Transcript Chapter 16

Chapter 16
Solubility and
Complex Ion
Equilibria
Chapter 16
Table of Contents
16.1
16.2
16.3
Solubility Equilibria and the Solubility Product
Precipitation and Qualitative Analysis
Equilibria Involving Complex Ions
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Section 16.1
Solubility Equilibria and the Solubility Product
Ca 5 (PO4 )3OH
Ca 5 (PO 4 )3  OH 
In the forward direction: the acidic water(containing
CO2) dissolves the underground limestone(CaCO3) ,
forming a cavern.
In the backward direction: the water drips from the
ceiling of the cave, and CO2 is lost to the air, causing
solid CaCO3 to form, producing stalactites(鐘乳石) and
stalagmites(石筍)
在石灰岩裏面，含有二氧化碳的水，滲入石灰岩隙縫中，會溶解其中
的碳酸鈣。這溶解了碳酸鈣的水，從洞頂上滴下來時，由於水分蒸發
、二氧化碳逸出，使被溶解的鈣質又變成固體(稱為固化)。由上而下
逐漸增長而成的，稱為「鐘乳石」。
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Section 16.1
Solubility Equilibria and the Solubility Product
藍湖洞(巴西)
巴西馬托格羅索地區藍湖洞。洞內的鐘乳石，石筍密密麻
麻排列，令人讚嘆不已。地質學家普遍認為是由於洞中藍湖的
存在，為鐘乳石的形成提供了必要的碳酸鹽等組成。
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Section 16.1
Solubility Equilibria and the Solubility Product
• Tooth decay: when food lodges between the teeth, acids
form that dissolve tooth enamel ( hydroxyapatite,
Ca5(PO4)3OH ).
Ca 5 (PO4 )3OH( s)
Ca 5 (PO4 )3 (aq)  OH 
Adding F- to drinking water, Since F- replaces the OH- in
Ca5(PO4)3OH to produce Ca5(PO4)3F, and CaF, both of
which are less soluble in acids than the original enamel.
Ca 5 (PO4 )3F( s)
CaF
Ca 5 (PO4 )3 (aq)  F
Ca 2+ + 2F-
Ksp
Ksp
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Section 16.1
Solubility Equilibria and the Solubility Product
Solubility Equilibria
• Solubility product (Ksp) – equilibrium constant;
has only one value for a given solid at a given
temperature.
• Solubility – an equilibrium position.
Bi2S3(s)
2Bi3+(aq) + 3S2–(aq)
2
2
K sp = Bi  S 
3+
3
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Section 16.1
Solubility Equilibria and the Solubility Product
Ksp
Values at
25 °C for
Common
Ionic
Solids
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Section 16.1
Solubility Equilibria and the Solubility Product
• The solubility product is an equilibrium constant and has
only one value for given solid at a given temperature.
• Solubility is an equilibrium position
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Section 16.1
Solubility Equilibria and the Solubility Product
Concept Check
In comparing several salts at a given
temperature, does a higher Ksp value always
mean a higher solubility?
Explain. If yes, explain and verify. If no,
provide a counter-example.
No
The salts must contain the same number of
ions to relate Ksp values to solubility directly,
For example, for a binary salt, Ksp = s2 (s =
solubility); for a ternary salt, Ksp = s3.(not
include coefficient of soluble salt)
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Section 16.1
Solubility Equilibria and the Solubility Product
Exercise
Calculate the solubility of silver chloride in
water. Ksp = 1.6 × 10–10 (AgCl
Ag++Cl- )
x
x
Ksp  x 2  x  1.6 1010
1.3×10-5
M
Calculate the solubility of silver phosphate in
water. Ksp = 1.8 × 10–18 (Ag3PO4 3Ag++PO4- )
3x
1.6×10-5
x
3
4
18
Ksp

(3
x
)
(
x
)

