CH 4: Chemical Reactions
Renee Y. Becker
Valencia Community College
CHM 1045
Solutions
• Solute – solid in liquid or lowest mass
quantity of substance
• Solvent- liquid solute is dissolved in or
highest mass quantity of substance
Solution Concentrations
• Concentration: allows us to measure out
a specific number of moles of a
compound by measuring the mass or
volume of a solution.
• Molarity(M) = Moles of Solute
Liters of Solution
moles = M•L
L = moles/M
Example: Solution Concentrations
• How many moles of solute are present
in 125 mL of 0.20 M NaHCO3?
Example: Solution Concentrations
• How many grams of solute would you
use to prepare 500.00 mL of 1.25 M
NaOH?
Solution Concentrations
• Dilution: the process of reducing a solution’s
concentration by adding more solvent.
Moles of solute(constant) = Molarity 
Volume
Mi • Vi = Mf • Vf
Mf = (Mi • Vi) / Vf
Vf = (Mi • Vi) / Mf
Example: Solution Concentrations
• What volume of 18.0 M H2SO4 is required to
prepare 250.0 mL of 0.500 M H2SO4?
Example: Solution Concentrations
• What is the final concentration if 75.0 mL of
3.50 M glucose is diluted to a volume of 400.0
mL?
Solution Stoichiometry
• Titration: a technique for determining the
concentration of a solution
– Standard solution: known concentration
– If you have a known volume of standard
solution and use it to titrate a known
volume of an unknown concentrated
solution you can calculate to find the
number of moles in the unknown and
therefore find it’s concentration
Titration
• When doing a titration you add titrant (standard
solution) to the analyte (unknown concentration
solution) until the endpoint or the equivalence
point is reached. This point is when you have
equal moles of titrant and analyte, from the
volume of the titrant and analyte used and the
molarity of the titrant, you can find the molarity of
the analyte
– Endpoint- based on an indicator
– Indicator- a substance that changes color in a specific
pH range
– Equivalence point- not based on an indicator, usually a
pH meter
– Use Manalyte• Vanalyte = Mtitrant • Vtitrant
Example: Solution Stoichiometry
• A 25.0 mL sample of vinegar (dilute
CH3CO2H) is titrated and found to react with
94.7 mL of 0.200 M NaOH. What is the
molarity of the acetic acid solution?
NaOH(aq) + CH3CO2H(aq)  CH3CO2Na(aq) + H2O(l)
Solution Stoichiometry
Oxidation–Reduction Reactions
• Assigning Oxidation Numbers: All atoms
have an “oxidation number” regardless of
whether it carries an ionic charge.
1. An atom in its elemental state has an
oxidation number of zero.
Elemental state as indicated by single elements with no
charge. Exception: diatomics H2 N2 O2 F2 Cl2 Br2 and I2
Oxidation–Reduction Reactions
2. An atom in a monatomic ion has an oxidation
number identical to its charge.
Oxidation–Reduction Reactions
3. An atom in a polyatomic ion or in a molecular compound
usually has the same oxidation number it would have if
it were a monatomic ion.
A. Hydrogen can be either +1 or –1.
B. Oxygen usually has an oxidation number of –2.
In peroxides, oxygen is –1.
C. Halogens usually have an oxidation number of –1.
• When bonded to oxygen, chlorine, bromine, and
iodine have positive oxidation numbers.
Oxidation–Reduction Reactions
4. The sum of the oxidation numbers must be
zero for a neutral compound and must be
equal to the net charge for a polyatomic ion.
A. H2SO4 neutral atom, no net charge
SO42- sulfate polyatomic ion
[SO4]2- [Sx O42-] = -2
X + -8 = -2
X = 6 so sulfur has an oxidation # of +6
Oxidation–Reduction Reactions
B. ClO4– , net charge of -1
[ClO4]-1 [Clx O42-] = -1
X + -8 = -1
X = 7 so the oxidation number of chloride is
+7
Example: Oxidation–Reduction Reactions
Assign oxidation numbers to each atom in the following
substances:
A. CdS
F. VOCl3
B. AlH3
G. HNO3
C. Na2Cr2O7
H. FeSO4
D. SnCl4
I. Fe2O3
E.
