CHAPTER 4 Types of Chemical Reactions and Solution Stoichiometry

Chapter 4
Types of Chemical Reactions and Solution
Water is the dissolving medium of the common
solvent: Some properties
Water is “bent” or v-shaped
The O—H bonds are covalent
Water is a polar molecule
Hydration occurs when salts dissolve in water
The Water Molecule is Polar
Polar Water Molecules Interact with the
Positive and Negative Ions of a Salt
A solute (substance being dissolved):
• Dissolves in water or other solvent
• Changes phase if different from the solvent
• Present in lesser amount (if the same phase as
the solvent)
A solvent:
• Retains its phase (if different from the solute)
• Present in greater amount (if the same phase as
the solute)
• Rule for predicting solubility:
Like dissolves Like
The Nature of Aqueous Solutions:
Strong and Weak Electrolytes
• Strong electrolytes: Conduct current very
efficiently. Strong electrolytes are completely
ionized when they are dissolved in water (soluble
salts, strong acids, strong bases).
• Weak electrolytes: Conduct only a small
current. Weak electrolytes are substances that
exhibit a small degree of ionization in water
(weak acids, weak bases).
• Non electrolytes: Permit no current to flow. Non
electrolytes are substances that dissolve in water
but do not produce any ions. (pure water, sugar
BaCI2 Dissolving
HCI (aq) is Completely Ionized
Acetic Acid in Water
The Composition of Solutions
To perform stoichiometric calculation in solution
we need• The nature of the reaction
• The amounts of chemicals present in the
solutions (concentration)
• The concentration of a solution can be expressed
by Molarity (M).
• Molarity (M) = moles of solute per volume of
solution in liters:
M = molarity = (moles of solute)/(liters of
Example: If you weigh out 5.84 g of NaCl and
dissolved it in 500. mL of water, what is the
molarity of the solution?
5.84 g NaCl x (1 mol NaCl)/(58.44 g NaCl) =
0.100 mol
500. mL x 1 L/1000 mL = 0.500 L
Molarity = (0.100 moles)/(0.500 L) = 0.200 M
Concentration of Ions
Example: Give the concentration of each type of ion in the
following solution:
a. 0.50 M Co(NO3)2 b. 1 M Fe(ClO4)3 c. 1.75 M Al2(SO4)3
a. Solution reaction: Co(NO3)2(s) H2O Co2+(aq) + 2NO3-(aq)
Each mole of Co(NO3)2 that is dissolved, the solution contains 1
mol Co2+ ions and 2 mol NO3- ions. Thus a solution of 0.50 M
Co(NO3)2 contains 0.50 M Co2+ and (2 x 0.50) M or 1.0 M NO3-.
b. Fe(ClO4)3(s) H2O Fe3+(aq) + 3ClO4-(aq)
1 M Fe(ClO4)3 contains 1 M Fe3+ ions and 3 M ClO4- ions.
c. Al2(SO4)3(s) H2O 2Al3+(aq) + 3SO42-(aq)
1.75 M Al2(SO4)3 contains (2 x 1.75) M or 3.50 M Al3+ ions and (3
x 1.75) M or 5.25 M SO42- ions.
Standard Solution
• A standard solution is a solution whose
concentration is accurately known.
Finding volume or weight:
• How many grams of K2Cr2O7 are required to
prepare 1.00 L of a 0.20 M solution?
M = moles/L
Moles = M x L = 0.200 M x 1.00 L = 0.200 mol
0.200 mol K2Cr2O7 x (294.20 g)/(1 mol) = 58.8 g
• What volume is needed to prepare a 1.20 M
solution from 10.0 g CaCl2 ? (mm = 111.0 g)
M = moles/L ==> L = moles/M
10.0 g CaCl2 x (1 mole CaCl2 )/111.0 g CaCl2)
= 0.090 moles
L = (0.090 moles)/(1.20 M) = 0.0750 L
= 75.0 mL
Preparation of a Standard Solution
• The desired molarity solutions are often
prepared from concentrated stock solutions
(routinely used solutions prepared in
concentrated form) by adding water. This
process is called dilution.
