
Vbase x Mbase = L x M = mol LiOH
From the balanced equation find the moles of acid needed:
2 LiOH(aq) + H2SO4 (aq) 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 mol H2SO4
Volume of acid:
Moles acid
Macid
moles
0.468 Mol L
Vacid = = = L H2SO4

73. A firefighter in a protective suit with an oxygen tank neutralizes an acid spill.

74. 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 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 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)

75. Potassium Hydrogenphthalate KHC8H4O4

76. 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 = mol KHP
Volume of base = Final buret reading - Initial buret reading
= ml ml = ml of base
one mole of acid = one mole of base; therefore moles of
acid will yield moles of base in a volume of ml.
moles L
molarity of base = = __________ moles per liter
molarity of base = ___________________ M

77. Like Example 4.12 (P )
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 ml of 1.5 M sodium hydroxide, a
strong base, and the remaining base titrated with hydrochloric acid
using ml of 1.80 M. What is the percentage of ascorbic acid?
Mol acid = L x 1.80 Mol/L = mol HCl
Mol base = L x 1.50 Mol/L = mol NaOH
The difference between the base and acid will be the moles of
ascorbic acid!
Reacted base = – = mol Ascorbic acid
Mass ascorbic acid = mol x 176g/mol = 1.95 g Ascorbic acid
1.95g
10.00g
% ascorbic acid = x 100% = ____________%

78. Figure 4.19: Reaction of solid sodium and gaseous chlorine to form solid sodium chloride.

80. Oxidation of copper metal by nitric acid.

83. Highest and Lowest oxidation numbers of
Chemically reactive main-group Elements 1A 2A 3A -4 4A 5A 6A 7A 1 +1 -1 H
Period +4 +4 +5 +6 +7 +1 +2 +3 -3 -2 -1 2 Li Be B C N O F 3 Na Mg Al Si P S Cl 4 K Ca Ga Ge As Se Br
non-metals 5 Rb Sr In Sn Sb Te I
metalloids 6 Cs Ba Tl Pb Bi Po At
metals 7 Fr Ra

84. Main Group Elements Period IA VIIIA H He 1 +1 -1 IIA IIIA IVA VA VIA
IIA
IIIA
IVA VA
VIA
VIIA Li Be B C N O F Ne 2
+4,+2
-1,-4
all from +1 +2 +3
-1,-2 -1 Na Mg Al Si P S -1 Cl Ar 3
+5,+3 -3
+6,+4
+2,-2
+7,+5
+3,+1 +1 +2 +3
+4,-4 Kr K Ca Ga Ge As Se -1 Br 4
+4,+2 -4
+5,+3 -3
+6,+4 -2
+7,+5
+3,+1 +2 +1 +2
+3, +2 Xe Rb Sr In Sn Sb Te -1 I 5
+3,+2 +1
+4,+2, -4
+5,+3 -3
+6,+4 -2
+7,+5
+3,+1
+6,+4 +2 +1 +2 Rn Cs Ba Tl Pb Bi Po -1 At 6
+6,+4
+2,-2
+7,+5
+3,+1 +1
+4,+2 +2 +2
+3,+1 +3

85. Transition Metals Possible Oxidation States VIIIB IIIB IVB VB VIB VIIBSc Ti V Cr +2 Mn Fe Co Ni Cu Zn
+4,+3 +2
+5,+4
+3+2
+6,+3 +2
+7,+6
+4,+3 +3
+3,+2
+3,+2 +2
+2,+1 +2 Y Zr Nb Mo Tc Ru Rh Pd Ag Cd
+5,+4 +2
+6,+5
+4,+3
+7,+5 +4
+8,+5
+4,+3 +3
+4,+3
+4,+3
+4,+2 +1 +2 La Hf Ta W Re +2 Os Ir Pt Au Hg
+5,+4 +3
+6,+5 +4
+7,+5 +4
+8,+6
+4,+3
+4,+3 +1 +3
+4,+3
+4,+2
+3,+1
+2,+1

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 ion’s 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.

