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Chapter 4
Types of Chemical
Reactions and Solution
Stoichiometry
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
Precipitation of
silver chromate by
adding potassium
chromate to a
solution of silver
nitrate.
K2CrO4 (aq) + 2 AgNO3 (aq)
Ag2CrO4 (s) + 2 KNO3 (aq)
Figure 4.1: A space-filling model of the water molecule.
Figure 4.2: Polar water molecules
interact with the positive and negative
ions of a salt, assisting with the
dissolving process.
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.
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 compound 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 through out the solution, and are conducting
electricity, or helping electrons to move through out the solution
Electrical Conductivity of Ionic Solutions
Figure 4.4:
Electrical
Conductivity
Figure 4.5:
HCL (aq) is
completely
ionized.
Figure 4.6:
An aqueous
solution of
sodium
hydroxide.
Figure 4.7:
Acetic acid
(HC2H3O2)
exists in water
mostly as
undissociated
molecules.
Figure 4.8:
The reaction of
NH3 in water.
Carbohydrates
Molecules that contain carbon and water!
H
CH2OH
C
HO
H
C
OHH C
C
OH
H
CxH2yOy
O
CH2OH O
H
C
C
O
H
C
OH
Sucrose
H
OH C
C
H
CH2OH
C12H22O11 , C12(H2O)11 a disaccharide
Molarity (Concentration of Solutions)= M
M=
Moles of Solute =
Liters of Solution
Moles
L
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%)
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
3.68g NH3 X
= 0.216 mol NH3
17.03g
Change the volume of the solution into liters:
75.7 ml X
1L
1000 mL
= 0.0757 L
Finally, we divide the number of moles of solute by the volume
of the solution:
Molarity =
0.216 mol NH3
0.0757 L
= ____________ M NH3
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)
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
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 x
2.5 mg NaCl x
1000 mg NaCl
1 mol NaCl = 4.28 x 10-5 mol
58.45g NaCl
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
Vx
= 4.28 x 10-5 mol NaCl
L solution
Solving for Volume gives:
-5
V = 4.28 x 10 mol NaCl = ______________________ L
0.14 mol NaCl
Or _________ ml of Blood!
L solution
Figure 4.9: Steps involved in the
preparation of a standard solution.
Like Example 4.4 (P 97)
A Chemist must prepare a 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:
1.00 L x
0.375 M (NH4)2CO3
L solution
= 0.375 M (NH4)2CO3
This amount can be converted to grams by using the molar mass:
94.07 g (NH4)2CO3
0.375 M (NH4)2CO3 x
= 35.276 g (NH4)2CO3
mol (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.
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.
1.58 g KMnO4 x 1 mole KMnO4 = 0.0100 moles KMnO4
158.04 g KMnO4
Molarity = 0.0100 moles KMnO4 =
0.250 liters
______________ M
Molarity of K+ ion = [K+] ion = [MnO4-] ion = _____________ M
Figure 4.10:
(a) A measuring pipette
(b) A volumetric pipette.
Figure 4.11: (a) A measuring pipette (b)
Water is added to the flask. (c) The
resulting solution is 1 M acetic acid.
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
Figure 4.13:
Reactant
solutions: (a)
Ba(NO3)3(aq)
Figure 4.13:
Reactant
solutions: (b)
K2CrO4(aq).
Figure 4.12:
When yellow
aqueous potassium
chromate is added
to a colorless
barium nitrate
solution, yellow
barium chromate
precipitates.
Figure 4.14: Reaction of K2CrO4 (aq) and
Ba(NO3)2 (aq).
Figure 4.15:
Precipitation
of silver
chloride by
mixing
solutions of
silver nitrate
and potassium
chloride.
Figure 4.16: Photos and molecular-level
representations illustrating the reaction of
KCL(aq) with AgNO3(aq) to form AgCl(s).
Table 4.1 (P102) Simple Rules for Solubility
of Salts in Water
1. Most nitrate (NO3-) salts are soluble.
2. Most salts of Na+, K+, and NH4+ are soluble.
3. Most chloride salts are soluble. Notable exceptions are AgCl,
PbCl2, and Hg2Cl2.
