10/18/2011
Reaction Classes
• Precipitation: synthesis of an ionic solid
– a solid precipitate forms when aqueous solutions of certain
ions are mixed
• Acid-Base: proton transfer reactions
– acid donates a proton to a base, forming a molecule (water or
another weak acid) and an aqueous salt
– Acid: proton-donor; Base: proton-acceptor
• Oxidation-Reduction: electron transfer reactions
– electron transfer from one species to another, causing a change
in the oxidation state of the two species
– OIL RIG: Oxidation Is Loss (of e-), Reduction Is Gain (of e-)
– includes combustion, the reaction of a substance with oxygen
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Precipitation Reactions
• Sometimes when we mix two solutions together, an
insoluble solid will form:
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
K2CrO4(aq) + Ba(NO3)2(aq)  BaCrO4(s) + 2 KNO3(aq)
• The solid, called a precipitate (or insoluble salt) is
insoluble in water.
• It is so insoluble that when its component ions find each
other in solution, they very rapidly get locked together in
large clumps, driving the reaction towards the products.
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10/18/2011
The Solubility of Ionic Compounds in Water
The solubility of ionic compounds in water depends upon the relative
strengths of the electrostatic forces between ions in the ionic
compound and the attractive forces between the ions and solvent
molecules (often water).
There is a tremendous range in the solubility of ionic compounds in
water. The solubility of so called “insoluble” compounds may be
several orders of magnitude less than ones that are called “soluble”
in water. For example, consider the solubility (in g/L) of the following
compounds in water at 20oC :
Solubility of NaCl
Solubility of MgCl2
Solubility of AlCl3
Solubility of PbCl2
Solubility of AgCl
Solubility of CuCl
= 365
= 542.5
= 699
=
9.9
=
0.009
=
0.0062
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Precipitation Reactions
• The main challenge with precipitation reactions is predicting
what solid (if any) will form.
• Example: Addition of potassium chromate to barium nitrate.
K2CrO4(aq) + Ba(NO3)3(aq)
What is the precipitate?
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10/18/2011
Precipitation Reactions
One approach: consider the ions present in the solutions:
K2CrO4 (aq)
Ba(NO3)2 (aq)
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Precipitation Reactions
• When combining ions, consider which precipitates could form
• Precipitates must have a net charge of zero
• Possible choices are BaCrO4 and KNO3
Clues: Potassium forms soluble salts.
Chromate is yellow.
BaCrO4 precipitates!
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Precipitation Reactions
BaCl2(aq)
Cl-
Ba2+
Cl-
AgNO3(aq)
NO3-
Ag+
Ba2+
Cl-
Cl-
Ag+
NO3-
BaCl2(aq) + AgNO3(aq)
Ag+
ClBa2+
Cl-
NO3
-
NO3Ag+
Cl-
Cl- Ba2+
2AgNO3(aq) + BaCl2(aq)  2AgCl(s) + Ba(NO3)2(aq)
Solid silver chloride Barium and nitrate
precipitates out of
ions are left in
solution.
aqueous solution.
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Precipitation Reactions
• In the previous example, we know that Ba(NO3)2 is not the
precipitate since in our first example this compound was
soluble in water.
• Therefore, one way to determine if a precipitate will form is
to simply study combinations of reactions where you know
that one of the reaction products is soluble.
• That is exactly how precipitation or “solubility” rules have
been determined…
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Solubility Rules
Know these!!!
In general, 1-4 are “solubles” and 5-6 are “insolubles.”
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Precipitation Reactions: Will a Precipitate form?
Example: If a solution containing potassium chloride is added to a
solution containing ammonium nitrate, will a precipitate form?
KCl(aq) + NH4NO3(aq) → K+(aq) + Cl-(aq) + NH4+(aq) + NO3-(aq)
Possible reaction products are KCl and NH4NO3, or NH4Cl and KNO3.
All are soluble, so there is no precipitate.
KCl(aq) + NH4NO3 (aq) = No Reaction!
Example: If a solution containing sodium iodide is added to a solution
containing lead(II) nitrate, will a precipitate form?