MnO4–
J. V2O3
Quiz
What is the oxidation number of arsenic in
AsO43- ?
• Electrolytes: Dissolve in
Electrolytes in Solution
water to produce ionic
solutions.
• Nonelectrolytes: Do not
form ions when they
dissolve in water.
a) NaCl sol’n conducts
electricity, completes circuit
(charged particles)
b) C6H12O6 does not
Electrolytes in Solution
• Dissociation:
The process by which a compound splits up to
form ions in the solution.
Electrolytes in Solution
• Strong Electrolyte: Total dissociation when
dissolved in water.
• Weak Electrolyte: Partial dissociation when
dissolved in water.
Types of Reactions
1. Precipitation
2. Acid-base neutralization
3. Oxidation-reduction (redox)
4. Metathesis Reactions (double
replacement)
Types of Chemical Reactions
• Precipitation Reactions: A process in which
an insoluble solid precipitate drops out of the
solution.
• Most precipitation reactions occur when the
anions and cations of two ionic compounds
change partners. (double replacement)
Pb(NO3)2(aq) + 2 KI(aq)  2 KNO3(aq) + PbI2(s)
Solubility Rules & Precipitation
• Allow you to predict whether a reactant or a
product is a precipitate.
• Soluble compounds are those which dissolve
to more than 0.01 M.
• There are three basic classes of salts:
Solubility Rules & Precipitation
1. Salts which are always soluble:
• All alkali metal salts: Cs+, Rb+, K+, Na+, Li+
• All ammonium ion (NH4+) salts
• All salts of the NO3–, ClO3–, ClO4–, C2H3O2–
, and HCO3– ions
Solubility Rules & Precipitation
2. Salts which are soluble with exceptions:
• Cl–, Br–, I– ion salts except with Ag+, Pb2+,
& Hg22+
• SO42– ion salts except with Ag+, Pb2+,
Hg22+, Ca2+, Sr2+, & Ba2+
Solubility Rules & Precipitation
3. Salts which are insoluble with exceptions:
• O2– & OH– ion salts except with the alkali
metal ions, and Ca2+, Sr2+, & Ba2+ ions
• CO32–, PO43–, S2–, CrO42–, & SO32– ion salts
except with the alkali metal ions and the
ammonium ion
• If not listed the compound is probably
insoluble
Example: Solubility Rules & Precipitation
• Predict the solubility of:
(a) CdCO3
(b) MgO
(d) PbSO4
(e) (NH4)3PO4
(c) Na2S
• Write the balanced reaction and predict whether a
precipitate will form for:
(a) NiCl2 (aq) + (NH4)2S (aq) 
(b) Na2CrO4 (aq) + Pb(NO3)2 (aq) 
(c) AgClO4 (aq) + CaBr2 (aq) 
Solubility Rules & Precipitation
Equations
• Molecular equation – as the reaction is
• Complete ionic equation – all broken up
into ions(only aqueous solutions)
• Net ionic equation – cancel out spectator
ions
Net Ionic Equations for Precipitation Reactions
• Write net ionic equation for the following reaction:
2 AgNO3(aq) + Na2CrO4(aq)  Ag2CrO4(s) + 2 NaNO3(aq)
1.
2.
3.
4.
Is it balanced? If not do it!
Separate all except solids into ions (complete ionic equation)
Cancel out spectator ions on both sides
Rewrite (net ionic equation)
Types of Chemical Reactions
• Acid–Base Neutralization: A process in
which an acid reacts with a base to yield
water plus an ionic compound called a salt.
• The driving force of this reaction is the
formation of the stable water molecule.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Acid–Base Concepts
• Arrhenius Acid:
A substance which dissociates in water to
form hydrogen ions (H+).
• Arrhenius Base:
A substance that dissociates in, or reacts
with, water to form hydroxide ions (OH–).
Limitations: Has to be an aqueous solution and doesn’t
account for the basicity of substances like NH3.
Acid–Base Concepts
• Brønsted Acid: Can donate protons (H+) to
another substance.
• Brønsted Base: Can accept protons (H+)
from another substance. (NH3)
Example : Conjugate acid-base pairs
For the following reactions label the acid,
base, conjugate acid, and conjugate base.
