2. Solubility equilibrium is base on the assumption that solids dissolve in water to give the basic particles from which they are formed.
Molecular solids dissolve to give individual aqueous molecules.
Ionic solids dissociate to give their respective positive and negative ions: .

3. The ions in formed from the dissociation of ionic solids can carry an electrical current. Salt solutions, therefore, are good conductors of electricity. Molecular solids, however, do not dissociate in water to give ions, so no electrical current can be carried.

4. Generally expressed in two ways: grams of solute per 100 g of water
Solubility
The ratio of the maximum amount of solute to the volume of solvent in which this solute can dissolve.
Generally expressed in two ways:
grams of solute per 100 g of water
moles of solute per Liter of solution

5. Definition of Solubility
A salt is considered soluble if it dissolves in water to give a solution with a concentration of at least 0.1 M at room temperature.
A salt is considered insoluble if the concentration of an aqueous solution is less than M at room temperature.
Salts with solubilities between M and 0.1 M are considered to be slightly soluble.

6. Solubility Rules Soluble: Dissolve - Do NOT form a solid precipitate.
alkali metal ions and ammonium ion: Li+, Na+, K+, NH4+ 
acetate ion: C2H3O21- 
nitrate ion: NO31- 
halide ions (X): Cl-, Br-, I-  (Exceptions:  AgX, HgX, and PbX2 are insoluble) 
sulfate ion: SO42-  (Exceptions:  SrSO4, BaSO4, and PbSO4 are insoluble;
                     AgSO4, CaSO4, and Hg2SO4 are slightly soluble) 

7. Solubility Rules
Insoluble: Do NOT Dissolve - Do form a solid precipitate.
carbonate ion: CO32- 
chromate ion: CrO42- 
phosphate ion: PO43- 
sulfide ion: S2-  (Exceptions:  CaS, SrS, and BaS are soluble) 
hydroxide ion: OH-  (Exceptions:  Sr(OH)2 and Ba(OH)2 are soluble; 
                     Ca(OH)2 is slightly soluble

8. Low solubility salts
Salts that have extremely low solubilities dissociate in water according to the principles of equilibrium. For example, the reaction for the dissociation of the s alt AgCl is:
Saturated solution - Contains the maximum concentration of ions that can exist in equilibrium with the solid salt at a given temperature

9. The reverse reaction for the dissolving of the salt would be the precipitation of the ions to form a solid:
The system has reached equilibrium when the rate at which AgCl dissolves is equal to the rate at which AgCl precipitates.

10. Solubility Equilibria
Solubility product (Ksp) – equilibrium constant; has only one value for a given solid at a given temperature.
Solubility – an equilibrium position.

11. Difference between liquid equilibrium and liquid solid equilibrium
NOTE: There is no denominator in the solubility product equilibrium constant. The key word to remember is PRODUCT which can remind you that you should have a multiplication (or product) of the concentrations of the ions. The reason that the solid reactant is not written is because its concentration effectively remains constant.

12. Solubility Equilibria- Effect of Stoichiometry
Coefficients from the balanced equations become exponents
Bi2S3(s) Bi3+(aq) + 3S2–(aq)
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13. 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.
 Unlike Ka and Kb for acids and bases, the relative values of Ksp cannot be used to predict the relative solubilities of salts if the salts being compared produce a different number of ions.
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.
Unlike Ka and Kb for acids and bases, the relative values of Ksp cannot be used to predict the relative solubilities of salts if the salts being compared produce a different number of ions.

14. Solubility of salt
Calculate the solubility of CaF2 in g/L (Ksp = 4.0 x 10-8)
First, write the BALANCED REACTION:
Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:
In the above equation, however, we have two unknowns, [Ca2+] and [F-]2. So, we have to write one in terms of the other using mole ratios. According to the balanced equation, for every one mole of Ca2+ formed, 2 moles of F- are formed. To simplify things a little, let's assign the the variable X for the solubility of the Ca2+:

15. If we SUBSTITUTE these values into the equilibrium expression,
we now only have one variable to worry about, X:
We can now SOLVE for X:

