Download CHAPTER 4: AQUEOUS REACTIONS AND SOLUTION

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CHAPTER 4: AQUEOUS
REACTIONS AND SOLUTION
STOICHIOMETRY
GENERAL PROPERTIES OF AQUEOUS
SOLUTIONS
A solution is a homogeneous mixture of two or
more substances.
A solution in which water is the dissolving
medium is called an aqueous solution.
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SOLUTE VS. SOLVENT
The substance present in the greatest quantity is
called the solvent.
Water is considered the universal solvent because of
its ability to dissolve many substances.
The other dissolved substances are called the solutes.
A solvent dissolves a solute.
ELECTROLYTES
A substance whose aqueous solutions contain ions is
called an electrolyte because it will allow electric
current to flow through it. Example: NaCl
A substance that does not form ions in solution is
called a nonelectrolyte. Example: C12H22O11
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IONIC COMPOUNDS IN WATER
When ionic compounds dissolve in water the ions
separate and are surrounded by water molecules.
Once the ions have separated, they can move and
conduct electricity. In the crystalline form they cannot
move or conduct electricity.
The solvation process helps stabilize the ions in
solution and prevents the cations and anions from
recombining.
IONIC COMPOUND IN WATER
Water molecules are polar and have a δ+ and δ- side.
The more electronegative oxygen atoms are attracted
to the positive ions.
The less electronegative hydrogen atoms are attracted
to the negative ions.
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Ionic Compounds: undergo dissociation: process by which
many ionic substances dissolve in water, the solvent
pulls the individual ions from the crystal and solvates
_
them.
{NaCl + H2O}
Polar water molecule
+
+
H2O
Dissociation
NaCl(s)
H2O
Na+(aq) + Cl-(aq)
Hydration of sodium chloride
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MOLECULAR COMPOUNDS IN WATER
When a molecular compound dissolves in water, the
solution usually consists of intact molecules
(neutral) dispersed throughout the system.
Consequently, most molecular compounds are
nonelectrolytes. A few exceptions include strong
acids like HCl.
Ionic (electrolyte)
Molecular (non electrolyte)
Molecular (covalent) Compounds:
mostly insoluble gases, except polar organic liquids containing O & N (polar:
acids, bases, alcohols, etc.)
Insoluble gases: NO2, CH4, CO2, O2, P2O5, N2, CO, etc.
Polar Covalent {carbon (C) chains containing H,O or N}: CH3OH, C6H12O6,
C6H5OH, etc.
C6H12O6(s)
C3H5OH(l)
H2O
H2O
C6H12O6 (aq)
C3H5OH (aq)
Dissolve without dissociating into ions
{Ethanol+Water}
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STRONG AND WEAK ELECTROLYTES
Strong electrolytes are those solutes that exist in
solution completely or nearly completely as ions.
Weak electrolytes are those solutes that exist in
solution mostly in the form of molecules with only a
small fraction in the form of ions.
ELECTROLYTES: STRONG AND WEAK
A strong electrolyte dissociates completely when
dissolved in water.
HCl (g)
H+(aq) + Cl-(aq)
H2O
A weak electrolyte only dissociates partially when
dissolved in water.
NH4OH(aq)
H2O
Acetic Acid,
HC2H3O2 (aq)
molecules
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H2O
NH4+(aq) + OH-(aq)
H+(aq) + C2H3O2-(aq)
ions
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ELECTROLYTES: A VISUAL
CHEMICAL EQUILIBRIUM
Chemical equilibrium is the state of balance in which
the relative numbers of each type of ion or molecule
in the reaction are constant over time.
Equilibrium is represented by a double arrow for the
yield sign. This shows that the reaction is occurring
in both directions.
Dynamic Equilibrium
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STRONG ELECTROLYTES
Remember that soluble ionic compounds are strong
electrolytes.
We consider ionic compounds as those composed of
metals and nonmetals – such as NaCl, FeSO4 and
Al(NO3)3 – or compounds containing the ammonium
ions, NH4+ - such as NH4Br and (NH4)2CO3.
