Download Lecture Sections 15.6 - Fulton County Schools

Chemistry
FIFTH EDITION
by Steven S. Zumdahl
University of Illinois
Chapter 15
Applications of Aqueous Equilibria
Schedule Chapter 15 on Website
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Section 15.6
Solubility Equilibria & the Solubility Product
The typical ionic solid dissolves in water and
Dissociates completely into separated hydrated
Anions and cations.
CaF2 (s)  Ca2+ (aq) + 2 F- (aq)
BUT once the ions are in solution, they can collide
and reform CaF2 (s).
THEREFORE, Equilibrium is reached!!!
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CaF2 (s)  Ca2+ (aq) + 2 F- (aq)
Equilibrium Expression:
K = [Ca2+] [F-]2
K = Ksp (Solubility Product)
[ ] = moles/L
CaF2 (s) not included in K.
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Solubility Product
•For solids dissolving to form aqueous
solutions.
•Bi2S3(s)  2Bi3+(aq) + 3S2(aq)
•Ksp = solubility product constant
•and
–
Ksp = [Bi3+]2[S2]3
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Ksp Values at 25C for Common Ionic
Solids
• See Table 15.4 on page 759.
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Solubility Product
•
“Solubility” = s = concentration of
Bi2S3 that dissolves, which equals
1/2[Bi3+] and 1/3[S2].
or [Bi3+] = 2s & [S2-] = 3s
•
Note: Ksp is constant (at a given
temperature)
•
s is variable (especially with a common
ion present)
–
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Important:
1. Experimentally determined solubility of an
ionic solid can be used to calculate its Ksp value.
2. Solubility of an ionic solid can be calculated if
its Ksp value is known.
3. Relative solubilities can be predicted by
comparing Ksp values only for salts
that produce the same total # of ions.
4. Solubility of a solid is lowered if the solution
already contains ions common to the solid.
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The pH of a solution can greatly affect
a salt’s solubility.
General Rule:
If the anion, X-, is an effective base
(i.e., HX is a weak acid)
Then, the salt, MX, will show increased
solubility in an acidic solution.
These salts are much more soluble in an acidic
solution than in pure water.
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Added H+ will react with the base X- forming
the conjugate acid.
As the base is removed, more of the salt will
dissolve to replenish the basic anion, X-.
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Let’s Do
81a & b, 85a, 87, 89 a & b, 91, 93, 95
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Section 15.7
Precipitation & Qualitative Analysis
So far solids dissolving in solutions
Now--Look at formation of solid from solution
That is, the REVERSE Process.
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Section 15.7
Precipitation & Qualitative Analysis
Calculate Q, ion product
Like Ksp , but use initial conc. rather than
equilibrium conc.
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CaF2 (s)  Ca2+ (aq) + 2 F- (aq)
Q = [Ca2+]o [F-]o2
Q = Ion Product
[ ] = moles/L
CaF2 (s) not included in Q.
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If Q > Ksp , then precipitation occurs.
If Q < Ksp , then no precipitation occurs.
If precipitation occurs, you can also do
calculations to determine equilibrium conc.
in solution after precipitation.
Let’s do: # 97 & 99
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Selective Precipitation
Method to separate a mixture of ions in which
reagents are added to precipitate single ions
or small groups of ions
Example:
Solution of Ba2+ and Ag+ ions
Add NaCl
only AgCl precipitates,
Ba2+ still in solution
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Qualititative Analysis Scheme
See Fig 15.11 & 15.12 on page 771
Read pages 770-772
Sulfide ion is often used to separate
metal ions. See page 770.
Let’s do: # 101
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Section 15.8
Equilibria Involving Complex Ions
•
Complex Ion: A charged species
consisting of a metal ion surrounded by ligands.
• Ligand - a Lewis base
- a molecule or ion having a lone
electron pair that can be donated to an empty
orbital on the metal ion to form a covalent bond.
• Common Ligands: H2O, NH3, Cl-, CNCopyright©2000 by Houghton
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Coordination Number: Number of ligands
attached to a metal ion.
(Most common are 6 , 4 and 2.)
Examples: Co(H2O)62+, CoCl42-, Ag(NH3)2+
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Metal ions can add ligands one at a time in steps.
+ NH3  Ag(NH3)+
K1 = 2.1 x 103
Ag(NH3)+ + NH3  Ag(NH3)2+
K2 = 8.2 x 103
Ag+
________________________________________________________________
Ag+
+ 2NH3  Ag(NH3)2+
K = K1 x K2
Formation (Stability) Constants: The
equilibrium constants characterizing the
stepwise addition of ligands to metal ions.
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Let’s do
Exercises # 103, 107, 109, 111, 115
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