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.

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) 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 +2
Zn(s) 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] S[+6] -2
S is Oxidized +6
O[0] O[-2]
S8 (s) O2 (g) SO3 (g)
O is Reduced
S8 is the reducing agent and O2 is the oxidizing agent
c) Assigning oxidation numbers:
C[+2] C[+4]
C is oxidized
Ni[+2] Ni[0]
Ni is Reduced -2 -2 -2 +2 +4 +2
NiO(s) CO(g) Ni(s) CO2 (g)
CO is the reducing agent and NiO is the oxidizing agent

94. Activity Series of the Metals
Strongly
reducing
Weakly
Li Li+ + e-
K K+ + e These elements react rapidly with aqueous H+ ions
Ba Ba e (acid) or with liquid H2O to release H2 gas.
Ca Ca e-
Na Na+ + e-
Mg Mg2+ + 2e-
Al Al e 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.
H 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) Fe3+(aq) + 3 Cu(s)
Mg(s) + Zn2+(aq) Mg2+(aq) + Zn(s)
2 K(s) + Sn2+(aq) K+(aq) + Sn(s)
Pt(s) + Ni2+(aq) N. R.
2 Al(s) + 6 H+(aq) Al3+(aq) + 3 H2 (g)
Au(s) + H+(aq) N. R.

96. Problem: Calculate the mass of metallic Iron that must be
added to liters of a solution containing
M of Pt2+(aq) ions in solution to reclaim all
of the Platinum.
Solution:
V x M = # moles
500.0L x 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 = 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
___ H2 (g) +___ O2 (g) ___ H2O(g) 2 2
- 1 e- +1
electrons lost must = electrons gained therefore multiply
Hydrogen reaction by 2!
and we are balanced!

99. REDOX Balancing Using Ox. No. Method - II+2
-1e- +3
Fe+2(aq) + MnO4-(aq) + H3O+(aq) Fe+3(aq) + Mn+2(aq) + H2O(aq)
+5 e- +2 +7
Multiply Fe+2 & Fe+3 by five to correct for the electrons gained by the
Manganese.
5 Fe+2(aq) + MnO4-(aq) + H3O+(aq) Fe+3(aq) + Mn+2(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 Fe+2(aq) + MnO4-(aq) +8 H3O+(aq) Fe+3(aq) + Mn+2(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
Fe+2(aq) + MnO4-(aq) Fe+3(aq) + Mn+2(aq) [acid solution]
Identify Oxidation and Reduction Half Reactions
Fe+2(aq) Fe+3(aq) + e [oxidation half-reaction]
MnO4-(aq) Mn+2(aq)
add H+ to the reactants and that will give water as a product!
MnO4-(aq) + 8H3O+(aq) +5e Mn+2(aq) + 12H2O(l)
[reduction half-reaction]
Sum the two half-reactions
{ Fe+2(aq) Fe+3(aq) +e- } x5
MnO4-(aq) + 8H3O+(aq) +5e Mn+2(aq) + 12H2O(l)
MnO4-(aq)+ 8H3O+(aq)+5e- +5Fe+2(aq) Fe+3(aq)+5e- + Mn+2(aq)+ 12H2O(l)

104. REDOX Balancing by Half-Reaction Method -II
MnO4-(aq) + SO32-(aq) MnO2 (s) + SO42-(aq) [basic solution]
Oxidation: SO 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) 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) SO42-(aq) + 2MnO2 (s) + H2O(l)

106. REDOX Balancing using Ox. No. Method - III+7
+ 3 e - +4
( Acidic Solution )
MnO4-(aq) + SO32-(aq) MnO2 (s) + SO42-(aq) +4
- 2 e - +6
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) 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
H+ion - hydronium ions,therefore water will be formed!
2 MnO4-(aq)+ 3 SO32-(aq)+2 H3O+(aq) 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) MnO2(s) + 3SO42-(aq) + H2O(l)
2MnO4-(aq) + 3SO32-(aq) + 2H+(aq) MnO2(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) MnO2(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) MnO2(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) MnO2(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
+8 e- -3
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) Zn2+(aq) + NH4+(aq) + 3 H2O(l)