4. Most sulfate salts are soluble. Notable exceptions are BaSO4,
PbSO4, and CaSO4.
5. Most hydroxide salts are only slightly soluble. The important
soluble hydroxides are NaOH, KOH, and Ca(OH)2
(marginally soluble).
6. Most sulfide (S2-), carbonate (CO32-), and phosphate (PO43-)
salts are only slightly soluble.
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
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 = C8H18
and / or
Benzene = C6H6
When a solution of Na2SO4 (aq) is added to
a solution of Pb(NO3)2, the white solid
PbSO4(s) forms.
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)
b)
c)
d)
e)
4.0 moles of sodium carbonate dissolved in water
46.5 g of rubidium fluoride dissolved in water
5.14 x 1021 formula units of iron (III) chloride dissolved in water
75.0 ml of 0.56M scandium bromide dissolved in water
7.8 moles of ammonium sulfate dissolved in water
a) Na2CO3 (s)
H2O
moles of Na+ = 4.0 moles Na2CO3 x
2 Na+(aq) + CO3-2(aq)
2 mol Na+
1 mol Na2CO3
= 8.0 moles Na+ and 4.0 moles of CO3-2 are present
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - II
b)
RbF(s)
H2O
Rb+(aq) + F -(aq)
moles of RbF = 46.5 g RbF x
1 mol RbF
104.47 g RbF
= 0.445 moles RbF
thus, 0.445 mol Rb+ and 0.445 mol F - are present
H2 O
c) FeCl3 (s)
Fe+3(aq) + 3 Cl -(aq)
moles of FeCl3 = 9.32 x 1021 formula units
x
= 0.0155 mol FeCl3
moles of
Cl -
1 mol FeCl3
6.022 x 1023 formula units FeCl3
3
mol
Cl
= 0.0155 mol FeCl3 x
= _________ mol Cl 1 mol FeCl3
and ____________ mol Fe+3 are also present.
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:
Moles of ScBr3 = 75.0 ml x 13L x0.56 mol ScBr3 = 0.042 mol ScBr3
10 ml
1L
3 mol Br Moles of Br = 0.042 mol ScBr3 x
= 0.126 mol Br 1 mol ScBr3
0.042 mol Sc+3 are also present
H2O
e) (NH4)2SO4 (s)
2 NH4+(aq) + SO4- 2(aq)
+
2
mol
NH
4
Moles of NH4 = 7.8 moles (NH4)2SO4 x
= ____ mol NH4+
1 mol(NH4)2SO4
+
and ______ mol SO4- 2 are also present.
Solid
Fe(OH)3
forms when
aqueous KOH
and Fe(NO3)3
are mixed.
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)
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)
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 NO3b) 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!
Figure 4.17:
Selective
precipitation of
Ag+, Ba2+, and
Fe3+ ions.
Species present,
Balanced net ionic
equation.
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:
2+
0.125
Mol
Pb
2
mol
I
2.50 L x
x
1.0 L soln.
1 mol Pb2+
The mass of sodium iodide is:
0.625 mol I- x 1 mol NaI x
1 mol I-
= 0.625 mol I-
149.9 g NaI
= __________ g NaI
1 mol NaI
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 ClSince 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
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 ILiter
0.00118
mol
I
Moles Lead ion =
2 mol I-/ mol Pb2+
= 0.000590 mol Pb2+
Mass of Lead = 207.2g Pb x 0.000590 moles = ____________ g Pb
mol Pb
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)
HI(g)
H2O
H3O+(aq) + I -(aq)
H+(aq) + I -(aq)
Strong Acids and the Molarity of H+ Ions in
Aqueous Solutions of Acids
Problem: In aqueous solutions, each molecule of sulfuric acid will
loose 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)
Moles H2SO4 = 155 g H2SO4 x 1 mole H2SO4 = 1.58 moles H2SO4
98.09 g H2SO4
-2
1.58
mol
SO
2
4 = 0.687 Molar in SO - 2
Molarity of SO4 =
4
2.30 l solution
Molarity of H+ = 2 x 0.687 mol H+ = 1.37 Molar in H+
The gravimetric procedure.(P109-110)
. H2O
1
mol
CaC
O
2
4
0.2920 g CaC2O4 . H2O X 146.12g CaC O . H O = 1.998 x 10-3 mol
2 4
2
CaC2O4 . H2O
2+
40.08 g Ca
-3
2+
-2g Ca2+
1.998 x 10 mol Ca X
=
8.009
x
10
1 mol Ca2+
8.009 x 10-2
Mass % Ca is:
x 100% = ______________%
0.4367 g
Species present,
Balanced net ionic
equation.
