Chapter 5
Chemical Reactions
Amedeo Avogadro
1776-1856.
Avogadro’s Law.
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Properties of Solutes in Aqueous Solution
There are two types of aqueous solutions:
Ionic Compounds Dissolved in Water
Molecular Compounds Dissolved in Water
A. Ionic Compounds Dissolved in Water
Ions dissociate in water:
M+X- (aq)
M+ (aq) + X- (aq)
In solution, each ion is surrounded by water molecules.
Transport of ions through solution causes flow of current.
These solutions conduct electricity.
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Ionic NaCl (Na+ and Cl-) dissolving in water
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Properties of Solutes in Aqueous Solution
B. Molecular Compounds in Water
Molecular compounds in water (e.g., CH3OH): no
ions are formed.
If there are no ions in solution, there is nothing to
transport electric charge.
These solutions do NOT conduct electricity.
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Molecular (non-ionic) CH3OH dissolving in water
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Properties of Solutes in Aqueous Solution
Ions in solution are called electrolytes.
Strong and Weak Electrolytes
Strong electrolytes:
completely (100%) dissociate in solution.
For example:
HCl (aq)
NaCl (aq)
H+ (aq) + Cl- (aq)
Na+ (aq) + Cl- (aq)
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Properties of Solutes in Aqueous Solution
Strong and Weak Electrolytes
Weak electrolytes: produce a small concentration of ions
when they dissolve. Only partially dissociated.
These ions exist in equilibrium with the unionized
substance.
For example:
CaCO3(s)
HC2H3O2 (aq)
Ca2+ (aq) + CO32- (aq)
H+ (aq) + C2H3O2- (aq)
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Metathesis (Exchange) Reaction
(also called “molecular reactions”)
Metathesis reactions involve swapping ions in solution:
AX + Z  AZ + X
AX + BY  AY + BX.
Metathesis reactions will lead to a change in
solution if one of two things occurs:
•an insoluble solid is formed (precipitate),
•an insoluble gas is formed.
Let’s first look at insoluble solid formation (precipitates)
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Metathesis Reactions: formation of precipitate
Molecular equation:
Pb(NO3)2(aq) +2 KI(aq) --->
PbI2(s ) + 2KNO3(aq)
Note solid (yellow) PbI2 is formed yellow
Ionic Equation (IE):
Pb2+(aq)+2NO3-(aq) + 2K+(aq) + 2I-(aq)
.
PbI2(s)
+ 2K+(aq) + 2NO3-(aq)
Net Ionic Equation (NIE)
Pb2+(aq) + 2 I-(aq) --> PbI2(s)
Spectator Ions:
There is no change in the K+ ions.
There is no change in the NO3- ions.
They are spectator ions
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Metathesis reactions: formation of water
Molecular equation: all species listed in their
molecular forms:
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Complete ionic equation: lists all ions:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)
 H2O(l) + Na+(aq) + Cl-(aq)
Net ionic equation:
Cross out ions common to both sides.
List only unique ions:
H+(aq) + OH-(aq)  H2O(l)
Note that only strong electrolytes are written in ionic form
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Metathesis Reactions
Formation of insoluble gas:
Addition of Mg metal to HCl acid:
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2 (aq)
Ionic reaction:
Mg(s) +2H+(aq) +2Cl- (aq)
H2(g) +Mg2+(aq) +2Cl- (aq)
Net ionic equation:
Mg(s) + 2 H+(aq)
H2(g) + Mg2+(aq)
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Write out NIE reaction; indicate spectator ions for:
(a) AgNO3 (aq) + KCl(aq)
(b)Na3PO4(aq) + CaCl2(aq)
AgCl(s) + KNO3 (aq)
NaCl(aq) + Ca3(PO4)2 (s)
(balance first!)
KCl(aq) + MgSO4(aq)
K2SO4(aq) + MgCl2(aq)
(balance first!)
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Need to know what’s soluble and what’s not
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Which of following is (are) soluble in water?
(a) K3PO4
(b) Pb(C2H3O2)2
(c) Ga(OH)3
(d) NaBr
(e) BaSO4
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Acids, Bases, and Salts
Acid = substance that ionizes to form H+
in solution (e.g. HCl, HNO3, CH3CO2H,
lemon, lime, vitamin C). Acids donate
H+ ions. H+ ions are protons.
Bases = substances that react with the
H+ ions formed by acids (e.g. NH3,
Drano™, Milk of Magnesia™). Bases
accept H+ ions.
Bases also donate hydroxide ions (OH-)
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Really important!
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Acids, Bases, and Salts
HCl(aq)
H+(aq) + Cl-(aq)
HNO3(aq)
HC2H3O2(aq)
(acetic acid)
H+(aq) + NO3- (aq)
H+(aq) +
C2H3O2- (aq)
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Strong and Weak Acids and Bases
Strong acids and bases are strong electrolytes.
They are completely ionized in solution.
Weak acids and bases are weak electrolytes.
They are partially ionized in solution.
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Acids, Bases, and Salts
Neutralization Reactions and Salts
Neutralization occurs when a solution of an acid and a
base are mixed:
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Notice we form a salt (NaCl) and water.
Salt = ionic compound whose cation (+) comes from a
base and anion (-) from an acid.
Neutralization between acid and metal hydroxide
produces water and a salt, e.g.,
HCl + NaOH
H2O + NaCl
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Introduction to Oxidation-Reduction
Reactions (REDOX)
Oxidation is loss of electron(s)
(think of LEO)
LEO the Lion
goes
GER!