109. REDOX Balancing by Half-Reaction Method-V
Given:
Zn(s) + H3O+(aq) + NO3-(aq) Zn2+(aq) + NH4+(aq)
Oxidation: Zn(s) Zn 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) Zn+2(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 Cr+3, 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) 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) H2O(l) CO2 (g) + 12 H3O+(aq) + 12 e -

111. REDOX Balancing by Half-Reaction Method - VI - B
Reduction: Cr2O72-(aq) Cr+3(aq)
Dichromate has two chromium atoms, therefore the products need to
have two Cr+3, 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) Cr+3(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 H3O+(aq) + Cr2O72-(aq) Cr+3(aq) + 21 H2O(l)
Adding the two equations will give the final equation:
Ox: C2H5OH(l) + 15 H2O(l) CO2 (g) + 12 H3O+(aq) + 12 e -
Rd:
[6e H3O+(aq) + Cr2O72-(aq) Cr+3(aq) + 21 H2O(l)] x 2
C2H5OH(l) + 16 H3O+(aq) + 2 Cr2O72-(aq) CO2 (g) + 4 Cr+3(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) 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) CN-(aq) + 4 Ag(s) Ag(CN)2-(aq) + 4 e -
Rd e - + O2 (g) + 2 H2O(l) OH -(aq)
8 CN -(aq) + 4 Ag(s) + O2 (g) + 2 H2O(l) Ag(CN)2-(aq) + 4 OH -(aq)

114. REDOX Balancing Using Ox. No. Method -V
-1 e - +1
Ag(s) + CN -(aq) + O2 (g) Ag(CN)2-(aq) + OH -(aq)
+ 2 e -
- 2
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) 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) Ag(CN)2-(aq) + 4 OH -(aq)

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
Ca+2.
b) calculate the Ca+2 ion conc.
Plan: a) Calculate the molarity of
Ca+2 in the H2SO4 solution.
b) Convert the Ca+2 concentration
into units of mg Ca+2/ 100 ml blood.
Volume (L) of KMnO4 Solution a)
M (mol/L)
Moles of KMnO4 b)
Molar ratio
Moles of CaC2O4 c)
Chemical Formulas
Moles of Ca+2

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 = L x 4.88 x 10- 4mol/L
Mol = 1.00 x mol 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 mol CaC2O4
c) Moles of Ca+2
1 mol Ca+2
Mol Ca+2 = 2.50 x mol CaC2O4 x =
1 mol CaC2O4
Mol Ca+2 = 2.50 x mol Ca+2

119. Redox Titration - Calculation Outline - II
Moles of Ca2+/ 1 ml of blood
multiply by a) Calc of mol Ca+2 per 100 ml
Moles of Ca2+/ 100 ml blood
M (g/mol) b) Calc of mass of Ca+2 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 Ca+2 per 100 ml Blood
Mol Ca+2
Mol Ca+2
= x 100 ml Blood =
100 ml Blood
1.00 ml Blood
Mol Ca+2
2.50 x mol Ca+2
= x 100 ml Blood =
100 ml Blood
1.00 ml Blood
Mol Ca+2
= x mol Ca+2
100 ml Blood
b) mass (g) of Ca+2
Mass Ca+2 = Mol Ca+2 x Mol Mass Ca/ mol =
Mass Ca+2 = 2.50 x 10 -4mol Ca+2 x 40.08g Ca/mol = g Ca+2
c) mass (mg) of Ca+2
Mass Ca+2 = g Ca+2 x 1000mg Ca+2/g Ca+2 = 10.0 mg Ca+2
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
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