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1411
Chapter 4
Aqueous Reactions and
Solution Stoichiometry
1
Solutions
• Solutions are defined as
homogeneous mixtures of two or
more pure substances.
• Aqueous solution – solution in which
water is the dissolving medium
• The solvent is present in greatest
abundance.
• All other substances are solutes;
they are dissolved in the solvent.
– Example: NaCl dissolved in water:
NaCl = solute
water = solvent
• Water can dissolve many ionic or
molecular compounds – “universal
solvent”
2
Dissociation
• When an ionic substance
dissolves in water, the
solvent separates the
individual ions from the
crystal.
• This process is called
dissociation.
3
Dissociation
• This occurs because water
is POLAR – it has poles: a
partial positive (δ+) end and
a partial negative (δ-) end
δO
H
δ+
H
δ+
O
H
•
Solvation – the surrounding of ions
by H2O molecules to help stabilize
the ions
- Keeps the ions from
recombining
NaCl(aq) 
Na+(aq)
+
Cl-(aq)
H
H
O
H
H
Na+
O
H
O
H
H
H
H
Cl-
O
H
H
O
H
H
H
H
O
4
O
Electrolytes
• An electrolyte is a substance that
dissociates into ions when
dissolved in water.
– May be strong or weak
• Strong electrolyte - completely
dissociates into ions
– Solutions conduct electricity well
H+(aq) + Cl-(aq)
Ex: HCl(aq)
• Weak electrolyte – partially dissociates
into ions, but some molecules remain
intact
– Solutions conduct electricity poorly
Ex: CH3COOH(aq)
CH3COO-(aq) + H+(aq)
5
Electrolytes
• A nonelectrolyte may
dissolve in water, but it
does not dissociate into
ions when it does so.
– Solutions do not conduct
electricity
6
Identifying Strong & Weak
Electrolytes & Nonelectrolytes
Strong electrolytes
1) Strong acids
2) Strong bases
3) Ionic compounds
These are
on the list
below
Weak electrolytes Any other
1) Weak acids
acid or
base NOT
2) Weak bases
on the list
below
Nonelectrolytes
Molecular
compounds, except
acids and bases
Memorize!
7
Solution Chemistry
• Pay attention to exactly what species are present in a
reaction mixture (i.e., solid, liquid, gas, aqueous
solution).
• If we are to understand reactivity, we must be aware of
just what is changing during the course of a reaction…
the driving force!
– The driving force is what makes the reaction react (Ex:
formation of a precipitate, gas, or liquid)
– Without a driving force, the reaction won’t occur
8
Precipitation Reactions
When one mixes ions
that form compounds
that are insoluble (as
could be predicted by
the solubility
guidelines), a precipitate
is formed.
9
Solubility Rules
Solubility of a substance is the amount that can be dissolved in a given quantity of
solvent at a given temperature.
Insoluble – the attraction between the ions in the solid is too great to separate the
ions to any significant extent; substance doesn’t dissolve
Memorize!
***All compounds of alkali metals (Group 1A) and of NH4+ ion are soluble ***
10
Solubility Rules
Classify the following as
soluble or insoluble:
a) CuCO3
b) Mg(OH)2
c) CaCl2
d) SrSO4
e) (NH4)3PO4
f) Fe2S3
11
Electrolytic Properties ≠ Solubility
Do not confuse extent of solubility with strong or weak
electrolyte
Examples:
• Acetic acid (CH3COOH or HC2H3O2): very soluble in
water but weak electrolyte because most remains in
form of molecule, not ions, in solution
• BaCl2: not very soluble in water but strong electrolyte
because what does dissolve dissociates completely
12
Metathesis (Exchange) Reactions
• Metathesis reactions: ions in the reactant
compounds exchange ions.
– They “exchange partners”
AB + CD  AD + CB
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
13
Metathesis (Exchange) Reactions
• To complete and balance a metathesis equation:
1.) Use the chemical formulas of the reactants to determine
the ions that are present
2.) Write the product formulas by combining the cation of one
reactant with the anion of the other. (Remember to use
the charges to come up with subscripts!!)