Reduction is gain of electron(s)
(think of GER)
BOTH must occur simultaneously. Something being
oxidized gives its electron(s) to something else being
reduced.
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21
Introduction to Oxidation ReductionReactions
Oxidation and Reduction
• When a metal undergoes corrosion it loses electrons to
form cations. This is oxidation. Example:
Ca(s) +2H+(aq)  Ca2+(aq) + H2(g)
Oxidized: atom, molecule, or ion becomes more positively
charged.
Oxidation is the loss of electrons.
Ca(s) is being oxidized to Ca2+
Reduced: atom, molecule, or ion becomes less positively
charged. Reduction is the gain of electrons.
H+ is being reduced to H2
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Oxidation of Metals by Acids and Salts
Metals are oxidized by acids to form salts:
Mg(s) +2HCl(aq)  MgCl2(aq) + H2(g)
During the reaction:
Mg(s) is oxidized to Mg2+ and
2H+(aq) is reduced to H2(g).
Metals can also be oxidized by other salts:
Fe(s) +Ni2+(aq)  Fe2+(aq) + Ni(s)
Notice that the Fe is oxidized to Fe2+ and
the Ni2+ is reduced to Ni.
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Introduction to Oxidation-Reduction
Reactions
Oxidation Number (O.N.) or Oxidation State
O.N. = 0 for atom in element form (C, Ag, O2, H2, etc)
= charge for any monoatomic ion
(e.g., +1 for Na+, -2 for S2- , +3 for Al3+)
= -2 nearly always for oxygen in compound or ion
= -1 always for fluorine in compound or ion
= +1 for hydrogen when bonded to non-metals
= -1 for hydrogen when bonded to metals
Sum of O.N. = 0 for neutral compound
Sum of O.N. = ion charge for polyatomic ion
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Introduction to Oxidation-Reduction
Reactions
Oxidation Number
Some examples; What is O.N. of underlined element?
HNO3
SO3
SO32-
KMnO4
MnO4-
C2O42-
NO2
NO3-
What occurs to N when a reaction goes from HNO3 to NO2?
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The Activity Series
Some metals are easily oxidized whereas others are not.
e.g., Fe is oxidized by Ni2+ but Ni is not oxidized by Fe2+
The reaction: Fe + Ni2+ Fe2+ + Ni occurs, but
Ni + Fe2+ Ni2+ + Fe does not
Activity series: a list of metals arranged in decreasing
ease of oxidation.
The higher the metal on the activity series, the more
active that metal, i.e., the more readily it is oxidized.
Any metal can be oxidized by the ions of elements below it
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Caution: the
Activity
series
is in opposite
order to the
Standard
Reduction
Potential
(SPR) list of
Ch 19!
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Based on Activity Series, what is outcome of each of
the following reactions:
(a) Al(s) + NiCl2(aq)
(b) Ag(s) + Pb(NO3)2 (aq)
(c) Cr(s) + NiSO4 (aq)
(d) Mn(s) + HBr(aq)
(e) H2(g) + CuCl2(aq)
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Solution Composition
Solvent: component present in greatest amount
Solute(s): component(s) present in smaller amount.
Water as solvent = aqueous solutions.
Change concentration by using different amounts of
solute and solvent.
Molarity (M): moles of solute per liter of solution
M = n/V or n = M x V
n=mol of solute
V = volume of solution in Liters
If we know: molarity and liters of solution, we can
calculate moles (and mass) of solute
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Molarity
Examples: what is molarity of:
(a) 3.7 mol solute in 600 mL solution
(b) 4.2 gram NaCl in 1.800 L solution
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Solution Composition
Molarity
Molarity: Moles of solute per liter of solution.
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Solution Composition
Dilution
We recognize that the number of moles of solute are the
same in dilute and concentrated solutions.
So:
MdiluteVdilute = moles = MconcVconc
Example: how do you make 350 mL of
a 0.200 M solution from a concentrated
(stock) solution which is 4.3 M ?
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Solution Stoichiometry and Chemical Analysis
For solution stoichiometric calculations, one needs moles
(recall from Ch. 3)
In solution stoichiometry, one calculates moles (n) using:
n=MxV
or
n = wt/MM
(MM=molar mass)
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Solution Stoichiometry and Chemical Analysis
Example: What mass of NaOH is required to precipitate all
of the Fe2+ ions from 25.0 mL of 0.500 M Fe(NO3)2 solution?
Reaction is:
NaOH + Fe(NO3)2
NaNO3 + Fe(OH)2
(not balanced)
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Solution Stoichiometry and Chemical Analysis
Titration is the addition of a standard
(known conc.) reagent from a buret to
another solution of of known volume but
unknown conc. until the equivalence point
is reached.
The equivalence point is determined by use of
an indicator, a substance that changes color
when the equivalence point is reached.
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Solution Stoichiometry and Chemical
Analysis
Titrations
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Solution Stoichiometry and Chemical Analysis
Titrations Example :
Reaction is: 2 HCl + Ba(OH)2
2H2O + BaCl2
What volume of 0.120 M HCl is needed to completely neutralize
50.0 mL of 0.101 M Ba(OH)2 ? (HCl is the titrant and is in
buret).
mol (Ba(OH)2) = M x V = 0.101 x 0.050 = 0.00505 mol
mol HCl = 2 x mol Ba(OH)2 = 2(.00505) = 0.0101 mol
Volume HCl = mol = 0.0101 mol= 0.0842 L = 84.2 mL
M
.120 M
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