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Lecture 4
Aqueous Reactions and Solution Stoichiometry
Text: Sections I, J, K, L, M
Solutions in which water is the dissolving medium are called aqueous solutions.
There are three major types of chemical processes occurring in aqueous solutions:
precipitation reactions
acid-base reactions
redox reactions
4.1
General Properties of Aqueous Solutions
solution – a homogeneous mixture of two or more substances.
solvent – usually the component that is present in greater quantity
solutes – the other substances in the solution
Electrolytic Properties
electrolyte – a substance whose aqueous solutions contain ions and hence conducts
electricity. Ex: NaCl
nonelectrolyte – a substance that does not form ions when is dissolves in water. Ex:
sugar
Ionic Compounds in Water
Ionic compounds dissociate into their component ions as they dissolve. NaCl  Na+ +
Cl–
Ch 4 Aqueous Reactions - 1 -
Molecular Compounds in Water
Most molecular compounds do not form ions when they dissolve in water; they are
nonelectrolytes.
Important exceptions are acids and compounds such as ammonia that react with water to
form ions.
Strong and Weak Electrolytes
Strong electrolytes exist in solution completely or nearly completely as ions.
Weak electrolytes produce small concentrations of ions when they dissolve.
Do not confuse the extent to which an electrolyte dissolves with whether it is a strong or
weak electrolyte.
CH3 COO– + H+
CH3COOH
[H  ][CH3COO  ]
Ka 
 1.8  10 5
[CH3COOH]
chemical equilibrium
4.2
Concentrations of Solutions
Molarity
Molarity 
moles solute
volume of soln in liters
Expressing the Concentration of an electrolyte
The concentration of an electrolyte in solution can be specified either in terms of the
compound used to make the solution or in terms of the ions that that the solution
contains.
Na2SO4 (aq)  2Na+ (aq) + SO42–.(aq)
Ex: 1.0 M Na2SO4 or 2.0 M Na+ or 1.0 M SO42–.
Ch 4 Aqueous Reactions - 2 -
Interconverting Molarity, Moles, and Volume
Use dimensional analysis
moles = liters  molarity
moles
moles = liters 
liters
Ch 4 Aqueous Reactions - 3 -
4.3
Precipitation Reactions
Reactions that result in the formation of an insoluble product are known as precipitation
reactions. (See figure I.1)
A precipitate is an insoluble solid formed by a reaction in solution.
AgBr (s) + KNO3(aq)
Ag(NO3) (aq) + KBr (aq)
Solubility Guidelines for Ionic Compounds
The solubility of a substance is the amount of that substance that can be dissolved in a
given quantity of solvent.
Any substance with the solubility of less than 0.001 mol/L will be referred to as
insoluble.
All common ionic compounds that contain the nitrate anion, NO3–, are soluble in water.
All common ionic compounds of the alkali metal ions (group 1A) and of the ammonium
ion, NH4+, are soluble in water.
Soluble Compounds
Important Exceptions (Cations)
Compounds containing (Anions) NO3-
None
C2H3O2-
None
Cl-
Ag+, Hg22+, Pb2+
Br-
Ag+, Hg22+, Pb2+
I-
Ag+, Hg22+, Pb2+
SO42-
Sr2+, Ba2+Ca2+
Insoluble
Important Exceptions
Compounds
(Cations)
Compounds
S2-
NH4+, the alkali metal cations, and Ca2+,Sr2+, Ba2+
CO32-
NH4+, the alkali metal cations
PO43-
NH4+, the alkali metal cations
OH-
NH4+, the alkali metal cations, and Ca2+,Sr2+, Ba2+
containing (Anions)
Ch 4 Aqueous Reactions - 4 -
Exchange (Metathesis) Reactions
AX  BY  AY + BX
AgNO 3aq  + KClaq   AgCls + KNO3 aq 
One of the following is needed to drive a metathesis reaction:

the formation of a precipitate (s)

the generation of a gas (g) (typically CO2 or H2S)

the production of a weak electrolyte (such as weak acids HCN, HC2H3O2 )

the production of a nonelectrolyte.
Ionic Equations
Molecular equation – shows complete chemical formulas of the reactants and products.