NaI (aq) + Pb(NO3)2 (aq) →
Na+(aq) + I (aq) + Pb2+(aq) + 2 NO3- (aq)
Lead(II) iodide is insoluble; therefore a precipitate will form.
2 NaI (aq) + Pb(NO3)2 (aq)
PbI2 (s) + 2 NaNO3 (aq)
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10/18/2011
Total and Net Ionic Equations
Conventional (molecular) equation: a bookkeeping of all species
present, and arranged for charge neutrality.
Pb(NO3)2(aq) + 2 NaI(aq)
PbI2(s) + 2 NaNO3(aq)
Total Ionic equation: all aqueous species are split up into their
component ions.
Pb2+(aq) + 2 NO3-(aq) +
2 Na2+(aq) + 2 I-(aq)
PbI2(s) + 2 Na+(aq) + 2 NO3-(aq)
Na+ and NO3- are
spectator ions in
Net Ionic equation: indicates exactly the
this
reaction…they
chemical change that occurs, and nothing more.
are not involved in
Pb2+(aq) + 2 I-(aq)
PbI2(s)
the chemical change.
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Predicting Products of Reactions
Write the complete ionic, net ionic, and conventional eqns.
Al(NO3)3(aq) +
Ba(OH)2(aq)
FeSO4(aq) +
CaCl2(aq) +
KCl(aq)
Na2SO4(aq)
Na2CrO4(aq) +
AlBr3(aq)
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Example
Write the complete ionic, net ionic and conventional balanced
chemical equations for the reaction of potassium sulfide and
nickel(II) nitrate.
First: Remember nomenclature and write the ions in solution.
K2S(aq) and Ni(NO3)2(aq) gives:
K+ (aq) S2-(aq) Ni2+(aq) NO3-(aq)
Next: Use the solubility rules to write the complete ionic eqn.
2 K+(aq) + S2-(aq) + Ni2+(aq) + 2 NO3-(aq)
2 K+(aq) + 2 NO3-(aq) + NiS(s)
Next: Cancel spectator ions on either side and write the net ionic eqn.
Ni2+(aq) + S2-(aq)
NiS(s)
Last: Write the conventional balanced equation.
K2S(aq) +
Ni(NO3)2(aq)
2 KNO3 (aq) +
NiS (s)
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Selective Precipitation
• We’ve seen that some ionic
compounds are soluble, while
others are not.
• We can use this behavior to
removes species selectively.
• Example: separating Ag+ from
Ba2+ and Fe3+.
• Notice that selective
precipitation is nothing more
than an application of the
solubility rules.
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Solubility and Stoichiometry - I
To determine the concentration of chloride ion in a sample of
groundwater, a chemist adds 1.0 mL of 1.00 M AgNO3(aq) to
100.0 mL of the sample. The resulting mass of AgCl
precipitate is 71.7 mg. What is the concentration of chloride
in the original sample (expressed in mol Cl- per L?)
Ag+(aq) + Cl-(aq)
Mass of AgCl

AgCl(s)
Moles of Cl-in
100.0 mL
Moles of AgCl
mol of Clper liter
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Example (cont)
? mol Cl- = 71.7 mg AgCl(s)
1g
1 mol AgCl
1 mol Cl-
1000 mg 143.32 g AgCl 1 mol AgCl
= 0.00050028 mol Cl? mol/L Cl- =
0.00050028 mol Cl- 1000 mL
100.0 mL
1L
= 0.00500 M Cl= 5.00 x 10–3 M Cl-
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Solubility and Stoichiometry - II
What mass of solid NaCl must be added to 1.50 L of a 0.100
M AgNO3 solution to precipitate all of the Ag+?
Net Ionic Rxn:
Ag+(aq) + Cl-(aq)  AgCl(s)
First: Determine the moles of Ag+ in solution.
0.100 mol AgNO3
1 L solution
1 mol Ag+
0.150 mol AgNO3 x
1 mol AgNO3
1.50 L solution x
= 0.150 mol AgNO3
= 0.150 mol Ag+
Then: Use stoichiometry of precipitation reaction to determine moles
Cl- needed, then the mass of NaCl.