27
x

1.8

10
M
x  1.6 105
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Section 16.1
Solubility Equilibria and the Solubility Product
Concept Check
How does the solubility of silver chloride in
water compare to that of silver chloride in an
acidic solution (made by adding nitric acid to
the solution)?
Explain.
The solubilities are the same.
The solubilities are the same. Since HCl is a strong acid,
it is completely dissociated in water.
There are no common ions between AgCl and HNO3.
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Section 16.1
Solubility Equilibria and the Solubility Product
Concept Check
How does the solubility of silver phosphate in
water compare to that of silver phosphate in an
acidic solution (made by adding nitric acid to
the solution)?
Explain.
The silver phosphate is more soluble in an acidic
solution.
This is because the phosphate ion is a relatively good base and
will react with the proton from the acid (essentially to completion).
The phosphate ion does not react nearly as well with water.
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Section 16.1
Solubility Equilibria and the Solubility Product
Concept Check
How does the Ksp of silver phosphate in water
compare to that of silver phosphate in an
acidic solution (made by adding nitric acid to
the solution)?
Explain.
The Ksp values are the same.
The Ksp values are the same (assuming the
temperature is constant).
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Section 16.1
Solubility Equilibria and the Solubility Product
Exercise
Calculate the solubility of AgCl in:
Ksp = 1.6 × 10–10
a) 100.0 mL of 4.00 x 10-3 M calcium chloride.
(8 103  x)( x)  1.6 1010  8 103 x  1.6 1010
2.0×10-8 M
b) 100.0 mL of 4.00 x 10-3 M calcium nitrate.
1.3×10-5 M
( x)( x)  1.6 1010  x  1.3 105
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Section 16.2
Atomic Masses
Precipitation
and Qualitative Analysis
Precipitation (Mixing Two Solutions of Ions)
MX(s)
M + + X-
Q=  M +   X - 
0
0
+