3.) Balance the equation
Mg(NO3)2 + Li3PO4 
14
Metathesis (Exchange) Reactions
• To complete and balance a metathesis equation:
1.) Use the chemical formulas of the reactants to determine
the ions that are present
2.) Write the product formulas by combining the cation of one
reactant with the anion of the other. (Remember to use the
charges to come up with subscripts!!)
3.) Balance the equation
Mg(NO3)2 + Li3PO4 
Mg2+
NO3- Li+ PO4315
Metathesis (Exchange) Reactions
• To complete and balance a metathesis equation:
1.) Use the chemical formulas of the reactants to determine
the ions that are present
2.) Write the product formulas by combining the cation of one
reactant with the anion of the other. (Remember to use the
charges to come up with subscripts!!)
3.) Balance the equation
Mg(NO3)2 + Li3PO4  Mg3(PO4)2 +
Mg2+
NO3- Li+ PO43-
LiNO3
16
Metathesis (Exchange) Reactions
• To complete and balance a metathesis equation:
1.) Use the chemical formulas of the reactants to determine
the ions that are present
2.) Write the product formulas by combining the cation of one
reactant with the anion of the other. (Remember to use the
charges to come up with subscripts!!)
3.) Balance the equation
3 Mg(NO3)2 + 2 Li3PO4  Mg3(PO4)2 + 6 LiNO3
Mg2+
NO3- Li+ PO4317
Predicting precipitation
• To predict whether a precipitate forms when
mixing aqueous solutions of 2 strong
electrolytes:
1) Write the balanced metathesis reaction
2) Use solubility rules to determine if any of the
products are insoluble
18
Predicting precipitation: Example
a)
Predict the products and write the balanced chemical equation for the reaction,
including the phases of all reactants and products. If no reaction, write NR.
Pb(NO3)2 mixed with Na2S
19
Molecular Equation
• The molecular equation lists the reactants and products in their
molecular form.
Molecular:
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
20
Ionic Equation
• A more accurate representation of the species that are found in
the reaction mixture is the ionic equation.
• In the ionic equation, all soluble, strong electrolytes are
dissociated into their ions.
– Remember, strong electrolytes = strong acids, strong bases, or ionic
compounds
Molecular:
Ionic:
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq)  AgCl (s) + K+ (aq) + NO3- (aq)
21
Net Ionic Equation
• Finally, the net ionic equation shows what is actually reacting.
• To form the net ionic equation, cross out anything that does not
change from the left side of the equation to the right.
• The only things left in the equation are those things that change
(i.e., react) during the course of the reaction.
Molecular:
Ionic:
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq)  AgCl (s) + K+ (aq) + NO3- (aq)
Net ionic:
Ag+(aq) + Cl-(aq)  AgCl (s)
22
Net Ionic Equation
• Those ions that didn’t change during the reaction (and so
were deleted from the net ionic equation) are called
spectator ions.
– In this example, the spectator ions are K+ and NO3– If all ions cancel, there is no reaction (no driving force).
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq)  AgCl (s) + K+ (aq) + NO3- (aq)
23
Writing Net Ionic Equations
1.
2.
3.
4.
Write a balanced molecular equation.
Write the ionic equation by dissociating all soluble,
strong electrolytes. (Remember to use coefficients to
indicate how many of each ion were in the molecular
equation.)
Cross out anything that remains unchanged from the
left side to the right side of the equation (spectator
ions).
Write the net ionic equation with the species that
remain. (Should already be balanced)
24
Example
• Write the balanced molecular, ionic, and net
ionic equation for the following, including
phases of each species. If no reaction, write
NR.