CH3CO2H(aq) + H2O(l)  H3O+(aq) + CH3CO2-(aq)
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
Quiz
Which of the following is a Bronsted-Lowry base
but not an Arrhenius base?
1.
2.
3.
4.
NaOH
NH3
Mg(OH)2
KOH
Acids and Bases
• Strong acid - st. electrolyte, almost completely
dissociates in water
– HCl, H2SO4, HNO3, HClO4, HI, HBr
• Weak acid - wk. electrolyte, does not dissociate well in
water
– HF, HCN, CH3CO2H
• Strong base - st. electrolyte, almost completely
dissociates in water
– Metal hydroxides
• Weak base - does not dissociate well in water
– NH3
Types of Chemical Reactions
• Metathesis Reactions: These are reactions
where two reactants just exchange parts.
(double replacement)
AX + BY  AY + BX
BaCl2(aq) + K2SO4(aq)  BaSO4(s) + 2 KCl(aq)
This is also a ppt reaction, if I ask you what type of
reaction is it, what is the best answer??
Types of Chemical Reactions
• Oxidation–Reduction (Redox) Reaction: A
process in which one or more electrons are
transferred between reaction partners.
• The driving force of this reaction is the
decrease in electrical potential.
Mg(s) + I2(g) 
MgI2(s)
Oxidation :
Mg0  Mg2+ + 2 electrons
Reduction:
I20 + 2 electrons  I21-
Quiz
Which of the following is not an acid-base
neutralization reaction?
1.
HCl(aq) + NaOH(s)  NaCl(aq) + H2O(l)
2.
2 HF(aq) + Mg(OH)2(aq)  MgF2(aq) + 2 H2O(l)
3.
Pb(NO3)2(aq) + 2 KI(aq)  PbI2 (s) + 2 KNO3(aq)
Oxidation–Reduction Reactions
• Redox reactions are those involving the
oxidation and reduction of species.
• Oxidation and reduction must occur together.
They cannot exist alone.
Fe2+ + Cu0  Fe0 + Cu2+
Reduced:
Oxidized:
Iron gained 2 electrons
Copper lost 2 electrons
Fe2+ + 2 e  Fe0
Cu0  Cu2+ + 2e
• Remember that electrons are negative so if you gain
electrons your oxidation # decreases and if you lose
electrons your oxidation # increases
Oxidation–Reduction Reactions
Fe2+ + Cu0  Fe0 + Cu2+
• Fe2+ gains electrons, is reduced, and we call it an
oxidizing agent
– Oxidizing agent is a species that can gain
electrons and this facilitates in the oxidation of
another species. (electron deficient)
• Cu0 loses electrons, is oxidized, and we call it a
reducing agent
– Reducing agent is a species that can lose
electrons and this facilitates in the reduction of
another species. (electron rich)
Quiz
Which is a reduction half reaction?
1. Fe  Fe2+ + 2e
2. Fe2+  Fe3+ + 1e
3. Fe  Fe3+ + 3e
4. Fe3+ + 1e  Fe2+
Example: Oxidation–Reduction Reactions
For each of the following, identify which species
is the reducing agent and which is the
oxidizing agent.
A) Ca(s) + 2 H+(aq)  Ca2+(aq) + H2(g)
B) 2 Fe2+(aq) + Cl2(aq)  2 Fe3+(aq) + 2 Cl–(aq)
C) SnO2(s) + 2 C(s)  Sn(s) + 2 CO(g)
Oxidation–Reduction Reactions
Balancing Redox Reactions
• Half-Reaction Method: Allows you to focus
on the transfer of electrons. This is important
when considering batteries and other aspects
of electrochemistry.
• The key to this method is to realize that the
overall reaction can be broken into two parts,
or half-reactions. (oxidation half and reduction
half)
Balancing Redox Reactions
• Balance the following using the half reaction
method
Mg(s) + N2(g)  Mg3N2(s)
Ca(s) + Cl(g)  CaCl2(s)
FeI3(aq) + Mg(s)  Fe(s) + MgI2(aq)
H2(g) + Ag+(aq)  Ag(s) + H+(aq)