SAMPLE EXERCISE 4.1
The diagram below represents an aqueous solution of
one of the following compounds: MgCl2, KCl, or
K2SO4. Which solution does the drawing best
represent?
The diagram shows twice as
many cations as anions,
consistent with the formulation
K2SO4.
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PRACTICE EXERCISE
If you were to draw diagrams like the previous one
representing aqueous solutions of each of the
following ionic compounds, how many anions would
you show if the diagram contained six cations?
a. NiSO4
6
b. Ca(NO3)2
12
c. Na3PO4
2
d. Al2(SO4)3
9
PRECIPITATION REACTIONS
Reactions that result in the formation of an
insoluble product are called precipitate reactions.
A precipitate is an insoluble solid formed by a
reaction in solution.
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PRECIPITATION REACTIONS
Aqueous solutions, reacting to produce a
precipitate (an insoluble compound). Example: KI(aq) +
Pb(NO3)2 (aq)
Predict the solubility of compounds in reaction:
Pb(NO3)2(aq) + 2KI(aq)
2 KNO3(aq)
PbI2(s)
Precipitate
(ppt)
Pb(NO3)2
KI
PbI2
SOLUBILITY
The solubility of a substance at a given temperature is
the amount of substance that can be dissolved in a
given quantity of solvent at a given temperature.
Most important solubility rule: Alkali metal ions and
ammonium ions are soluble in water.
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SOLUBILITY – ABILITY TO BE DISSOLVED
Soluble (aq)
a substance which easily dissolves in water
Slightly Soluble (s)
a substance that only dissolves a tiny bit in
water
Insoluble (s)
substance that does NOT dissolve in water
substance remains separate from the H2O
molecules (no interaction)
substances are insoluble when the ions
attract so strongly that they CANNOT be
pulled apart by H2O molecules
SOLUBILITY: A VISUAL
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SOLUBILITY RULES
To know if an ionic substance is soluble or not, you
must consult the solubility rules
lists which ions are typically soluble
and insoluble
also lists exceptions to the rules
Note: Acids are not explicitly listed in
the rules because all acids are
aqueous solutions (thus soluble)
SOLUBILITY RULES: for Ionic Compounds (Salts)
1. All salts of alkali metals (IA) are soluble.
2. All NH4+ salts are soluble.
3. All salts containing the anions: NO3-, ClO3-, ClO4-,
(C2H3O2-) are soluble.
4. All Cl-, Br-, and I- are soluble except for Hg22+, Ag+,
and Pb2+ salts.
5. All SO42- are soluble except for Pb2+, Ba2+, and Sr2+.
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Solubility Rules for Common Ionic Compounds in
Water
Soluble Ionic Compounds
Nitrates, NO3Acetates, C2H3O2Chlorides, ClBromides, BrIodides, ISulfates, SO42Alkali metal cations
Ammonium, NH4+
Exceptions: these are insoluble w/ respective anions
none
none
Ag+, Hg22+ (2 Hg atoms), Pb2+
Ag+, Hg22+, Pb2+
Ag+, Hg22+, Pb2+
Ag+, Hg22+, Pb2+, Sr2+, Ba2+
none
none
Insoluble Ionic Compounds
Sulfides, S2Carbonates, CO32Phosphates, PO43Hydroxides, OH-
Exceptions: these are soluble w/ respective anions
alkali metal cations, NH4+, Ca2+, Sr2+, Ba2+
alkali metal cations, NH4+
alkali metal cations, NH4+
alkali metal cations, NH4+, Ca2+, Sr2+, Ba2+
SAMPLE EXERCISE 4.2
Classify the following ionic compounds as
soluble or insoluble in water:
a. Sodium carbonate: Na2CO3,
Soluble
b. Lead (II) sulfate : PbSO4
Insoluble
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PRACTICE EXERCISE
Classify the following compounds as soluble or
insoluble in water:
a.