Ksp=  M   X 
• Q > Ksp; precipitation occurs and will continue
until the concentrations are reduced to the point
that they satisfy Ksp.
• Q < Ksp; no precipitation occurs.
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Section 16.2
Atomic Masses
Precipitation
and Qualitative Analysis
Selective Precipitation (Mixtures of Metal Ions)
• Use a reagent whose anion forms a precipitate
with only one or a few of the metal ions in the
mixture.
• Example:
 Solution contains Ba2+ and Ag+ ions.
 Adding NaCl will form a precipitate with Ag+
(AgCl), while still leaving Ba2+ in solution.
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Section 16.2
Atomic Masses
Precipitation
and Qualitative Analysis
Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
+
-
H 2S
H + HS
HS-
H + + S2-
CuS(Ksp=8.5 10-45 )
HgS(Ksp=1.6 10-54 )
MnS(Ksp=2.3 10-13 )
NiS(Ksp=3 10-21 )
• At a low pH, [S2–] is relatively low and only the
very insoluble HgS and CuS precipitate.
• When OH– is added to lower [H+], the value of
[S2–] increases, and MnS and NiS precipitate.
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Section 16.2
Atomic Masses
Precipitation
and Qualitative Analysis
Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
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Section 16.2
Atomic Masses
Precipitation
and Qualitative Analysis
Qualitative Analysis
• Group 1- insoluble chloride (Ag+,Pb2+,Hg22+)
• Group 2- Sulfides insoluble in acidic solution ( Hg2+,
Cd2+, Bi3+, Cu2+, Sn4+)
• Group 3- Sulfides insoluble in basic solution ( Co2+, Zn2+,
Mn2+, Ni2+, Fe2+) and Cr(OH)3, Al(OH)3)
• Group 4- Insoluble carbonates ( Group 2A cations form
insoluble carbonates by the addition of CO32-. For
example, Ba2+, Ca2+ and Mg2+)
• Group 5- Alkali metal and ammonium ions
(Group 1A cations and NH4+,identified by the
characteristic colors they produce by flame test)
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Section 16.2
Atomic Masses
Precipitation
and Qualitative Analysis
Separating the Common Cations by Selective Precipitation
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Section 16.2
Precipitation
Atomic
Masses
and Qualitative Analysis
Flame Test
Potassium
Sodium
Ken O'Donoghue
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Section 16.2
Atomic Masses
Precipitation
and Qualitative Analysis
CdS
Cr(OH)3
Al(OH)3 Ni(OH)2
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Section 16.3
The Mole Involving Complex Ions
Equilibria
Complex Ion Equilibria
• Charged species consisting of a metal ion
surrounded by ligands.
 Ligand: Lewis base(molecule or ion having a
lone pair that can donated to an empty
orbital on the metal ion to form covalence
bond)
• Formation (stability) constant.
 Equilibrium constant for each step of the
formation of a complex ion by the addition of
an individual ligand to a metal ion or complex
ion in aqueous solution.
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Section 16.3
The Mole Involving Complex Ions
Equilibria
Aqueous Ammonia Added
to Silver Chloride
Silver Chloride Dissolves to
Form Ag(NH3)2+ (aq) and Cl- (aq)
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Section 16.3
The Mole Involving Complex Ions
Equilibria
Complex Ion Equilibria
Be2+(aq) + F–(aq)
BeF+(aq)
K1 = 7.9 x 104
BeF+(aq) + F–(aq)
BeF2(aq)
K2 = 5.8 x 103
BeF2(aq) + F–(aq)
BeF3– (aq)
K3 = 6.1 x 102
BeF3– (aq) + F–(aq)
BeF42– (aq)
K4 = 2.7 x 101
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Section 16.3
The Mole Involving Complex Ions
Equilibria
Example 16.8(page761)
Ag + + S2O32Ag(S2O3 )- + (S2O3 )2-
Ag(S2O3 )Ag(S2O3 )23-
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Section 16.3
The Mole Involving Complex Ions
Equilibria
Complex Ions and Solubility
• Two strategies for dissolving a water–insoluble
ionic solid.
 If the anion of the solid is a good base, the
solubility is greatly increased by acidifying
the solution.
 In cases where the anion is not sufficiently
basic, the ionic solid often can be dissolved
in a solution containing a ligand that forms
stable complex ions with its cation.
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Section 16.3
The Mole Involving Complex Ions
Equilibria
Concept Check
Calculate the solubility of silver chloride in 10.0
M ammonia given the following information:
Ksp (AgCl) = 1.6 x 10–10
Ag+ + NH3
AgNH3+
AgNH3+ + NH3
Ag(NH3)2+
0.48 M(page 762)
K = 2.1 x 103
K = 8.2 x 103
Calculate the concentration of NH3 in the final
equilibrium mixture.
9.0 M
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Section 16.3
The Mole Involving Complex Ions
Equilibria
Ag + (aq) + Cl- (aq)
AgCl(s)
Ag + + NH3
K = 2.1  103
AgNH3+
Ag(NH3 ) + + NH3
Ksp = 1.6  10-10
Ag(NH3 ) 2+ K = 8.2  103
 Ag total dissolved  Ag +   Ag(NH3 )+   Ag(NH3 )2+ 
AgCl(s) + 2NH3
10-2x
Ag(NH3 )2+ + Clx
K = Ksp  K1  K 2  2.8 103
x
x2
x
3
K=2.8 10 
 2.8 10 
 x  0.48M
2
(10-2x)
10  2 x
3
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Section 16.3
The Mole Involving Complex Ions
Equilibria
Figure 16.3:
Separation
of Group 1 Ions
PbCl2  heat
Pb2+ + 2Cl-
Hg 2Cl2 (s)+2NH3 (aq)  HgNH 2Cl( s)+
Hg(l )+NH 4 + (aq)+Cl2H+ (aq)+Ag(NH3 )2+ (aq) +Cl- (aq) 
2NH4 +(aq)+AgCl(s)
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