CH 4: Chemical Reactions
Renee Y. Becker
Valencia Community College
CHM 1045
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Solutions
• Solute – solid in liquid or lowest mass quantity
of substance
• Solvent- liquid solute is dissolved in or highest
mass quantity of substance
2
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
3
Example 1: Solution Concentrations
• How many moles of solute are present in
125 mL of 0.20 M NaHCO3?
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Example 2: Solution Concentrations
• How many grams of solute would you use
to prepare 500.00 mL of 1.25 M NaOH?
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Solution Concentrations
• Dilution: the process of reducing a solution’s
concentration by adding more solvent.
Moles of solute(constant) = Molarity x Volume
Mi • Vi = Mf • Vf
Vf = (Mi • Vi) / Mf
Mf = (Mi • Vi) / Vf
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Example 3: Solution Concentrations
• What volume of 18.0 M H2SO4 is required to
prepare 250.0 mL of 0.500 M H2SO4?
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Example 4: 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?
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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
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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
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Example 5: Solution Stoichiometry
• A 25.0 mL sample of vinegar (dilute CH3CO2H) is
titrated and found to react with 94.7 mL of a
0.200 M NaOH. What is the molarity of the acetic
acid solution?
NaOH(aq) + CH3CO2H(aq)  CH3CO2Na(aq) + H2O(l)
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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 I212
Oxidation–Reduction Reactions
2. An atom in a monatomic ion has an oxidation
number identical to its charge.
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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.
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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
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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
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Example 6: Oxidation–Reduction Reactions
Assign oxidation numbers to each atom in the following:
A. CdS
F. VOCl3
B. AlH3
G. HNO3
C. Na2Cr2O7
H. FeSO4
D. SnCl4
I. Fe2O3
E. MnO4–
J. V2O3
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• 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
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Electrolytes in Solution
• Dissociation:
The process by which a compound splits up to form
ions in the solution.
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Electrolytes in Solution
• Strong Electrolyte: Total dissociation when dissolved in water.
• Weak Electrolyte: Partial dissociation when dissolved in
water.
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Types of Reactions
1. Precipitation
2. Acid-base neutralization
3. Oxidation-reduction (redox)
4. Double replacement
5. Single replacement
6. Combination
7. Decomposition
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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)
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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:
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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
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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+
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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
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Example 7: Solubility Rules & Precipitation
• Predict the solubility of the following in water:
(a) CdCO3
(b) MgO
(c) Na2S
(d) PbSO4
(e) (NH4)3PO4
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Example 8: Solubility Rules & Precipitation
• 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) 
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Equations
• Molecular equation – Balanced reaction
2 FeBr3(aq) + 3 Pb(NO3)2(aq)  2 Fe(NO3)3(aq) + 3 PbBr2(s)
• Complete ionic equation – All broken up into ions
(only aqueous solutions)
2 Fe3+(aq) + 6 Br-(aq) + 3 Pb2+(aq) + 6 NO3-(aq)  2 Fe3+(aq) + 6 NO3-(aq) + 3 PbBr2(s)
• Net ionic equation – Cancel out spectator ions
3 Pb2+(aq) + 6 Br-(aq)  3 PbBr2(s)
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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.
Is it balanced? If not do it! (molecular equation)
2.
Separate all aqueous sol’n into ions (complete ionic equation)
3.
Cancel out spectator ions on both sides
4.
Rewrite (net ionic equation)
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Example 9:
• Write the ME, CIE, and NIE for the following
reaction
Na2CrO4 (aq) + Pb(NO3)2 (aq)  NaNO3(aq) + PbCrO4(s)
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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)
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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.
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Acid–Base Concepts
• Brønsted Acid: Can donate protons (H+) to
another substance.
• Brønsted Base: Can accept protons (H+) from
another substance. (NH3)
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Example 10: 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)
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Acid–Base Concepts
• Lewis Acid: Electron pair acceptor.
Al3+, H+, BF3.
• Lewis Base: Electron pair donor.
H2O, NH3, O2–.
• Bond formed is called
a coordinate bond or dative bond.
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Example 11
Which of the following is a Bronsted-Lowry base
but not an Arrhenius base?
1.
2.
3.
4.