1.) AlCl3(aq) + K3PO4(aq) 
2.) NaCl(aq) + Zn(CH3COO)2(aq) 
25
Acids
• Acids – substance that ionizes in
aqueous solution to form hydrogen
ion (H+), or a proton
• So we say acids are proton donors
– Monoprotic acids yield one H+ per
molecule of acid
• Ex: HNO3, HCl
HCl(aq)  H+(aq) + Cl-(aq)
– Diprotic acids yield two H+ ions per
molecule of acid
• Ex: H2SO4
H2SO4(aq)  H+(aq) + HSO4-(aq)
HSO4 -(aq)
H+(aq) + SO42-(aq)
26
Bases
• Bases are substances that
accept (react with) H+ ions
• Bases produce hydroxide
inos (OH-) when dissolved
in water
– Some include OH- in
formula:
• NaOH, Ca(OH)2
– Some don’t include OH- in
formula:
• NH3 (only one you should
worry about for now)
27
Acid-Base Reactions
In an acid-base
reaction, the acid
donates a proton (H+)
to the base.
28
Neutralization Reactions
In an acid-base (neutralization) reaction, the acid donates a proton
(H+) to the base.
Generally, when solutions of an acid and a base are combined, the
products are a salt (ionic compound) and water.
- Water is neutral on the pH scale, unlike acids or bases, hence the
name neutralization reaction
- The formation of liquid water is the driving force for these reactions
HBr (aq) + LiOH (aq)  LiBr (aq) + H2O (l)
HC2H3O2 (aq) + NaOH (aq)  NaC2H3O2 (aq) + H2O (l)
29
Neutralization Reactions
When a strong acid reacts with a strong base, the net ionic
equation is…
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)
30
Gas-Forming Reactions
• Some metathesis reactions do not give the
product expected.
• In this reaction, the expected product, H2CO3,
decomposes to give a gaseous product, CO2.
CaCO3 (s) + 2HCl (aq)  CaCl2 (aq) + H2CO3 (aq)
CaCO3 (s) + 2HCl (aq)  CaCl2 (aq) + CO2 (g) + H2O (l)
Any time you form carbonic acid,
H2CO3, it is unstable and will
decompose to give H2O(l) and CO2(g)
31
Gas-Forming Reactions
When a carbonate or bicarbonate reacts with an
acid, the products are a salt, carbon dioxide, and
water.
CaCO3 (s) + HCl (aq)  CaCl2 (aq) + CO2 (g) + H2O (l)
NaHCO3 (aq) + HBr (aq)  NaBr (aq) + CO2 (g) + H2O (l)
Antacids!
32
Gas-Forming Reactions
• This reaction gives the predicted product, but you
had better carry it out in the hood, or you will be
very unpopular!
• But just as in the previous examples, a gas is
formed as a product of this reaction.
Na2S (aq) + H2SO4 (aq)  Na2SO4 (aq) + H2S (g)
• In these cases of gas-forming reactions, the
driving force of each reaction is the formation of
gas!
33
Examples
• Write balanced molecular, ionic, and net ionic equations for the
following pairs of reactants. Include states of matter: s, l, g, aq.
a) Li2CO3 (aq) + HNO3 (aq) 
b) KOH (aq) + SnCl2 (aq) 
c) (NH4)2S (aq) + HBr (aq) 
d) Zn(OH)2 (s) + H2SO4 (aq) 
34
Examples
• Write balanced molecular, ionic, and net ionic equations for the following pairs of
reactants. Include states of matter: s, l, g, aq.
a)
Li2CO3(aq) + 2HNO3(aq)  2LiNO3(aq) + H2O(l) + CO2(g)
2Li+(aq) + CO32-(aq) + 2H+(aq) + 2NO3-(aq)  2Li+(aq) + 2NO3-(aq) + H2O(l) + CO2(g)
CO32-(aq) + 2H+(aq)  H2O(l) + CO2(g)
b)
2KOH(aq) + SnCl2(aq)  2KCl (aq) + Sn(OH)2 (s)
2K+(aq) + 2OH-(aq) + Sn2+(aq) + 2Cl-(aq)  2K+(aq) + 2Cl-(aq) + Sn(OH)2 (s)
2OH-(aq) + Sn2+(aq)  Sn(OH)2 (s)
c)
(NH4)2S(aq) + 2HBr(aq)  2NH4Br (aq) + H2S(g)
2NH4+(aq) + S2-(aq) + 2H+(aq) + 2Br-(aq)  2NH4+(aq) + 2Br-(aq) + H2S(g)
S2-(aq) + 2H+(aq)  H2S(g)
d)
Zn(OH)2(s) + H2SO4(aq)  ZnSO4(aq) + 2H2O(l)
Zn(OH)2(s) + 2H+(aq) + SO42-(aq)  Zn2+(aq) + SO42-(aq) + 2H2O(l)
Zn(OH)2(s) + 2H+(aq)  Zn2+(aq) + H2O(l)
35
Oxidation-Reduction (Redox)
Reactions
• An oxidation occurs
when an atom or ion
loses electrons.