Cobalt (II) hydroxide: Insoluble
b.
Barium nitrate: Soluble
c.
Ammonium phosphate: Soluble
EXCHANGE REACTIONS
Reactions in which positive ions and negative ions
appear to exchange partners conform to the following
generic equation:
AX + BY
AY + BX
Such reactions are called exchange reactions, double
displacement reactions, or metathesis reactions.
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SOLUBILITY RULES - PRACTICE
Are the following soluble or insoluble?
Be(C2H3O2)2 soluble
MgS
insoluble
BaSO4
insoluble
K3PO4
soluble
STEPS FOR WRITING DOUBLE
DISPLACEMENT REACTIONS
1.
2.
3.
4.
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Write the formulas for the reactants (balance
charges).
Write the formulas for the products by switching the
cations of the two reactants (balance charges).
Balance the equation.
Include states of
matter when needed.
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SAMPLE EXERCISE 4.3
a) Predict the identity of the precipitate that forms
when solutions of BaCl2 and K2SO4 are mixed.
BaSO4 is insoluble and will precipitate from solution.
KCl is soluble.
b) Write the balanced chemical equation for the
reaction.
PRACTICE EXERCISE
a) What compound precipitates when solutions of
Fe2(SO4)3 and LiOH are mixed?
Fe(OH)3
b) Write a balanced equation for the reaction.
Fe2(SO4)3(aq) + 6 LiOH(aq)
2Fe(OH)3(s) +3Li2SO4(aq)
c) Will a precipitate form when solutions of Ba(NO3)2
and KOH are mixed?
no (both possible products, Ba(OH)2 and KNO3, are
water soluble)
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COMPLETE IONIC EQUATIONS
o
o
o
o
A molecular equation is an equation showing the
complete chemical formulas of the reactants and
products without showing dissociated ions.
Ex:
Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)
A complete ionic equations is an equation that
shows all soluble strong electrolytes as ions.
Ex:
Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) PbI2(s) +
2K+(aq) + 2NO3-(aq)
NET IONIC EQUATIONS
Ions that appear in identical forms on both sides of
the yield sign are called spectator ions because they
do not actually react.
A net ionic equation is an equation in which the
spectator ions have been removed.
Complete Ionic Equation
Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)
PbI2(s) + 2K+(aq) +
2NO3 (aq)
Net Ionic Equation
Pb2+(aq) + 2I-(aq) PbI2(s)
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NET IONIC EQUATIONS
In a net ionic equation, the numbers of atoms and the
charges are conserved in the reaction.
If every ion in a complete ionic equation is a spectator,
then no reaction occurs.
STEPS FOR WRITING NET IONIC EQUATIONS
1.
2.
3.
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Write the balanced equation.
Write the complete ionic equation which shows strong
electrolytes (ionic compounds and strong acids)
written as ions.
Identify and cancel spectator ions.
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WAYS OF EXPRESSING PRECIPITATION
REACTIONS
There are three different:
(1) Molecular Equations
AgNO3 (aq) + KCl (aq) → AgCl (s) + KNO3 (aq)
(2) Ionic Equations
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq) →
AgCl (s) + K+ (aq) + NO3- (aq)
(3) Net Ionic Equations
Ag+(aq) + Cl-(aq) → AgCl (s)
SAMPLE EXERCISE 4.4
Write the net ionic equation for the precipitation
reaction that occurs when solutions of calcium chloride
and sodium carbonate are mixed.
Molecular eq.
CaCl2(aq) + Na2CO3(aq)
CaCO3(s) + 2NaCl (aq)
Ionic Eq.
Ca2+(aq) + 2 Cl- (aq) + 2Na+(aq) + CO32-(aq)
CaCO3(s) + 2Na+(aq) + 2Cl- (aq)
Net Ionic eq.
Ca2+(aq) + CO32-(aq)
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CaCO3(s)
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PRACTICE EXERCISE
Write the net ionic equation for the precipitation reaction
that occurs when aqueous solutions of silver (I) nitrate
and potassium phosphate are mixed.