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)
Acid + Base
=
MX(aq) + H2O(L)
Salt + Water
Selected Acids and Bases
Acids
Bases
Strong
Hydrochloric, HCl
Hydrobromic, HBr
Hydroiodoic, HI
Nitric acid, HNO3
Sulfuric acid, H2SO4
Perchloric acid, HClO4
Strong
Sodium hydroxide, NaOH
Potassium hydroxide, KOH
Calcium hydroxide, Ca(OH)2
Strontium hydroxide, Sr(OH)2
Barium hydroxide, Ba(OH)2
Weak
Hydrofluoric, HF
Phosphoric acid, H3PO4
Acidic acid, CH3COOH
(or HC2H3O2)
Weak
Ammonia, NH3
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)
Writing Balanced Equations for
Neutralization Reactions - II
b) LiOH(aq) + HNO3
LiNO3 (aq) + H2O(l)
(aq)
Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq)
Li+(aq) + NO3-(aq) + H2O(l)
OH -(aq) + H+(aq)
c)
Ba(OH)2 (aq) + H2SO4 (aq)
H2O(l)
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)
Figure 4.18: The titration of
an acid with a base.
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 = 0.0268 mol LiOH
From the balance 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
Vacid =
=
Macid
0.0134 moles
0.468 Mol = 0.0286 L H2SO4
L
A firefighter in a protective suit with an
oxygen tank neutralizes an acid spill.
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)
Potassium Hydrogenphthalate KHC8H4O4
O
O
C
C
C
K+
O K+
O
O
O
H
C
H+
O
O
Finding the Concentration of Base from an
Acid - Base Titration - II
moles KHP = 50.00 mg KHP x 1.00 g
204.2 g KHP
1000 mg
1 mol KHP
= 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.
molarity of base = 0.00024486 moles = __________ moles per liter
0.03332 L
molarity of base = ___________________ M
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
% ascorbic acid = 10.00g x 100% = ____________%
Figure 4.19: Reaction of solid sodium and
gaseous chlorine to form solid sodium
chloride.
Oxidation
of copper
metal by
nitric acid.
Highest and Lowest oxidation numbers of
Chemically reactive main-group Elements
+1
+1
-1
1 H
non-metals
1A 2A
+1 +2
3A 4A 5A 6A 7A
+3 +4-4 +5-3 +6-2 +7-1
Li
B
C
N
O
F
3 Na Mg
Al
Si
P
S
Cl
4
Ga Ge As Se
Br
5 Rb Sr
In
I
6 Cs
Tl Pb
2
Be
K Ca
Sn Sb Te
metalloids
metals
Ba
7 Fr Ra
Bi
Po At
Period IA
H
1
+1 -1
Li
2
Main Group Elements
IVA
VA
VIA
C
N
O
+4,+2 all from
-1,-4 +5 -3 -1,-2
IIA
Be
IIIA
B
+1
+2
+3
3
Na
+1
Mg
+2
Al
+3
4
K
+1
Ca
+2
Ga
Ge
+4,+2
+3, +2 -4
5
Rb
+1
Sr
+2
6
Cs
+1
Ba
+2
Xe
In
Sn
Sb
Te -1 I
+3,+2 +4,+2, +5,+3 +6,+4 +7,+5 +6,+4
+1
-4
-3
-2
+3,+1 +2
Rn
Tl
Pb
Bi
Po -1 At
+6,+4 +7,+5
+2
+3,+1 +4,+2
+3
+2,-2 +3,+1
Si
P
+4,-4 +5,+3
-3
VIIA
F
VIIIA
He
Ne
-1
S -1 Cl
+6,+4 +7,+5
+2,-2 +3,+1
Ar
As
Se -1 Br
+5,+3 +6,+4 +7,+5
-3
-2
+3,+1
Kr
+2
Transition Metals
Possible Oxidation States
VIIIB
IIIB IVB VB VIB VIIB
IB IIB
Sc
Ti
V
Cr +2Mn Fe
Co
Ni
Cu Zn
+3 +4,+3 +5,+4 +6,+3 +7,+6 +3,+2 +3,+2 +2 +2,+1 +2
+2
+3+2 +2
+4,+3
Y
Zr
Nb Mo Tc Ru Rh Pd
Ag
+5,+4 +6,+5 +7,+5 +8,+5
+3 +4,+3
+4,+3 +4,+2 +1
+2
+4,+3 +4
+4,+3
Cd
+2
La
Hf
Ta
W
Re +2Os Ir
Pt
Au Hg
+3 +4,+3 +5,+4 +6,+5 +7,+5 +8,+6 +4,+3 +4,+2 +3,+1 +2,+1
+3
+4
+4
+4,+3 +1
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.