AgNO 3aq + KClaq  AgCls + KNO3 aq 
Complete ionic equation
Ag +aq  + NO3 – aq  + K +aq  + Cl – aq  
AgCls + K+aq  + NO3 – aq 
Spectator ions – present but play no role in the reaction.
Net ionic equation
Ag+aq + Cl – aq  AgCl s
1. Write a balanced molecular equation for the reaction.
2. Rewrite the equation to show the ions that form in solution when each soluble strong
electrolyte dissociates or ionizes into its component ions. Only dissolved strong
electrolytes are written in ionic form.
Ch 4 Aqueous Reactions - 5 -
3. Identify and cancel spectator ions that occur on both sides of the equation.
(J)
4.4
Acid-Base Reactions
Acids are substances that are able to ionize in aqueous solutions to form a hydrogen ion
(H+) and thereby increase the concentration of H+(aq) ions.
H 2SO 4aq   H aq   HSO4  aq 
HSO4  aq  H  aq  SO 4 2aq 
Bases are substances that accept (react with) H+ ions.
H aq   OH aq   H2 Ol 
NH 3aq  H 2Ol   NH4  aq  OH  aq
Bases increase the concentration of OH–(aq) ions.
Strong and Weak Acids and Bases
Acids and bases that are strong electrolytes are called strong acids and strong bases.
Those that are weak electrolytes are weak acids and weak bases.
Molecular Formula
Name of acid
Anion
Name of Anion
HCl (Hydrogen chloride gas)
Hydrochloric acid
Cl-
Chloride
HBr (Hydrogen bromide gas)
Hydrobromic acid
Br-
Bromide
HI (Hydrogen Iodide gas)
Hydroiodic acid
I-
Iodide
HNO3
Nitric acid
NO3-
Nitrate
HClO3
Chloric acid
ClO3-
Chlorate
HClO4
Perchloric acid
ClO4-
Perchlorate
H2SO4
Sulfuric acid
SO42-
Sulfate
Ch 4 Aqueous Reactions - 6 -
Identifying Strong and Weak Electrolytes
Back to Figures I.2, I.3, I.4
Neutralization Reactions and Salts
A neutralization reaction occurs when a solution of an acid and that of a base are
mixed.
HClaq   NaOHaq   H 2Ol  + NaClaq 
acid base 
water
salt 
net ionic
H aq   OH aq   H2 Ol 
Extra, not in textbook, but you should know
Acid-Base Reactions with Gas Formation
2HClaq  Na 2Saq  H 2Sg  2NaClaq
2H  aq   S2aq   H2 Sg 
Carbonates and Bicarbonates
HClaq  NaHCO3aq  H2CO3aq  NaCl aq
H 2CO 3aq   H 2Ol  CO 2g 
H+aq  + HCO3  aq   H 2Ol  CO 2g 
Ch 4 Aqueous Reactions - 7 -
(Section K)
4.5
Oxidation –Reduction Reactions
Loss of electrons by a substance is called oxidation.
The gain of electrons by a substance is reduction.
Oxidation of one substance is always accompanied by the reduction of another as
electrons are transferred between them.
LEO-GER
Loss of Electrons-Oxidation
Gain of Electrons-Reduction
OIL RIG
Oxidation Is Loss of electrons
Reduction Is Gain of electrons
Oxidation Numbers
The oxidation number of an atom in a substance is the actual charge of the atom if it is a
monatomic ion.
1. For an atom in its elemental form, the oxidation number is always zero.
2. For any monatomic ion, the oxidation number equals the charge on the ion.
3. Nonmetals usually have negative oxidation numbers, although they can sometimes be
positive.
a. The oxidation number of oxygen is usually –2 in both ionic and molecular
compounds. The major exception is in compounds called peroxides, which
contain O22- ion, giving each oxygen an oxidation number of –1.
b. The Oxidation number of hydrogen is +1 when bonded to nonmetals and –1
when bonded to metals.
c. The oxidation number of fluorine is –1 in all compounds. The other halogens
have an oxidation number of –1 in most binary compounds. When combined
with oxygen, as in oxyanions, they have positive oxidation states.
4. The sum of the oxidation numbers of all atoms in a neutral compound is zero. The
sum of the oxidation numbers in a polyatomic ion equals the charge of the ion.
Maximum oxidation number = group number.
Ch 4 Aqueous Reactions - 8 -
Minimum oxidation number = group number – 8.
ex:
NO3– Oxidation number of N is +5.