• A reduction occurs
when an atom or ion
gains electrons.
• One cannot occur
without the other.
“OIL RIG”
Oxidation Is Loss (of electrons)
Reduction Is Gain (of electrons)
36
Oxidation Numbers
Oxidation numbers are for bookkeeping of
electrons.
To determine if an oxidation-reduction
reaction has occurred, we assign an oxidation
number (or oxidation state) to each element
in a neutral compound or charged entity.
37
Rules for Assigning Oxidation Numbers
1. Elements in their elemental form have an
oxidation number of 0.
• Examples: Na, O2, H2, Cu, Ag, etc… all have oxidation numbers
of 0
Note: the remaining rules are for elements when they are part of a compound
2. The oxidation number of a monatomic ion is the
same as its charge.
• Examples:
Na+
ClO2Cu2+
oxidation number = +1
oxidation number = -1
oxidation number = -2
oxdiation number = +2
38
Rules for Assigning Oxidation Numbers
3. Oxygen has an oxidation number of −2, except in
the peroxide ion O22- in which it has an oxidation
number of −1.
4. Hydrogen is −1 when bonded to a metal, +1 when
bonded to a nonmetal.
5. Fluorine always has an oxidation number of −1.
• The other halogens have an oxidation number of −1
when they are monatomic ions; they can have positive
oxidation numbers, however, most notably in
oxyanions.
39
Rules for Assigning Oxidation Numbers
4. The sum of the oxidation numbers…
- in a neutral compound is 0.
• Ex: MgCl2
Mg2+
Cl-
oxidation number = +2
oxidation number = -1
(+2) + 2(-1) = 0
- in a polyatomic ion is the charge on the ion.
• Ex: SO42-
Sum of oxidation numbers is -2
We can use the sum to figure out oxidation number of S:
We know oxidation number of O is -2.
Make x = oxidation # of S and set up an algebraic
equation:
x + 4(-2) = -2
x = +6 (oxidation number of S)
40
Examples: Oxidation Numbers
• Assign oxidation numbers to each element in
the following examples:
a)
b)
c)
d)
e)
f)
g)
H2
MoS2
ClO2Ag
H2O2
Li3PO4
CaH2
41
Examples: Oxidation Numbers
• Assign oxidation numbers to each element in
the following examples:
a)
b)
c)
d)
e)
f)
g)
H2
MoS2
ClO2Ag
H2O2
Li3PO4
CaH2
H=0
S = -2
O = -2
Ag = 0
O = -1
Li = +1
Ca = +2
Mo = +4
Cl = +3
H = +1
O = -2
H = -1
P = +5
42
Displacement Reactions
A + BX  AX + B
• In displacement reactions, ion B oxidizes an element A.
• The ion B, then, is reduced to elemental form.
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
43
Displacement Reactions
A + BX  AX + B
• In displacement reactions, ion B oxidizes an element A.
• The ion B, then, is reduced to elemental form.
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
O
•
•
•
+1 -1
+2 -1
O
Oxidation number of Mg increases from 0 to +2
Mg is losing electrons so Mg is being oxidized!
Oxidation number of H decreases from +1 to 0
H is gaining electrons so H is being reduced!