• Dilution with water does not alter the numbers
of moles of solute present.
Moles of solute before dilution
= moles of solute after dilution
M1V1 = M2V2
(a) A Measuring Pipet
(b) A Volumetric (transfer) Pipet
Example: What volume of 16 M sulfuric acid
must be used to prepare 1.5 L of 0.10 M H2SO4
V1 = Volume before dilution = ?
M1 = Concentration before dilution = 16 M
V2 = Volume after dilution = 1.5 L
M2 = Concentration after dilution = 0.10 M
M1V1 = M2V2
V1 =M2V2 / M1 = (0.10 M x 1.5 L)/ 16 M
= 0.0094 L = 9.4 mL
Types of Chemical Reactions
Solution Reactions are classified as follows:
• Precipitation reactions
• Acid – base reactions
• Oxidation – reduction reactions
Other classification:
• Formation reaction
• Decomposition reaction
• Single replacement
• Double replacement
Precipitation Reactions
When two solutions are mixed, an insoluble
substance (solid) sometimes forms and separates
from the solution. Such a reaction is called a
precipitation reaction, and the solid that forms
is called a precipitate.
Example: K2CrO4(aq) + Ba(NO3)2(aq)  Products
Contains the ions: K+, CrO42-, Ba2+, NO3• Product must contain both anions and cations
• Most ionic products contain one type of cation
and one type of anion
… Precipitation Reactions continued…
The possible combinations areK2CrO4, KNO3, BaCrO4, Ba(NO3)2
The only real possibility for the solid yellow
products are- KNO3 or BaCrO4
Both K+ and NO3- ions are colorless but CrO42ion is yellow, so the yellow solid is BaCrO4.
K2CrO4(aq) + Ba(NO3)2(aq)  BaCrO4(s) +
The Reaction of K2CrO4(aq) and Ba(NO3)2(aq)
Describing Reactions in Solutions
• Regular balanced equation or molecular equation: (gives
the overall reaction stoichiometry but not the actual
forms of reactants and products)
K2CrO4(aq) + Ba(NO3)2(aq)  BaCrO4(s) + 2KNO3(aq)
• Complete ionic equation: (represents as ions all
reactants and products that are strong electrolyte)
2K+(aq) + CrO42-(aq) + Ba2+(aq) + 2NO3-(aq)
 BaCrO4(s) + 2K+(aq) + 2NO3-(aq)
• Net ionic equation: (Ions participate in reaction)
Ba2+(aq) + CrO42-(aq)  BaCrO4(s)
• Spectator ions: The ions that do not
participate directly in the reaction, present in
solution both before and after the reaction.
K+ and NO3- ions.
2K+(aq) + CrO42-(aq) + Ba2+(aq) + 2NO3-(aq)
 BaCrO4(s) + 2K+(aq) + 2NO3-(aq)
Simple Rules for Solubility
1.Most nitrate (NO3) salts are soluble.
2.Most alkali (group 1A) salts and NH4+ are
3.Most Cl, Br, and I salts are soluble (NOT
Ag+, Pb2+, Hg22+)
4.Most sulfate salts are soluble (NOT BaSO4,
PbSO4, HgSO4, CaSO4)
5.Most OH salts are only slightly soluble (NaOH,
KOH are soluble, Ba(OH)2, Ca(OH)2 are
marginally soluble)
6.Most S2, CO32, CrO42, PO43 salts are only
slightly soluble.
Stoichiometry of Precipitation Reactions
Step 1: Identify the species present in the
combined solution, and determine what reaction
Step 2: Write the balanced net ionic equation for
the reaction.
Step 3: Calculate the moles of the reactants (use
volume and molarity).
Step 4: Determine which reactant is limiting.
Step 5: Calculate the moles of product or products,
as required.
Step 6: Convert to grams or other units, as
Example: Calculate the mass of solid NaCl that
must be added to 1.50 L of a 0.100 M AgNO3
solution to precipitate all the Ag+ ions in the
form of AgCl.