NH3
Oxidation number of N is –3.
Oxidation of Metals by Acids and Salts
A + BX  AX + B
Zns   2HBraq  ZnBr2 aq   H 2g 
Mns   PbNO 3 2aq   MnNO 3 2aq  Pbs 
These reactions are called displacement reactions because the ion in solution is displace
or replaced through oxidation of an element.
Whenever one substance is oxidized, some other substance must be reduced.
The Activity Series
Any metal on the list can be oxidized by the ions of elements below it.
Cus  + 2Ag +aq   Cu 2+aq  + 2Ags 
TABLE: Activity Series of Metals in Aqueous Solution (to be used, not memorized)
Metal
Oxidation Reaction
Lithium
Li(s)  Li+(aq) + e–
Potassium
K(s)  K+(aq) + e–
Barium
Ba(s)  Ba2+(aq) + 2e–
Calcium
Ca(s)  Ca2+(aq) + 2e–
Sodium
Na(s)  Na+(aq) + e–
Magnesium
Mg(s)  Mg2+(aq) + 2e–
Aluminum
Al(s)  Al3+(aq) + 3e–
Manganese
Mn(s)  Mn2+(aq) + 2e–
Ch 4 Aqueous Reactions - 9 -
Zinc
Zn(s)  Zn2+(aq) + 2e–
Chromium
Cr(s)  Cr3+(aq) + 3e–
Iron
Fe(s)  Fe2+(aq) + 2e–
Cobalt
Co(s)  Co2+(aq) + 2e–
Nickel
Ni(s)  Ni2+(aq) + 2e–
Tin
Sn(s)  Sn2+(aq) + 2e–
Lead
Pb(s)  Pb2+(aq) + 2e–
Hydrogen
H2(g) 2H+(aq) + 2e–
Copper
Cu(s)  Cu2+(aq) + 2e–
Silver
Ag(s)  Ag+(aq) + e–
Mercury
Hg(l)  Hg2+(aq) + 2e–
Platinum
Pt(s)  Pt2+(aq) + 2e–
Gold
Au(s)  Au3+(aq) + 3e–
Ch 4 Aqueous Reactions - 10 -
(L)
3.6 Quantitative Information from Balanced Equations
2 H2(g)
+
O2(g)

2 H2O(g)
2 molecules
1 molecule
2 molecules
2 pair molecules
1 pair molecule
2 pair molecules
2 dozen molecules
1 dozen molecule
2 dozen molecules
2(6x1023 molecules)
6x1023 molecules
2(6x1023 molecules)
2 moles
1 mole
2 moles
The coefficients in a balanced chemical equation can be interpreted both as the relative
numbers of molecules (or formula units) involved in the reaction and as the relative
numbers of moles.
2 mol H2 = 1 mol O2 = 2 mol H2O
They are stoichiometrically equivalent.
How much H2O can be produced from 1.57 mol O2?
2 mol H2O 
  3.14 mol H2O
1.57 mol O2  
 1 mol O2 
Combustion of butane, C4H10.
2 C4 H10g   13 O2g  8 CO2g  10 H 2Og 
If we burn 1.00 g of butane, what mass of CO2 can be produced?
Use “box method”.
Ch 4 Aqueous Reactions - 11 -
1.00 g butane X
1 mol butane X 8 mol CO2 X 44.01 g CO2 =
58.14 g butane 4 mol butane
1 mol CO2
= 1.51 g CO2
4.6
Solution Stoichiometry and Chemical Analysis
(Molarity as a conversion factor)
Figures L.2, L.3
Toolbox L.2
Titration – the use of a solution of known concentration, called a standard solution, that
undergoes a specific chemical reaction of known stoichiometry with the solution of
unknown concentration.
Equivalence point – the point at which stoichiometrically equivalent amounts of
quantities are brought together (also called stoichiometric point).
Indicators are used in acid-base reactions to show the end point of a titration, which
coincides closely with the equivalence point.
Ch 4 Aqueous Reactions - 12 -
Example: A 25.0 mL sample of a solution of calcium hydroxide was titrated with 38.0
mL of 6.00M HCl(aq). Calculate the concentration of the Ca(OH)2 solution.