Method to remove chlorinated organic molecules from Ground water
Fe(s) + RCl(aq) + H+(aq)
Fe2+(aq) + RH(aq) + Cl-(aq)
Figure 4.20:
A summary of an
oxidationreduction process,
in which M is
oxidized and X is
reduced.
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
0
Zn(s) + 2 HCl(aq)
-1
0
+2
ZnCl2 (aq) + H2 (g)
HCl is the oxidizing agent, and Zn is the reducing agent!
Recognizing Oxidizing and Reducing Agents - II
b) Assigning oxidation numbers:
0
0
S8 (s) + 12 O2 (g)
-2
S [0]
S[+6]
S is Oxidized
8 SO3 (g)
O[0]
O[-2]
O is Reduced
+6
S8 is the reducing agent and O2 is the oxidizing agent
c) Assigning oxidation numbers:
C[+2]
C[+4]
C is oxidized
-2
+2
NiO(s)
-2
+2
+ CO(g)
Ni[+2]
Ni[0]
Ni is Reduced
0
Ni(s)
-2
+4
+ CO2 (g)
CO is the reducing agent and NiO is the oxidizing agent
Activity Series of the Metals
Strongly
reducing
Weakly
reducing
Li
K
Ba
Ca
Na
Li+ + eK + + eBa2+ + 2 eCa2+ + 2 eNa+ + e-
Mg
Al
Mn
Zn
Cr
Fe
Mg2+ + 2eAl3+ + 3eMn2+ + 2eZn2+ + 2eCr3+ + 3eFe2+ + 2e-
Co
Ni
Sn
Co2+ + 2eNi2+ + 2eSn2+ + 2e-
H2
2 H+ + 2e-
Cu
Ag
Hg
Pt
Au
Cu2+ + 2eAg+ + eHg2+ + 2ePt2+ + 2eAu3+ + 3e-
These elements react rapidly with aqueous H+ ions
(acid) or with liquid H2O to release H2 gas.
These elements react with aqueous H+ ions or with
steam to release H2 gas.
These elements react with aqueous H+ ions to
release H2 gas.
These elements do not react with aqueous H+ ions
to release H2 gas.
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.
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+
2 mol Fe
x
= 0.133 mol Fe
2+
3 mol Pt
0.133 mol Fe x
55.85 g Fe
= ____________ g Fe
mol Fe
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.
REDOX Balancing using Ox. No. Method - I
+2 e-
0
___
2 H2 (g) +___ O2 (g)
0
- 1 e-
-2
___
2 H2O(g)
+1
electrons lost must = electrons gained therefore multiply
Hydrogen reaction by 2!
and we are balanced!
REDOX Balancing Using Ox. No. Method - II
+2
-1e-
Fe+2(aq) + MnO4-(aq) + H3O+(aq)
+3
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)
5 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)
5 Fe+3(aq) + Mn+2(aq) +12 H2O(aq)
Balancing Oxidation-Reduction Equations
Occurring in Acidic Solution by the HalfReaction Method.