NaOH
NH3
Mg(OH)2
KOH
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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
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Acid–Base Concepts
Other Weak bases – trimethyl ammonia N(CH3)3, C5H5N pyridine,
ammonium hydroxide NH4OH, H2O water
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ME, CIE, NIE for Acids/Bases
Strong Acid Strong Base
ME: HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Complete Ionic Equation:
H+ + Cl- + Na+ + OH-  H2O(l) + Na+ + ClNet Ionic Equation:
H+ + OH-  H2O(l)
or
H3O+ + OH-  2 H2O(l)
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ME, CIE, NIE for Acids/Bases
Weak Acid Strong Base
ME: HF(aq) + NaOH(aq)  H2O(l) + NaF(aq)
Complete Ionic Equation:
HF + Na+ + OH-  H2O(l) + Na+ + FNet Ionic Equation:
HF + OH-  H2O(l) + F-
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Example 12: ME, CIE, NIE for Acids/Bases
Write ME, CIE and NIE for the following:
(a) NaOH(aq) + CH3CO2H(aq) 
(b) HCl(aq) + NH3(aq) 
• NaOH strong base will dissociate well
• CH3CO2H weak acid doesn’t dissociate well
• HCl is a strong acid and therefore a strong electrolyte
• NH3 is a weak base and is a weak electrolyte
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Types of Chemical Reactions
• Double Replacement: These are reactions where
two reactants just exchange parts. (double
displacement)
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??
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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  I2144
Example 12:
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)
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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: Iron gained 2 electrons
Oxidized: 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
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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)
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Example 13:
Which is a reduction half reaction?
1. Fe  Fe2+ + 2e
2. Fe2+  Fe3+ + 1e
3. Fe  Fe3+ + 3e
4. Fe3+ + 1e  Fe2+
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Example 14: 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)
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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 halfreactions. (oxidation half and reduction half)
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Balancing Redox Reactions
Balance for an acidic solution:
MnO4–(aq) + Br–(aq)  Mn2+(aq) + Br2(aq)
1. Determine oxidation and reduction half-reactions:
Oxidation half-reaction: Br–(aq)  Br20(aq)
Reduction half-reaction: MnO4–(aq)  Mn2+(aq)
2. Balance for atoms other than H and O:
Oxidation: 2 Br–(aq)  Br2(aq)
Reduction: MnO4–(aq)  Mn2+(aq)
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Balancing Redox Reactions
3. Balance for oxygen by adding H2O to the side
with less oxygen
Oxidation: 2 Br–(aq)  Br2(aq)
Reduction: MnO4–(aq)  Mn2+(aq) + 4 H2O(l)
4. Balance for hydrogen by adding H+ to the side
with less hydrogens
Oxidation: 2 Br–(aq)  Br2(aq)
Reduction: MnO4–(aq) + 8 H+(aq)  Mn2+(aq) + 4 H2O(l)
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Balancing Redox Reactions
5. Balance for charge by adding electrons (e–):
Oxidation: 2 Br–(aq)  Br2(aq) + 2 e–
Reduction: MnO4–(aq) + 8 H+(aq) + 5 e–  Mn2+(aq) + 4 H2O(l)
6. Balance for numbers of electrons by multiplying:
Oxidation: 5[2 Br–(aq)  Br2(aq) + 2 e–]
Reduction: 2[MnO4–(aq) + 8 H+(aq) + 5 e–  Mn2+(aq) + 4 H2O(l)]
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Balancing Redox Reactions
7. Combine and cancel to form one equation:
Oxidation: 10 Br–(aq)  5 Br2(aq) + 10 e–
Reduction: 2 MnO4–(aq) + 16 H+(aq) + 10 e– 2 Mn2+(aq) + 8 H2O(l)
2 MnO4–(aq) + 10 Br–(aq) + 16 H+(aq) 2 Mn2+(aq) + 5 Br2(aq) + 8 H2O(l)
We will not be balancing in basic solutions!! (until CHM 1046)
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Example 15: Balancing Redox Reactions
Balance the following in an acidic sol’n
NO3–(aq) + Cu(s)  NO(g) + Cu2+ (aq)
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Types of Reactions
• A single-replacement reaction is a a reaction where a
more active metal displaces another, less active metal
in a compound.
• If a metal precedes another in the activity series, it
will undergo a single-replacement
reaction:
Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)
FeSO4(aq) + Cu(s) → NR
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Activity Series
• Metals that are most reactive appear first in the
activity series.
• Metals that are least reactive appear last in the
activity series.
• The relative activity series is:
Li > K > Ba > Sr > Ca > Na > Mg >
Al > Mn > Zn > Fe > Cd > Co > Ni >
Sn > Pb > (H) > Cu > Ag > Hg > Au
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Types of Reactions
• A combination reaction is a reaction where
two simpler substances are combined into a
more complex compound.
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Types of Reactions
• In a decomposition reaction, a single
compound is broken down into simpler
substances.
• Heat or light is usually required to start a
decomposition reaction. Ionic compounds
containing oxygen often decompose into a
metal and oxygen gas.
2 HgO(s) → 2 Hg(l) + O2(g)
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