ACID-BASE REACTIONS
Acids and bases are also common electrolytes.
Acids are substances that ionize in aqueous solutions
to form hydrogen ions.
Since hydrogen ions consist of only 1 proton, acids are
often called proton donors.
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Bases
ACIDS
Substances that
produce H+ ions
when dissolved in
water
(Arrhenius).
Substances that
produce OH− ions
when dissolved in
water
(Arrhenius).
Ex)
HNO3
H+ + NO3–
Proton donor (L&B)
HCl (g)
Ex)
NaOH
Na+ + OH–
Proton Acceptor (L&B)
NH3 (g)
STRONG VS. WEAK
Both strong & weak acids and bases exist
Strong acids & bases
completely in solution
dissociate
Exist as 100% ions
Strong electrolytes & conductors
Weak acids & bases only partially
dissociate in solution
Few ions exits
Weak electrolytes & conductors
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STRONG VS. WEAK ELECTROLYTES
Strong Acid:
HCl
BASES
The strong bases are :
•
•
•
•
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Alkali metals (IA)
Hydroxides
Barium Hydroxide
Strontium Hydroxide
(weaker: Ammonium,
Calcium Hydroxides)
Weak Acid:
HNO2
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Acids
Monoprotic acids contain one ionizable hydrogen.
Example: HCl
Diprotic acids contain two ionizable hydrogens.
Example: H2SO4
Triprotic acids contain three ionizable hydrogens.
Example: H3PO4
Most multiprotic acids only lose one hydrogen to a
reasonable extent.
Citric acid
Acids
Acids taste sour.
Some acids also contain hydrogens that will not
ionize.
Normally the hydrogen(s) that will ionize are listed
at the from of the chemical formula.
Example: HC2H3O2
Sometimes the hydrogen that will ionize is listed at
the end of a COOH group.
Example: CH3COOH
Diprotic acid
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Triprotic acid
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Bases
Bases taste bitter.
Bases are substances that accept (react with )
hydrogen ions.
Bases produce hydroxide ions (OH-) when they
dissolve in water.
Bases do not have to contain the hydroxide ion.
Examples: NaOH vs. NH3
NaOH(aq)
Na+ (aq) + OH- (aq)
Strong vs. Weak
Strong acids and bases are strong electrolytes so they
fully ionize in solution.
Weak acids and bases are weak electrolytes so they
only partially ionize in solution.
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The Strong Acids
You need to memorize the strong acids.
HCl (aq)
HBr (aq)
HI (aq)
HClO3 (aq)
HClO4 (aq)
HNO3 (aq)
H2SO4 (aq)
hydrochloric acid
hydrobromic acid
hydroiodic acid
chloric acid
perchloric acid
nitric acid
sulfuric acid
STRONG BASES
You need to memorize the strong bases.
Group 1 metal hydroxides: LiOH, NaOH, KOH, RbOH,
and CsOH
Heavy Group 2 metal hydroxides: Ca(OH)2, Sr(OH)2,
and Ba(OH)2
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SAMPLE EXERCISE 4.5
The following diagrams represent aqueous solutions
of three acids (HX, HY, and HZ) with water molecules
omitted for clarity. Rank them from strongest to
weakest.
PRACTICE EXERCISE
Imagine a diagram showing 10 Na+ ions and 10 OHions. If this solution were mixed with the one
pictured on the previous question for HY, what
would the diagram look like that represents the
solution after any possible reaction? (H+ ions will
react with OH- ions to form H2O.)
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STRONG VS. WEAK ELECTROLYTE
Strong
Electrolyte
Weak
Electrolyte
Non-electrolyte
Ionic
All
None
None
Molecular
Strong Acids
Weak acids and
weak bases
All other
compounds
SAMPLE EXERCISE 4.6
Classify the following substances as a strong
electrolyte, weak electrolyte, or nonelectrolyte: CaCl2,
HNO3, C2H5OH (ethanol), HCOOH (formic acid),
KOH.
