1. Solubility and Complex Ion Equilibria
Chapter 16
Solubility and Complex Ion Equilibria

2. 16.1 Solubility Equilibria and the Solubility Product
Precipitation and Qualitative Analysis
Equilibria Involving Complex Ions
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3. 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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4. 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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5. Adding F- to drinking water, Since F- replaces the OH- in
Tooth decay: when food lodges between the teeth, acids form that dissolve tooth enamel ( hydroxyapatite, Ca5(PO4)3OH ).
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.
Ksp
Ksp
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6. 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) Bi3+(aq) + 3S2–(aq)
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7. Ksp Values at 25 °C for Common Ionic Solids

8. Solubility is an equilibrium position
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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9. Explain. If yes, explain and verify. If no, provide a counter-example.
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
No. In order to relate Ksp values to solubility directly, the salts must contain the same number of ions. For example, for a binary salt, Ksp = s2 (s = solubility); for a ternary salt, Ksp = 4s3.
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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10. Exercise
Calculate the solubility of silver chloride in water. Ksp = 1.6 × 10–10 (AgCl Ag++Cl- )
1.3×10-5 M
Calculate the solubility of silver phosphate in water. Ksp = 1.8 × 10–18 (Ag3PO Ag++PO4- )
1.6×10-5 M
x x
a) 1.3×10-5 M
b) 1.6×10-5 M
3x x
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11. The solubilities are the same.
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.
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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12. 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.
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. This is an example of the effect of LeChâtelier's principle on the position of the solubility equilibrium.
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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13. The Ksp values are the same.
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).
The Ksp values are the same (assuming the temperature is constant).
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14. 100.0 mL of 4.00 x 10-3 M calcium chloride. 2.0×10-8 M
Exercise
Calculate the solubility of AgCl in:
Ksp = 1.6 × 10–10
100.0 mL of 4.00 x 10-3 M calcium chloride.
2.0×10-8 M
100.0 mL of 4.00 x 10-3 M calcium nitrate.
1.3×10-5 M
a) 2.0×10-8 M Note: [Cl-] in CaCl2 is twice the [CaCl2] given mL is not used in the calculation.
b) 1.3×10-5 M
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15. Precipitation (Mixing Two Solutions of Ions)
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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16. 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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17. Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
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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18. Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
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19. Group 1- insoluble chloride (Ag+,Pb2+,Hg22+)
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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20. Separating the Common Cations by Selective Precipitation
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21. Flame Test
Potassium
Sodium
Ken O'Donoghue

22. CdS
Cr(OH)3
Al(OH)3
Ni(OH)2

23. 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.

24. Aqueous Ammonia Added to Silver Chloride
Silver Chloride Dissolves to Form Ag(NH3)2+ (aq) and Cl- (aq)

25. 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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26. Example 16.8(page761)
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27. 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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28. Calculate the concentration of NH3 in the final equilibrium mixture.
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+ + NH AgNH K = 2.1 x 103
AgNH3+ + NH Ag(NH3)2+ K = 8.2 x 103
0.48 M(page 762)
Calculate the concentration of NH3 in the final equilibrium mixture.
9.0 M
a) 0.48 M
b) 9.0 M
This problem is discussed at length in the text.
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30. Figure 16.3: Separation of Group 1 Ions