0.150 mol Ag+ x
1 mol Clx 1 mol NaCl
x 58.4 g NaCl = 8.76 g NaCl
1 mol Ag+
1 mol Cl1 mol NaCl
Solubility and Stoichiometry - III
We can also think about limiting reactants in precipitation
reactions…
Calculate the mass of AgCl formed when 0.500 L of 1.20 M AgNO3
is mixed with 0.250 L of 0.350 M NaCl.
First: Determine the moles of Ag+ and Cl-.
1.20 mol AgNO3
1 L solution
0.250 L solution x 0.350 mol NaCl
1 L solution
0.500 L solution x
= 0.600 mol AgNO3
= 0.0875 mol NaCl
LR
Then: Because it is a 1:1 rxn, the LR is easy to identify…NaCl is the
LR and the 0.0875 moles will dictate the mass of AgCl produced.
1 mol AgCl 143.3 g AgCl
0.0875 mol Cl- x
x
= 12.5 g AgCl
1 mol Cl1 mol AgCl
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10/18/2011
Reaction Classes
• Precipitation: synthesis of an ionic solid
– a solid precipitate forms when aqueous solutions of certain
ions are mixed
• Acid-Base: proton transfer reactions
– acid donates a proton to a base, forming a molecule (water or
another weak acid) and an aqueous salt
– Acid: proton-donor; Base: proton-acceptor
• Oxidation-Reduction: electron transfer reactions
– electron transfer from one species to another, causing a change
in the oxidation state of the two species
– OIL RIG: Oxidation Is Loss (of e-), Reduction Is Gain (of e-)
– includes combustion, the reaction of a substance with oxygen
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Acid-Base Rxns
• Bronsted-Lowry Theory: acid/base reactions are proton-transfer
processes.
– acid is proton-donor (H+ ion donor).
– base is proton- acceptor (H+ ion acceptor).
B
+
H—A
B—H
A-
+
• When an acid gives its proton to water, water is acting as a base.
H2O
+
H—A
H2O—H
+
A-
• When a base accepts a proton from water, water is acting as an
acid.
HO—H
+
B
HO
+
H—B+
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H3O+ is called the hydronium ion
Strong and Weak Acids
• Strong acids (think “strong electrolyte”) undergo complete
ionization.
– For HCl, essentially all the HCl molecules split up into ions.
HCl(aq) + H2O(l)
H3O+(aq) + Cl-(aq)
Dominant species
• Weak acids (think “weak electrolyte”) undergo incomplete
ionization.
– For HF, only a very few H+ and F- ions exist in solution…the
reverse reaction dominates the chemistry
– Weak acids are like insoluble salts…they don’t like to dissociate
very much.
HF(aq) + H2O(l)
H3O+(aq) + F-(aq)
Dominant species
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Strong and Weak Acids
• Strong acids to know:
– Two well-known oxyacids: nitric (HNO3), sulfuric (H2SO4)
– Group 7 acids: hydrochloric (HCl), hydrobromic (HBr),
hydroiodic (HI)
– Two chlorine oxyacids: chloric (HClO3), perchloric (HClO4)
Any other acids we see in this course will be weak.
• Polyprotic acids lose only their first H+ easily, producing H3O+(aq) and
an acid anion:
H2SO4(aq) + H2O(l)  HSO4-(aq) + H3O+(aq)
The acid anion of sulfuric acid, hydrogen sulfate anion, is also an acid,
but it does not dissociate as readily as sulfuric acid…it is a weak acid.