Oxidation number of Cl remains the same
Cl is a spectator ion
44
Displacement Reactions
In this reaction,
silver ions oxidize
copper metal.
Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s)
45
Displacement Reactions
The reverse reaction,
however, does not
occur.
Cu2+ (aq) + 2 Ag (s) 
x Cu (s) + 2 Ag+ (aq)
46
Activity Series
• Different metals vary in how easily
they are oxidized (how easily they
lose electrons)
• Activity series – list of metals
arranged in order of decreasing ease
of oxidation
- Allows us to predict whether a
metal will be oxidized or not
• Any neutral metal (or hydrogen) on
the list can be oxidized by the ions
of elements below it
• That is why silver ion oxidized copper
metal, but copper ion could not
Do NOT need to memorize activity series,
oxidize silver metal
but you should know how to use it if given
this table.
47
Activity Series
• Examples:
Predict the outcome of the following
reactions. If no reaction occurs, write
NR. If a reaction occurs, write a
balanced equation.
a) Fe(s) + MgCl2(aq)
b) Cr(s) + Pb(NO3)2(aq)
c) Na(s) + HCl(aq) 
48
Activity Series
• Examples:
Predict the outcome of the following
reactions. If no reaction occurs, write
NR. If a reaction occurs, write a
balanced equation.
a) Fe(s) + MgCl2(aq) NR
b) 2Cr(s) + 3Pb(NO3)2(aq)
2Cr(NO3)3(aq) + 3Pb(s)
c) 2Na(s) + 2HCl(aq)  2NaCl(aq) + H2(g)
49
Molarity
• Concentration – tells how much solute is dissolved in a
solvent
• Molarity is one way to measure the concentration of a
solution.
moles of solute
Molarity (M) =
volume of solution in liters
Ex: 2.1 M MgCl2 = 2.1 mol MgCl2/L MgCl2 solution
(read as:) “2.1 molar solution of MgCl2”
• Because the units of molarity are mol/L, we can use molarity as a
conversion factor in stoichiometric calculations to interconvert:
moles of solute
liters of solution
2.1 mol MgCl2
1 L solution
or
1 L solution
2.1 mol MgCl2
50
Using Molarities in
Stoichiometric Calculations
51
Molarity: Examples
1.
Calculate the molarity of 142 mL of an aqueous solution containing 13.31 g
NaNO3.
1.10 M
2.
Calculate the number of moles of Cl- ions in 32.3 mL of 0.45 M CaCl2.
0.029 mol
3.
Consider the reaction of aluminum chloride with silver acetate. How many
milliliters of 0.250 M AlCl3 would be needed to react completely with
13.3 mL
20.0mL of 0.500 M AgC2H3O2 solution?
52
Dilution: Adding water
If an original solution is diluted, the molarity of the new
solution can be determined from the equation:
M1  V1 = M 2  V2
moles of solute before dilution = moles of solute after dilution
(because all we are doing is adding solvent to the original solution)
where M1 and M2 are the molarity of the initial (concentrated) and
final (dilute) solutions, respectively, and V1 and V2 are the volumes of
the two solutions.
53
Dilution: Example
If a stock solution of 25.0 mL of 1.00 M HCl is diluted to a
total of 100.0 mL, what is the final concentration of the HCl
solution?
M1  V1 = M 2  V2
M2 = M1V1
V2
M2 = (1.00 M)(0.0250 L)
0.1000 L
M2 = 0.250 M
54
Titration
Titration is an analytical
technique in which one
can calculate the
concentration of a
solute in a solution.
55
Titration
• A sample of known
concentration (standard
solution) is added to solution
of unknown concentration.
• The point at which
stoichiometrically equivalent
quantities are reached is the
equivalence point.
• For acid-base reactions, the
equivalence point is observed
with an indicator.
– Ex: Phenolphthalein is colorless
in acidic solutions but turns
pink in basic solutions.
56
Titration Example
In a titration, 23.25 mL of 0.105 M NaOH was needed to react
with 21.45 mL of H2SO4 solution. What is the molarity of the acid?
0.0569 M
57