Strong:
Weak:
Non:
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CaCl2, KOH, HNO3
HCOOH
C2H5OH
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PRACTICE EXERCISE
Consider solutions in which 0.1 mol of each of the
following compounds is dissolved in 1L of water:
Ca(NO3)2, C6H12O6 (glucose), CH3COONa (sodium
acetate), and CH3COOH (acetic acid). Rank the
solutions in order of increasing electrical conductivity
based on the fact that the greater the number of ions
in solution the greater the conductivity.
NEUTRALIZATION REACTIONS
When a solution of an acid and a solution of a base are
mixed, a neutralization reaction occurs.
The products of the reaction are water and a salt.
A salt is any ionic compound whose cation comes from
a base and whose anion comes from an acid.
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NEUTRALIZATION REACTIONS
(ARRHENIUS).
Generally, when solutions of an acid and a base are
combined, the products are a salt and water.
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
H+ (aq) + Cl- (aq) + Na+ (aq) + OH-(aq) →
Na+ (aq) + Cl- (aq) + H2O (l)
H+ (aq) + OH- (aq) →H2O (l)
NEUTRALIZATION REACTIONS
Observe the reaction between a
weak base, Milk of Magnesia,
Mg(OH)2 (s) and a strong acid HCl
(aq).
How would you write the net ionic
equation for such a reaction?
{Movie}
Mg(OH)2(s) + 2HCl (aq)
Mg(OH)2(s) + 2H+(aq) + 2Cl-(aq)
MgCl2(aq) + 2H2O (l)
Mg2+(aq) + 2Cl-(aq) + 2H2O (l)
Mg(OH)2(s) + 2H+(aq)
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Mg2+(aq) + 2H2O (l)
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SAMPLE EXERCISE 4.7
a.
b.
Write a balanced equation for the reaction
between aqueous solutions of acetic acid
(CH3COOH) and barium hydroxide.
Write the net ionic equation for this reaction.
PRACTICE EXERCISE
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a.
Write a balanced equation for the reaction of
carbonic acid (H2CO3) and potassium hydroxide.
b.
Write the net ionic equation for this reaction.
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GAS FORMATION
The sulfide ion, bicarbonate and the
carbonate ion
will react with acids to form gases.
Example: S22HCl(aq) + Na2S(aq)
H2S(g) + 2NaCl(aq)
Example: HCO32HCl(aq) + NaHCO3(aq) NaCl(aq) + H2CO3(aq)
H2CO3(aq) H2O(l) + CO2(g)
So …
HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g)
OXIDATION-REDUCTION REACTIONS
Oxidation-reduction reactions (redox reactions) are
reactions in which electrons are transferred between
reactants.
Oxidation is loss of electrons (Increase in ON).
Reduction is gain of electrons (Decrese in ON).
Oxidation and reduction always occur together.
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RULES FOR OXIDATION NUMBERS
1.
2.
3.
4.
5.
٣٢
For an atom in its elemental form, the oxidation
number is always zero.
Ex: Na = 0 S8 = 0
N2 = 0
F2 = 0 P4 = 0
O2 = 0
For any monatomic ion the oxidation number
equals the charge on the ion.
Ex: Mg2+ = +2,
N3- = -3 Al3+ = +3
The oxidation number of hydrogen is usually +1
when bonded to nonmetals and -1 when bonded
to metals
Ex: HCl H = +1
LiH
H = -1
The oxidation number of oxygen is usually -2 (except
in peroxide, O22-, where it is -1).
Example: CO2
O = -2
H2O2 O = -1
The oxidation number of fluorine is -1 in all
compounds. The other halogens have an oxidation
number of -1 except when combined with oxygen in
oxyanions.
Example : NaF
F = -1
HBr
Br = -1
ClO
Cl = +1
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The sum of the oxidation numbers of all atoms
in a neutral compound is zero.