Selected Acids and Bases
Acids
Strong: H+(aq) + A-(aq)
hydrochloric, HCl
hydrobromic, HBr
hydroiodoic, HI
nitric acid, HNO3
sulfuric acid, H2SO4
perchloric acid, HClO4
chloric acid, HClO4
Weak
hydrofluoric, HF
phosphoric acid, H3PO4
acetic acid, CH3COOH
(or HC2H3O2)
Bases
Strong: M+(aq) + OH-(aq)
lithium hydroxide, LiOH
sodium hydroxide, NaOH
potassium hydroxide, KOH
calcium hydroxide, Ca(OH)2
strontium hydroxide, Sr(OH)2
barium hydroxide, Ba(OH)2
*(M is Group I or II metal)
Weak
ammonia, NH3
accepts proton from water to make
NH4+(aq) and OH-(aq)
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Relative Acid Strength
• NOTE: The terms strong and weak refer only to how many
acid (HA) molecules dissociate to H+ and A-.
• They have nothing to do with how corrosive the acid is.
• They also have nothing to do with concentration.
Acid burn resulting from
contact with HF, hyrdofluoric
acid (a weak acid!).
Predicting Products of Acid/Base Rxns
Key to predicting the outcome of an acid/base reaction is
identifying what is the acid and what is the base.
2 HClO4(aq) +
HCN(aq) +
Mg(OH)2(aq)
2 H2O(l) + Mg(ClO4)2(aq)
NaOH(aq)
H2O(l)
CH3COOH(aq) +
KF(aq) +
KOH(aq)
H2O(l)
H2PO4-(aq) +2 CH3COONa(aq)
+
H2O(l)
+
HF(aq)
+
NaCN(aq)
CH3COOK(aq)
KOH(aq)
Na2HPO4(aq)+2 CH3COOH(aq)
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10/18/2011
Writing Balanced Equations for
Neutralization Reactions - I
Write balanced descriptive equations (molecular, total ionic,
and net ionic) for the following chemical reactions:
a) calcium hydroxide(aq) and hydroiodic acid(aq)
b) lithium hydroxide(aq) and nitric acid(aq)
c) barium hydroxide(aq) and sulfuric acid(aq)
These are all strong acids and bases, therefore, the products will be
water and the corresponding salts.
a)
Ca(OH)2 (aq) + 2HI(aq)
CaI2 (aq) + 2H2O(l)
Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq) 
Ca2+(aq) + 2 I -(aq) + 2 H2O(l)
2 OH -(aq) + 2 H+(aq)
2 H2O(l)
Writing Balanced Equations for
Neutralization Reactions - II
b) LiOH(aq) + HNO3
LiNO3 (aq) + H2O(l)
(aq)
Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq)

Li+(aq) + NO3-(aq) + H2O(l)
OH -(aq) + H+(aq)
H2O(l)
c) Ba(OH)2 (aq) + H2SO4 (aq)
BaSO4 (s) + 2 H2O(l)
Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42-(aq) 
Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42-(aq)

BaSO4 (s) + 2 H2O(l)
BaSO4 (s) + 2 H2O(l)
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Acid/Base and Stoichiometry
What volume of 0.125 M HCl is needed to neutralize 200.0 mL of
a 0.00955 M Ca(OH)2 solution?
Calculate the number of moles of base:
Vbase x Mbase = 0.2000 L x 0.00955 M = 0.00191 mol Ca(OH) 2
From the balanced equation find the moles of acid needed:
Ca(OH)2(aq) + 2 HCl (aq)
2 H2O(l) + CaCl2 (aq)
Since there are two hydroxide ions per molecule of base and one
proton per molecule of acid, we need twice as much acid as we
have base: 0.00382 mol HCl
Volume of acid:
moles acid
0.00382 mol
Vacid =
=
= 0.0306 L HCl
Macid
0.125 mol
L
Acid/Base and Limiting Reagent
75.0 mL of 0.250 M HCl is added to 225.0 mL of 0.0550 M Ba(OH)2
solution. What is the concentration of the excess H+ or OH- ions
left in this solution?
2 HCl(aq) + Ba(OH)2(aq)
75.0 mL
225.0 mL
0.250 M
0.0550 M
 2 H2O(l) + BaCl2(aq)
1. How many moles of each reactant?
2. Which is limiting?
3. How much H+ or OH- is left over?
4. What is the new volume?
5. What is the concentration of the excess reactant?
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