2 HCl + Ca(OH)2  CaCl2 + 2H2O
38.0 mL HCl X 1 L X 6.00 mol HCl X 1 mol Ca(OH)2 = 0.114 mol Ca(OH)2
103 mL
L
2 mol HCl
25.0mL X ( 1 L / 1000 mL) = 0.025 mL
0.114 mol Ca(OH)2 / 0.025 L = 4.56 M Ca(OH)2
Ch 4 Aqueous Reactions - 13 -
(M)
4.7 Theoretical Yields and Limiting Reactants
Theoretical Yields
The quantity or product that is calculated to form when all of the mentionned reactant
reacts is called the theoretical yield.
The amount actually obtained in a reaction is called the actual yield. The major physical
reasons why the actual yield is lower than the theoretical yield are:
1) competing reactions (e.g. CO and C formed along with CO2)
2) precision of experimental apparatus.
Human error or manipulation can contribute to lower actual yield but are considered to be
non-physical reasons.
The percent yield of a reaction relates the actual yield to the theoretical (calculated) yield:
Percent Yield =
actual yield
 100%
theoretical yield
Example: Aluminum sulfide, Al2S3 reacts with water to form aluminum hydroxide,
Al(OH)3 and hydrogen sulfide gas, H2S according to the following chemical equation:
Al2S3 (s) + 6 H2O (l)  2 Al(OH)3 (s) + 3 H2S (g)
Calculate the percent yield when 45.7 g of Al(OH)3 are recuperated after 50.0 g of Al2S3
are completely reacted.
Ch 4 Aqueous Reactions - 14 -
LR, the Limiting Reagent
The reagent that is completely consumed in a reaction is called the limiting reactant or
limiting reagent, because it determines, or limits, the amount of product formed.
The other reactants are sometimes called excess reactants or excess reagents.
TIP Whenever the amounts of two or more reactants are given, you must
do calculations for each reactant to determine whether each is completely
consumed!!! (You can also calculate afterwards how much of the other
reactants are left-over.)
There are (at least) two ways that you can use to calculate whether two
reactants are completely consumed. (Toolbox M.1) Use whichever you
prefer:
Method 1: Determine whether there is enough of one reactant to react with
the other.
Method 2: Determine how much of a product each reactant would give. The
smaller amount of product comes from the limiting reagent.
Example: 100.0 g of ethylene oxide, C2H4O and 150.0 g of oxygen are
reacted to give carbon dioxide, CO2 and water, H2O. Calculate the maximum
mass of CO2 that can be produced with these reactants (and any left over
reactants).
Method 1:
2 C2H4O + 5 O2  4 CO2 + 4 H2O
Step 1 Determine the moles of each reactant
n (C2H4O) = mass (C2H4O) / gFW (C2H4O)
= 100.0 g(C2H4O ) / 44.06 g·mol-1(C2H4O)
= 2.270 mol(C2H4O)
n (O2) = mass (O2) / gFW (O2)
= 150.0 g(O2) / 32.00 g·mol-1(O2)
= 4.688 mol(O2)
Step 2 Apply stoiciometric relation to one of the reactants
Ch 4 Aqueous Reactions - 15 -
2 mol C2H4O = 5 mol O2
2.270 mol(C2H4O) X (5 mol O2 / 2 mol C2H4O) = 5.675 mol O2
This means 2.270 mol(C2H4O) need 5.675 mol O2 to completely react
Step 3 Compare what is needed to what is available.
4.688 mol(O2) available but 5.675 mol O2 needed to react
Since less mol(O2) is available than is needed, O2 is the Limiting Reagent
Step 4 Continue calculations with the LR
4.688 mol(O2) X (4 mol CO2 / 5 mol O2) X 44.01 g·mol-1(CO2)
= 165.1 g (CO2)
To calculate left-over C2H4O:
= used-up (C2H4O) - available (C2H4O)
The “used-up (C2H4O)” is calculated from the LR
4.688 mol(O2) X (2 mol C2H4O / 5 mol O2) X 44.06 g·mol-1(C2H4O)
= 82.62 g (C2H4O) used-up
Left-over = 100.0 g (C2H4O) available - 82.62 g (C2H4O) used-up
Left-over = 17.4 g (C2H4O)
Ch 4 Aqueous Reactions - 16 -
Method 2:
Ch 4 Aqueous Reactions - 17 -