Example:
P2O5 = 0
O = -2, so P = +5
7. The sum of the oxidation numbers in a
polyatomic ion equals the charge of the ion.
Example:
CO32- = 0 O = -2, so C = +4
NH4 = +1 H = +1, so N = -3
8. For compounds contain elements of:
G1A = +1, G2A = +2, Al = +3
6.
SAMPLE EXERCISE 4.8
Determine the oxidation number of sulfur in each
of the following:
a. H2S
2(+1) + S = 0
S = -2
b. S8
S=0
c. SCl2
S + 2(-1) = 0
S= +2
d. Na2SO3 Na 2(+1) + S + 3(-2) = 0
S = +4
e. SO42S+4(-2) = -2
S =+6
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PRACTICE EXERCISE
What is the oxidation state of each element in the
following:
a. P2O5
b. NaH
c. Cr2O72d. SnBr4
e. BaO2
OXIDATION OF METALS
Many metals undergo displacement reactions
with acids to form salts and hydrogen gas.
Write the oxidation numbers for each element to
determine which element is oxidized and which is
reduced.
Since Mg goes from 0 to +2, it is oxidized. Since H
goes from +1 to 0, it is reduced.
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SAMPLE EXERCISE 4.9
Write the balance equation and net ionic equation for
the reaction of aluminum with hydrobromic acid.
o
+1
+3
0
PRACTICE EXERCISE
٣٥
a.
Write a balanced equation and net ionic equation for
the reaction between magnesium and cobalt (II)
sulfate.
b.
What is oxidized and what is reduced in the
reaction?
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THE ACTIVITY SERIES
A list of metals arranged in order of decreasing ease
of oxidation is called an activity series.
Active metals are at the top of the series and are
most easily oxidized.
Noble metals are at the bottom of the series are not
very reactive.
Any metal on the list can be
oxidized by the ions of
elements below it.
ACTIVITY SERIES
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USING THE ACTIVITY SERIES EXAMPLES
1) magnesium sulfate+ zinc metal
2) iron metal + copper (II) nitrate
Fe (s) + Cu(NO3)2 (aq)
NR
Cu (s) + Fe(NO3)2
We can also write CIE & NIE!!! (only aqueous dissociate)
CIE:
Fe (s) + Cu2+ (aq) + 2 NO3– (aq)
Cu (s) + Fe2+ (aq) + 2 NO3– (aq)
NIE:
Fe (s) + Cu2+ (aq)
Cu (s) + Fe2+ (aq)
METALS REACTING W/ACIDS
Only metals listed ABOVE hydrogen in the series are
able to react with acids
If the reaction does occur, H2 gas is produced.
Example #1:
Ni (s) + 2 HCl (aq)
NiCl2(aq) + H2 (g)
Ni is above H2, so the rxn occurs!
Example #2:
Cu (s) + 2HCl (aq)
NR
Cu is below H2, so no rxn occurs!
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ACTIVITY SERIES OF HALOGENS:
The idea is the same here…
Fluorine is the most reactive
Iodine is the least reactive
Only a MORE reactive halogen can replace a
LESS reactive one
In these single replacement rxns, the single
element (halogen gas) must be listed
ABOVE the halogen anion in the 2nd
reactant for a rxn to occur
Cl2 +
I2 +
ACTIVITY SERIES
OF HALOGENS:
Fluorine
Chlorine
Bromine
Iodine
2NaBr (aq) 2NaCl (aq) + Br2 (g)
2KCl
NR
SAMPLE EXERCISE 4.10
Will an aqueous solution of iron (II) chloride oxidize
magnesium metal? If so, write the balanced equation
and net ionic equation for the reaction.
٣٨
Mg (s)
+
Mg(s) +
Fe2+
FeCl2 (aq)
MgCl2(aq) +
Mg2+(aq) +
Fe(s)
Fe (s)
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PRACTICE EXERCISE
Which of the following metals will be oxidized by
Pb(NO3)2: Zn, Cu, Fe?