Ag+ NO-3 Na+ ClAg+(aq) + Cl-(aq)  AgCl(s)
1.50 L x (0.100 mol Ag+) = 0.150 mol Ag+
Ag+ and Cl- react in a 1:1 ratio, 0.150 mol Clions and 0.150 mol NaCl x (58.45 g NaCl)/(1
mol NaCl) = 8.77 g NaCl
Acid-Base Reactions
• Acid is a substance that produces H+ ions when
dissolved in water
• Base is substances that produces OH- ions
• An acid is a proton donor
• A base is a proton acceptor
Balanced equation: HCl + NaOH  NaCl + H2O
Complete ionic equation: H+ + Cl- + Na+ + OH- 
Na+ + Cl- + H2O
Net ionic equation: H+ + OH- H2O
Neutralization Reaction
Example: What volume of a 0.100 M HCl solution is
needed to neutralize 25.0 mL of 0.350 M NaOH?
The species available for reaction are – H+, Cl-, Na+, OHNaCl is soluble, so Na+, Cl- are spectator ions, net ionic
H+(aq)+ OH-(aq)  H2O(l)
Moles of OH- ions: 25.0 mL x (1L)/(1000 mL) x (0.350
mol OH- )/(L NaOH) = 8.75 x 10-3 mol OHH+and OH- ions react in a 1:1 ratio, 8.75 X 10-3 mol H+
ions is required to neutralize OH- ions present.
V x (0.100 mol H+)/L = 8.75 x 10-3 mol H+
V = [(8.75 x 10-3 mol H+)]/[(0.100 mol H+)/L] =
8.75 x 10-2 L x (1000 mL)/1L = 87.5 mL
Acid-Base Titration
• Acid-base reaction is called a neutralization reaction (the
acid and base are destroyed to leave a “neutral” solution).
• Volumetric analysis is a technique for determining the
amount of a certain substance by doing a titration.
• A titration involves delivery of a measured volume of a
solution of known concentration (the titrant) into a solution
containing the substance being analyzed (the analyte).
• Equivalence point or stoichiometric point: enough titrant
added to react exactly with the analyte.
• Indicator: A substance added at the beginning of the
titration that changes color at or near the equivalence point.
• End point: The point where the indicator actually changes
Oxidation-Reduction Reactions
• In oxidation-reduction (redox) reaction, one
species loses electrons while another species gains
• Both processed must occur simultaneously. We
can’t have a net ionic reaction that contain only
oxidation or only reduction.
• The number of electron lost by the oxidation
process must equal the number gained by the
reduction process. We cannot create or destroy
Cu  Cu2+ + 2e- (oxidation)
2Ag+ + 2e-  2Ag (reduction)
Cu + 2Ag+  Cu2+ +2Ag (net ionic reaction)
Oxidation States
Oxidation states or oxidation numbers provides
a way to keep track of electrons (accounting
for electrons) in redox reactions.
• Rules for Assigning Oxidation States:
1. The oxidation state of an atom in an element is
O (element in its elemental form is O).
2. The oxidation state of a monatomic ion is the
same as its charge.
3. In its compound, fluorine is always assigned an
oxidation state of –1.
…Oxidation States continued…
4. Oxygen is assigned an oxidation state –2 in
covalent compound (except in peroxides
where it is –1)
5. Hydrogen is assigned an oxidation state +1 in
its covalent compounds with non metals.
6. Sum of the oxidation states must be zero for
an electrically neutral compound, for an ion
equals the charge of the ion.
Assigning Oxidation States
Example: Assign oxidation states to all atoms in the
a. CO2  O = -2 (for each oxygen); C = +4
b. SF6  F = -1 (for each fluorine); S = +6
c. NO3-  O = -2 (each oxygen); N = +5
d. MnO4-  O = -2 (each oxygen); Mn = +7
e. HSO3-  O = -2 (each oxygen); H = +1, S = +4
f. H2O  H = +1 (each); O = -2
g. Li3N  Li = +1 (each); N = -3
• Oxidation is an increase in oxidation state or a loss of
• Reduction is a decrease in oxidation state or a gain of
• Oxidizing Agent is the electron acceptor (reduced).