Balancing
OxidationReduction
Equations
Occurring in
Basic
Solution by
the HalfReaction
Method.
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.
H+ + OH -
H2O
BASE : You may add either OH -, or water to either side of the
chemical equation.
H+ + OH H+ + H 2O
H2 O
H 3O+
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)
MnO4-(aq)
Fe+3(aq) + e-
[oxidation half-reaction]
Mn+2(aq)
add H+ to the reactants and that will give water as a product!
MnO4-(aq) + 8H3O+(aq) +5eSum the two half-reactions
Mn+2(aq) + 12H2O(l)
[reduction half-reaction]
{ Fe+2(aq)
MnO4-(aq) + 8H3O+(aq) +5eMnO4-(aq)+ 8H3O+(aq)+5e- +5Fe+2(aq)
Fe+3(aq) +e- } x5
Mn+2(aq) + 12H2O(l)
5Fe+3(aq)+5e- + Mn+2(aq)+ 12H2O(l)
REDOX Balancing by Half-Reaction Method -II
MnO4-(aq) + SO32-(aq)
MnO2 (s) + SO42-(aq) [basic solution]
Oxidation:
SO32SO42-(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 looses 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)
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 eMnO2 (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)
REDOX Balancing using Ox. No. Method - III
+7
+ 3 e-
MnO4-(aq) + SO32-(aq)
+4
( Acidic Solution )
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)
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
H+ion - 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)
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)
To convert to base, add two OH- to each side of the equation:
2MnO4-(aq)+ 3SO32-(aq)+2 H2O(l)
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)
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)
4 Zn2+(aq) + NH4+(aq) + 3 H2O(l)
REDOX Balancing by Half-Reaction Method-V
Given:
Oxidation:
Zn(s) + H3O+(aq) + NO3-(aq)
Zn(s)
Zn2+(aq) + NH4+(aq)
Zn2+ + 2 e-
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)
Reduction:
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 Zn+2(aq) + NH4+(aq) + 13 H2O(l)
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)
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 -
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 there by
accepting 6 electrons, so we add 6 electrons to the reactant side.
6e - + 14 H3O+(aq) + Cr2O72-(aq)
2 Cr+3(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 Cr+3(aq) + 21 H2O(l)] x 2
C2H5OH(l) + 16 H3O+(aq) + 2 Cr2O72-(aq)
2 CO2 (g) + 4 Cr+3(aq) + 27 H2O(l)
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 looses 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)
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)
REDOX Balancing Using Ox. No. Method -V
0
-1 e -
Ag(s) + CN -(aq) + O2 (g)
+1
Ag(CN)2-(aq) + OH -(aq)
0
+ 2 e-2
To balance electrons we must put a 4 in front of the Ag, since each
oxygen looses 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)
Redox Titration- Calculation outline - I
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
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.
Figure 4.21:
Permanganate
being
introduced into
a flask of
reducing agent.
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
-6
Mol CaC2O4 = 1.00 x 10 mol KMnO4 x
=
2 mol KMnO4
Mol CaC2O4 = 2.50 x 10 -6 mol CaC2O4
c) Moles of Ca+2
Mol
Ca+2
= 2.50 x
10 -6
mol CaC2O4 x
Mol Ca+2 = 2.50 x 10 -6 mol Ca+2
1 mol Ca+2
=
1 mol CaC2O4
Redox Titration - Calculation Outline - II
Moles of Ca2+/ 1 ml of blood
multiply by 100
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
Mass (mg) of Ca2+ / 100 ml blood
c) convert g to mg!
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
-6 mol Ca+2
Mol Ca+2
2.50
x
10
=
x 100 ml Blood =
100 ml Blood
1.00 ml Blood
Mol Ca+2
= 2.50 x 10 -4 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 = 0.0100 g Ca+2
c) mass (mg) of Ca+2
Mass Ca+2 = 0.0100g Ca+2 x 1000mg Ca+2/g Ca+2 = 10.0 mg Ca+2
100 ml Blood
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
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