16. EXERCISE!
Calculate the solubility of silver chloride in water. Ksp = 1.6 × 10–10
[Ag+]=[Cl-] so you solubility = √Ksp
1.3×10-5 M
Calculate the solubility of silver phosphate in water. Ksp = 1.8 × 10–18 =[Ag+]*[PO4+3]3
[Ag+]=3x [PO4+3]=x
1.6×10-5 M
a) 1.3×10-5 M
b) 1.6×10-5 M
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17. We assigned X as the solubility of the Ca2+ which is equal
to the solubility of the salt, CaF2. 
However, our units right now are in molarity (mol/L),
so we have to convert to grams:

18. Calculate Ksp from Solubility
The solubility of AgCl in pure water is 1.3 x 10-5 M. Calculate the value of Ksp.
First, write the BALANCED REACTION:
Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:
It is given in the problem that the solubility of AgCl is 1.3 x Since the mole ratio of AgCl to both Ag+ and Cl- is 1:1, the solubility of each of the ions is equal to the solubility of AgCl. SUBSTITUTE the solubility given into the equilibrium expression to get Ksp:
1.3 x 10-5 M X
1.3 x 10-5 M

19. Calculate the solubility of AgCl in:
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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20. Ion Product (Qsp)
The product of the concentrations of the ions at any moment in time (not necessarily at equilibrium).
Imagine we have a saturated solution of AgCl. The equilibrium reaction for the dissociation of this salt is:

21. 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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22. Common Ion Effect
The decrease in the solubility of a salt that occurs when the salt is dissolved in a solution that already contains another source of one of its ions.  For example, if AgCl is added to a NaCl solution (which contains the common ion, Cl-) the solubility of the AgCl decreases.

23. Effect of adding a common ion
[Ag+] is equal to [Cl-] at equilibrium because the mole ratio of Ag+ to Cl- is 1:1. What would happen to the solution if a tiny bit of AgNO3 (a soluble salt) were added?
Since AgNO3 is soluble, it dissociates completely to give Ag+ and NO3- ions. There would now be two sources of the Ag+ ion, from the AgCl and from the AgNO3:

24. Adding AgNO3 increases the Ag+ concentration and the solution is no longer at equilibrium.  The ion product (Qsp) at that moment is bigger than the solubility product (Ksp).  The reaction will eventually return to equilibrium but when it does, the [Ag+] is no longer equal to the [Cl-].  Instead, the [Ag+] will be larger than the [Cl-].

25. Let's go back to the saturated AgCl solution
Let's go back to the saturated AgCl solution. What would happen this time if a tiny bit of NaCl (a soluble salt) were added? Since NaCl is soluble, it dissociates completely to give Na+ and Cl- ions. There would now be two sources of the Cl- ion, from AgCl and from NaCl:
Adding NaCl increases the Cl- concentration and the solution is no longer at equilibrium. The ion product (Qsp) at that moment is bigger than the solubility product (Ksp). The reaction will eventually return to equilibrium but when it does, the [Ag+] is no longer equal to the [Cl-]. Instead, the [Cl-] will be larger than the [Ag+].

26. 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.
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27. The silver phosphate is more soluble in an acidic solution.
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.
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28. 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).
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29. Effect of pH on Solubility
Sometimes it is necessary to account for other reactions aqueous ions might undergo.
For example, if the anion is the conjugate base of a weak acid, it will react with H3O+.
You should expect the solubility to be affected by pH. By adding and complexing out ions you can affect the pH of solution which could affect ppt reactions. 2

30. Consider the following equilibrium.
H2O
Because the oxalate ion is conjugate base, it will react with H3O+ (added acid to lower pH).
H2O
According to Le Chatelier’s principle, as C2O42- ion is removed by the reaction with H3O+, more calcium oxalate dissolves (increase solubility).
Therefore, you expect calcium oxalate to be more soluble in acidic solution (lower pH) than in pure water. The acidity will react with the oxalate and shift the equil toward the right and allow more calcium oxalate to dissolve. 2

31. 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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32. 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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33. Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
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34. Separating the Common Cations by Selective Precipitation
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35. Complex Ion Equilibria
Charged species consisting of a metal ion surrounded by ligands.
Ligand: Lewis base
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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36. Complex-Ion Equilibria
For example, the silver ion, Ag+, can react with ammonia to form the Ag(NH3)2+ ion. 2