CONCENTRATIONS OF SOLUTIONS
Scientists use the term concentration to designate the
amount of solute dissolved in a given quantity of solvent
or solution.
Molarity (M) expresses the concentration of a solution as
the number of moles of solute in a liter of solution.
M = mol/Vl
Units: M, mol/L or mol L-
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SAMPLE EXERCISE 4.11
Calculate the molarity of a solution made by dissolving
23.4g of sodium sulfate in enough water to form 125mL
of solution.
PRACTICE EXERCISE
Calculate the molarity of a solution made by
dissolving 5.00g of glucose in sufficient water to form
exactly 100mL of solution.
Answer: 0.278 M
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IONIC COMPOUNDS
When calculating concentration, remember that ionic
compounds break into ions in solutions.
Example: MgCl2 breaks into
Mg2+ and 2Cl-
NaCl (aq)
0.1M
Na+ (aq) + Cl- (aq)
0.1M
0.1M
Na2SO4 (aq)
1M
2Na +(aq) + SO42- (aq)
2M
1M
SAMPLE EXERCISE 4.12
What are the molar concentrations of each of the
ions present in a 0.025M aqueous solution of
calcium nitrate?
Ca (NO3)2 (aq)
0.025 M
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Ca2+ (aq) + 2NO3- (aq)
0.025 M
2 x 0.025M
= 0.05 M
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PRACTICE EXERCISE
What is the molar concentration of K+ ions in a
0.015M solution of potassium carbonate?
Answer : 0.030 M
MOLARITY
Since molarity equals mol/L, it can be used as a
conversion factor.
Example: 0.3 M NaOH
ie: 0.3 mol NaOH
1 L soln
0.3mol NaOH
1000 mL soln
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SAMPLE EXERCISE 4.13
How many grams of Na2SO4 are required to make
0.350L of 0.500M Na2SO4?
PRACTICE EXERCISE
a) How many grams of Na2SO4 are there in 15mL of
0.50M Na2SO4?
Answers: 1.1 g
b) How many milliliters of 0.50M Na2SO4 solution are
needed to provide 0.038 mol of this salt?
Answer: 76 mL
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DILUTION
Stock solutions are solutions that are purchased at a
specific concentration, but sometimes a different
concentration is needed for an experiment.
Solutions of lower concentration can then be obtained
by adding water, a process called dilution.
When solvent is added to dilute a solution, the
number of moles of solute remains the same.
M1 x V1
=
M2 x V2
(before dilution)
(after dilution)
SAMPLE EXERCISE 4.14
How many milliliters of 3.0M H2SO4 are needed to
make 450mL of 0.10M H2SO4?
M1 x V1
=
M2 x V2
3 x V1 =
450 x 0.10
V1= 15 mL
Note: Acid added to water for dilution
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PRACTICE EXERCISE
a) What volume of 2.50M lead (II) nitrate solution
contains 0.0500mol of Pb2+? Answer: 0.0200 L= 20 ml
b) How many milliliters of 5.0M K2Cr2O7 solution must
be diluted to prepare 250mL of 0.10M solution?
Answer: 5.0 mL
c)
If 10.0mL of a 10.0M stock solution of NaOH is
diluted to 250mL, what is the concentration of the
resulting stock solution? Answer: 0.40 mL
EXAMPLE
If 50.0 mL of 0.10 M NaOHaq is added to 30 mL of
0.25 M NaOHaq, Calculate the molarity of the final
solution.
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PRACTICE EXERCISE
If 75.0 mL of 0.15 M NaCl(aq) is added to 50 mL of 0.30
M CaCl2(aq), Calculate the molarity of the final
solution.
SOLUTION STOICHIOMETRY AND
CHEMICAL ANALYSIS
You can use molar mass and molarity to solve
stoichiometry problems.
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SAMPLE EXERCISE 4.15
How many grams of Ca(OH)2 are needed to neutralize
25.0mL of 0.100M HNO3?
