Chapter Outline
 8.1 Solutions and Their Concentrations
 8.2 Dilutions
 8.3 Electrolytes and Nonelectrolytes
 8.4 Acids, Bases, and Neutralization Reactions
 8.5 Precipitation Reactions
 8.6 Oxidation-Reduction Reactions
 8.7 Titrations
 8.8 Ion Exchange
Double-Displacement Reactions Again
Precipitation reactions in this section follow the same
pattern as acid-base reactions:
(+) (-)
(+) (-)
(+) (-)
(+) (-)
AB + CD → AD + CB
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Types of Solutions
 Saturated solution:
• Contains the maximum amount of solute that can
dissolve in a given volume at a given temperature.
• Solubility = g solute/100 mL solvent.
 Supersaturated solution:
• Contains more dissolved solute than predicted at a
given temperature.
A Saturated Solution Example
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Supersaturated Solution
Sodium acetate precipitates from a supersaturated solution.
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 Precipitate:
• Solid product formed from reactants in
solution.
• AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s)
 Can predict formation of precipitates
based on solubility “rules.”
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We learned in Ch 4, p. 134, that the electrostatic energy (strength) of an
ionic bond depended on the product of the charges, divided by the
internuclear distance. The solubility of ionic compounds generally
follows the trend that the greater the charge, and the shorter the
internuclear distance, the more insoluble the compound is.
α (alpha) means “is proportional to”
Q+ = charge on the cation
Q- = charge on the anion
d = internuclear distance
-
+
𝑄+ 𝑥 𝑄−
𝐸 α
𝑑
d
cations
+1
compounds
alkali metals
NH4+
+2
+3
Al3+
anions
compounds
solubility
-1
NO3-
Always soluble
CH3COO-2
SO42-
Soluble with exceptions
-2
S2-,
O2-
Insoluble with
exceptions
-3
PO43-
Insoluble with
exceptions
Always
soluble
Soluble
w/exceptions
and Ag+
Add to table
Insoluble
w/exceptions
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Table 8.3 - Soluble Compounds
 Soluble Cations:
• Group I ions (alkali metals) and NH4+
 Soluble Anions:
• NO3− and CH3COO− (acetate)
• Halides (Group 17)
»
Exceptions: Cu+, Hg22+, Ag+, Pb2+
• Sulfates (SO42−)
»
Exceptions: Hg22+, Ag+, Pb2+, Ca2+, Ba2+, Sr2+
Add to table
Table 8.3 - Insoluble Compounds
 All hydroxides (OH−) except:
• Group IA = Li, Na etc
• Group IIA = Ca(OH)2, Sr(OH)2, and Ba(OH)2
 All sulfides (S2−) except:
• Group IA and NH4+
• All carbonates (CO32−) except IA, NH4+
 All phosphates (PO43−) except IA, NH4+
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Mnemonic device to help remember the
exceptions to the soluble compounds -
Cute HAPpy halides
HAPpy sulfates "2"
Cl-, Br-, I-
Cu+
Ca2+
Sr2+
Hg22+
Pb2+
Ag+
Hg22+
Pb2+
Ag+
Ba2+
from
Group 2
Mnemonic device to help remember the insoluble
compounds -
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S2OHO2-
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CrO42-
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CO32-
PO43-
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Applying the Solubility Rules – Will a Precipitate Form?
Does a precipitate form when sodium chloride is mixed with silver nitrate?
If so, write the net ionic equation for the formation of the precipitate:
Cute HAPpy halides
Cl-, Br-, I-
Cu+ Hg22+ Ag+ Pb2+
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
Na+(aq) + Ag+(aq) + Cl-(aq) + NO3-(aq)  AgCl(s) + Na+(aq) + NO3-(aq)
Ag+(aq) + Cl-(aq)  AgCl(s)
Formation of an Insoluble Product
NaI(aq) + Pb(NO3)2(aq) →
The soluble compounds break up
into ions and recombine -
K+(aq)
I-(aq)
Pb2+(aq)
NO3-(aq)
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Sample Exercise 8.6
A precipitate forms when aqueous solutions of ammonium sulfate and
barium chloride are mixed. Write the net ionic equation for the reaction.
(NH4)2SO4(aq) + BaCl2(aq)  2 NH4Cl(aq) + BaSO4(s)
2NH4+(aq) + SO42-(aq) + Ba2+(aq) + 2Cl-(aq)  BaSO4(s) + 2NH4+(aq) + 2Cl-(aq)
Ba2+(aq) + SO42-(aq)  BaSO4(s)
Sample Exercise 8.7 – Prediction the Mass of a Precipitate
Barium sulfate is used to enhance Xray imaging of the upper and lower GI tract. In
upper GI imaging, patients drink a suspension of solid barium sulfate in H 2O. The
compound is not toxic because of its limited solubility. To make pure barium sulfate,
a precipitation reaction is employed: aqueous solutions of soluble barium nitrate
and sodium sulfate are mixed together, and solid barium sulfate is separated by
filtration. How many grams of barium sulfate (MW = 233.40) will be produced if
exactly 1.00 L of 1.55 M barium nitrate is reacted with excess sodium sulfate?
Ba(NO3)2(aq) + Na2SO4(aq)  BaSO4(s) + 2 NaNO3(aq)
V = 1.00 L
1.55 M
Plan:
g BaSO4 =
Excess
no LR
?g
MW = 233.40
ratio
M x V = mol Ba(NO3)2  mol BaSO4  g BaSO4
233.40 g BaSO4
1.55 mol
1 mol BaSO4
x 1.00 L x
x
= 362 g BaSO4
1 mol BaSO4
1 mol Ba(NO3)2
L
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Sample Exercise 8.8 – Calculating Solute Concentration
from Mass of Precipitate
To determine the concentration of chloride ion in a 100.0 mL sample of
groundwater, a chemist adds a large enough volume of a solution of AgNO 3
to precipitate all the Cl- ions as silver chloride. The mass of the resulting
precipitate is 71.7 mg. What is the Cl- concentration in the sample in mg/L?
AgNO3(aq) + Cl-(aq)
excess
no LR

AgCl(s) + NO3-(aq)
V = 100.0 mL 71.7 mg
= 0.100 L
= 0.0717 g
? mg/L
ratio
Plan: mg AgCl  mol AgCl  mol Cl- g  mg/L
mg/L Cl- =
0.0717 g AgCl
0.100 L
= 0.177 g Cl-/mL
x
1 mol AgCl
143.32 g AgCl
x
1 mol Cl1 mol AgCl
x
35.45 g Cl1 mol Cl-
= 177 g Cl-/mL
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