37. A complex is defined as a compound containing complex ions.
A ligand is a Lewis base (makes electron pair available) that bonds to a metal ion to form a complex ion. Lewis Acid is the cation. 2

38. Complex-Ion Formation
The aqueous silver ion forms a complex ion with ammonia in steps.
When you add these equations, you get the overall equation for the formation of Ag(NH3)2+. 2

39. The formation constant, Kf, is the equilibrium constant for the formation of a complex ion from the aqueous metal ion and the ligands.
The formation constant for Ag(NH3)2+ is:
The value of Kf for Ag(NH3)2+ is 1.7 x 107. 2

40. Complex-Ion Formation
The large value means that the complex ion is quite stable.
When a large amount of NH3 is added to a solution of Ag+, you expect most of the Ag+ ion to react to form the complex ion (large Kf - equil lies far to right).
Handle calculations same way as any other K 2

41. Complex-Ion Formation
Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, Majority of figures/tables are from the Ebbing lecture outline.
Complex-Ion Formation
The dissociation constant, Kd , is the reciprocal, or inverse, value of Kf.
The equation for the dissociation of Ag(NH3)2+ is
The equilibrium constant equation is

42. Equilibrium Calculations with Kf
What is the concentration of Ag+(aq) ion in 1.00 liters of solution that is M AgNO3 and 1.00 M NH3? The Kf for Ag(NH3)2+ is 1.7 x 107.

43. Complex Ion Equilibria
Be2+(aq) + F–(aq) BeF+(aq) K1 = 7.9 × 104 BeF+(aq) + F–(aq) BeF2(aq) K2 = 5.8 × 103 BeF2(aq) + F–(aq) BeF3– (aq) K3 = 6.1 × 102 BeF3– (aq) + F–(aq) BeF42– (aq) K4 = 2.7 × 101
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44. 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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45. solubility and complex ions

46. Complex Ions Are Formed From Lewis Acid Metal Interactions

47. Ksp (AgCl) = 1.6 × 10–10 0.48 M CONCEPT CHECK!
Calculate the solubility of silver chloride in 10.0 M ammonia given the following information:
Ksp (AgCl) = 1.6 × 10–10
Ag+ + NH AgNH K = 2.1 × 103
AgNH3+ + NH Ag(NH3)2+ K = 8.2 × 103
0.48 M
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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48. Combining the two equilibria
Ag+ + NH3 <=> AgNH K = 2.1 × 103
AgNH3+ + NH3 <=> Ag(NH3)2+ K = 8.2 × 103
Ag+ + 2NH3 <=> [Ag(NH3)2]
Kf= 2.1 × 103 x 8.2 × 103 = 1.6 x 107

49. AgCl(s) + 2 NH3(aq) <=> [Ag (NH3)2]+ + Cl-
where K = [Ag (NH3)2] [Cl-] / [NH3]2 is a combination of 3 equilibriums:
A) AgCl(s) <=> Ag+ + Cl-
whose Ksp = [Ag+] [Cl-] = Ksp = 1.8 X 10-10
B) Ag+ + 2NH3 <=> [Ag(NH3)2]
whose Kf = [Ag(NH3)2] / [Ag+] [NH3]2 = 1.6 X 10^7

50. if multiplying the ratio (A) time (B) gives your ratio:
([Ag+] [Cl-]) * ([Ag(NH3)2] / [Ag+] [NH3]2 )
=[Ag (NH3)2] [Cl-]/[NH3]2
then multiplying the Ksp times the Kf gives your K
(1.8 X 10-10) (1.6 X 107) = 2.88 X 10-3
(with a constant that large, we need to use the quadratic)

51. Set Up a Rice Table K = [Ag (NH3)2] [Cl-] / [NH3]2
AgCl(s)
2 NH3(aq)
Ag (NH3)2+
Cl- 10
-2X X
K = [Ag (NH3)2] [Cl-] / [NH3]2
2.88 X 10-3 = [X] [X] / [10 - 2X]2