2NHO3 (aq) + Ca(OH)2(aq)
0.1 M, 25.0 ml
Ca(NO3)2 (aq) + H2O(l)
g ??
PRACTICE EXERCISE
a) How many grams of NaOH are needed to neutralize
20.0mL of 0.150M H2SO4 solution? Answer:0.240 g
b) How many liters of 0.500M HCl are needed to react
completely with 0.100 mol of Pb(NO3)2, forming a
precipitate of PbCl2? Answer: 0.400L
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TITRATIONS
To determine the concentration of a particular solute
in a solution, chemists often carry out a titration,
which involves combining a sample of the solution
with a reagent solution of known concentration,
called a standard solution.
Balanced chemical equation is
needed to calculate the unknown con.
TITRATIONS
The point at which stoichiometrically equivalent
quantities are brought together is known as the
equivalence point.
In acid-base titrations, dyes known as acid-base
indicators are used to identify the equivalence point.
The color change of the indicator signals the end
point of the titration, which usually coincides very
nearly with the equivalence point.
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• Suppose we know the molarity of a NaOH solution and
we want to find the molarity of an HCl solution.
• We know:
– molarity of NaOH, volume of HCl.
• What do we want?
– Molarity of HCl.
• What do we do?
– Take a known volume of the HCl solution, measure
the mL of NaOH required to react completely with
the HCl.
TITRATION: EXAMPLE: NEUTRALIZATION REACTION
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IMPRTANT NOTES:
Or
SAMPLE EXERCISE 4.16
a) How many grams of chloride ion are in a sample of
the water if 20.2mL of 0.100M Ag+ is needed to react
with all the chloride in the sample?
Ag+(aq)
+
Cl-(aq)
AgCl(s)
20.0 mL, 0.1 M
Solution
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?g
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If the sample has a mass of 10.0g, what percent Cldoes it contain?
PRACTICE EXERCISE
a) A solution that contains Fe2+ is titrated with 47.20mL
of 0.02240M MnO4-. How many moles of MnO4- were
added to the solution?
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Answers: 1.057 × 10–3 mol MnO4–
b) How many moles of Fe2+ were in the sample?
Answer: 5.286 × 10–3 mol Fe2+
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PRACTICE EXERCISE
c) How many grams of iron were in the sample?
Answer: 0.2952 g
d) If the sample had a mass of 0.8890g, what is the
percentage of iron in the sample? Answer: 33.21%
SAMPLE EXERCISE 4.17
The concentration of NaOH is normally in the range of 3
to 6M. The NaOH is analyzed periodically. In one such
analysis, 45.7mL of 0.500M H2SO4 is required to
neutralize a 20.0mL sample of NaOH solution. What is
the concentration of the NaOH solution?
45.7mL of 0.500M
20.0mL, M??
Note: The eq. must be balanced
Then apply the neutralization law:
acid
M2 = 2.28 M
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base
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PRACTICE EXERCISE
What is the molarity of an NaOH solution if 48.0mL is
needed to neutralize 35.0mL of 0.144M H2SO4?
Answer: 0.210 M
SAMPLE INTEGRATIVE EXERCISE
A sample of 70.5mg of potassium phosphate is added to
15.0mL of 0.050M silver nitrate, resulting in the
formation of a precipitate.
a. Write the molecular equation for the reaction.
K3PO4(aq) + 3AgNO3(aq)
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Ag3PO4(s) 3KNO3(aq)
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b. What is the limiting reagent in the reaction?
K3PO4(aq) + 3AgNO3(aq)
Ag3PO4(s) 3KNO3(aq)
70.5 mg
15.0mL, 0.050M
moles K3PO4 :
Then /1 = 3.32x10-4
moles AgNO3 : =
Then /3 = 2.5x10-4
(LR)
SAMPLE INTEGRATIVE EXERCISE
c) Calculate the theoretical yield, in grams, of the
precipitate that forms.
3AgNO3(aq)
Ag3PO4(s)
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