• Reducing Agent is the electron donor (oxidized)
PbO(s) + CO(g)  Pb(s) + CO2(g)
+2 -2 +2 -2
+4 -2 (each)
Pb  reduced, C  oxidized, PbO  oxidizing agent,
CO  reducing agent
A Summary of an Oxidation-Reduction Process
Balancing Oxidation-Reduction Equations
Redox reactions in acidic solution
Step 1: Write separate equation for the oxidation and
reduction half reactions.
Step 2: For each half reactiona. Balance all the elements except hydrogen and oxygen
b. Balance oxygen using H2O
c. Balance hydrogen using H+
d. Balance the charge using electrons
Step 3: If necessary, multiply one or both balanced halfreactions by an integer to equalize the number of
electrons transferred
Step 4: Add the half reactions and cancel identical species
Step 5: Check that the elements and charges are balanced
Balancing Oxidation-Reduction Reactions
Example: Balance the equation for the reaction
between permanganate and iron (II) ion in acidic
MnO4-(aq) + Fe2+(aq) Acid Fe3+(aq) + Mn2+(aq)
Half reactions: MnO4-  Mn2+ (reduction)
Fe2+  Fe3+ (oxidation)
Balance half reactions: MnO4-  Mn2+
a. Manganese is balanced
b. Balance oxygen by adding 4H2O to the right
side of the equation
c. Balance hydrogen by adding 8H+ to the left side
8H+(aq) + MnO4-(aq)  Mn2+(aq) + 4H2O
d. Equalize the charge by adding electrons
5e- + 8H+(aq) + MnO4-(aq)  Mn2+(aq) + 4H2O(l)
Fe2+(aq)  Fe3+(aq) + e- x 5
5 Fe2+(aq)  5 Fe3+(aq) + 5eAdd half reactions –
5Fe2+(aq) + MnO-4(aq) + 8H+(aq)
 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
Check that elements and charges are balanced.
Balancing Oxidation-Reduction
Reactions in Basic Solution
• Step 1: Use the half reaction method as
specified for acidic solution.
• Step 2: Add a number of OH- ions that is equal
to the number of H+ ions (to both sides).
• Step 3: Form H2O by combining H+ and OH-.
• Step 4: Check that elements and charges are
ClO- + CrO2-  Cl - + CrO42- (basic solution)
-1 +6
First half reaction:
ClO- +2e-  ClClO- + 2e-  Cl- + H2O
ClO- + 2e- +2H+  Cl- + H2O
Second half reaction:
CrO2-  CrO42CrO2-  CrO42- + 3eCrO2- + 2H2O  CrO42- + 3eCrO2- + 2H2O  CrO42- + 3e- + 4H+
Multiply first half reaction by 3 and second half reaction
by 2 then add two half reactions.
3ClO- + 6e- + 6H+  3Cl- + 3H2O
2CrO2- + 4H2O  2CrO42- + 6e- + 8H+
3ClO- + 2CrO2- + H2O  3Cl- + 2CrO42- + 2H+
Add OH- to both sides to remove H+
3ClO- + 2CrO2- + H2O + 2OH-  3Cl- + 2CrO42- + 2H+ + 2OH3ClO- + 2CrO2- + H2O + 2OH-  3Cl- + 2CrO42- + 2H2O
3ClO- + 2CrO2- + 2OH-  3Cl- + 2CrO42- + H2O
Like dissolves Like
Strong, weak and non electrolytes
Molarity: moles of solute/Liters of solution
Standard Solution: concentration accurately known
Dilution: M1V1 = M2V2
Precipitation, acid-base and oxidation reduction reactions
Spectator ions
Acids: produces H+ or proton donor
Bases: produces OH- and proton acceptor
Acid-Base titration: Neutralization reaction
Oxidation-Reduction reactions
Oxidation States
Balancing redox equations

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