2. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. (Sec )
LO 6.21 The student can predict the solubility of a salt, or rank the solubility of salts, given the relevant Ksp values. (Sec 16.1)
LO 6.22 The student can interpret data regarding solubility of salts to determine, or rank, the relevant Ksp values. (Sec 16.1)
LO 6.23 The student can interpret data regarding the relative solubility of salts in terms of factors (common ions, pH) that influence the solubility. (Sec )

3. AP Learning Objectives, Margin Notes and References
LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.
LO 6.21 The student can predict the solubility of a salt, or rank the solubility of salts, given the relevant Ksp values.
LO 6.22 The student can interpret data regarding solubility of salts to determine, or rank, the relevant Ksp values.
LO 6.23 The student can interpret data regarding the relative solubility of salts in terms of factors (common ions, pH) that influence the solubility.

4. Solubility Equilibria
Because ionic compounds are strong electrolytes, they dissociate completely to the extent that they dissolve.
When an equilibrium equation is written, the solid is the reactant and the ions in solution are the products.
The equilibrium constant expression is called the solubility-product constant. It is represented as Ksp.

5. BaSO4(s) ⇌ Ba2+(aq) + SO42–(aq) The equilibrium constant expression is
Solubility Product
For example:
BaSO4(s) ⇌ Ba2+(aq) + SO42–(aq)
The equilibrium constant expression is
Ksp = [Ba2+][SO42]
Another example:
Ba3(PO4)2(s) ⇌ 3 Ba2+(aq) + 2 PO43–(aq)
Ksp = [Ba2+]3[PO43]2
Neither the amount of excess solid nor the size of the particles will affect either Ksp or the position of equilibrium.

6. Solubility Equilibria
Solubility product (Ksp) – equilibrium constant; has only one value for a given solid at a given temperature. (not affected by equilibrium position)
Solubility – an equilibrium position.
Bi2S3(s) Bi3+(aq) + 3S2–(aq)
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7. Solubility vs. Solubility Product
Ksp is not the same as solubility.
Solubility is the quantity of a substance that dissolves to form a saturated solution (reach equilibrium)
Common units for solubility:
Grams per liter (g/L)
Moles per liter (mol/L)

8. Calculating Solubility from Ksp
The Ksp for CaF2 is 3.9 × 10–11 at 25 °C. What is its molar solubility?
Follow the same format as before:
CaF2(s) ⇌ Ca2+(aq) + 2 F–(aq)
Ksp = [Ca2+][F–]2 = 3.9 × 10–11
CaF2(s)
[Ca2+](M)
[F–](M)
Initial concentration
(M)
---
Change in concentration +s
+2s
Equilibrium concentration s 2s

9. Example (completed)
Solve: Substitute the equilibrium concentration values from the table into the solubility-product equation:
3.9 × 10–11 = (s) (2s)2 = 4s3
s = 2.1 × 10–4 M
(If you want the answer in g/L, multiply by molar mass; this would give g/L.)

10. CONCEPT CHECK! Relative Solubilities
In comparing several different 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. 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.
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11. EXERCISE!
Writing Solubility-Product (Ksp) Expressions
Which of these expressions correctly expresses the solubility-product constant for Ag3PO4 in water? (a) [Ag][PO4], (b) [Ag+][PO43–], (c) [Ag+]3[PO43–], (d) [Ag+][PO43–]3, (e) [Ag+]3[PO43–]3.
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12. Calculating Ksp from Solubility
EXERCISE!
Calculating Ksp from Solubility
Solid silver chromate is added to pure water at 25 °C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 × 10–4 M. Assuming that the Ag2CrO4 solution is saturated and that there are no other important equilibria involving Ag+ or CrO42– ions in the solution, calculate Ksp for this compound.
Ksp = [Ag+]2[CrO42–] = (1.3 × 10–4)2(6.5 × 10–5) = 1.1 × 10–12
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13. EXERCISE!
Calculate the solubility of silver chloride in water. Ksp = 1.6 × 10–10
1.3×10-5 M
Calculate the solubility of silver phosphate in water. Ksp = 1.8 × 10–18
1.6×10-5 M
a) 1.3×10-5 M
b) 1.6×10-5 M
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14. Factors Affecting Solubility
The Common-Ion Effect
If one of the ions in a solution equilibrium is already dissolved in the solution, the solubility of the salt will decrease.
If either calcium ions or fluoride ions are present, then calcium fluoride will be less soluble.

15. Calculating Solubility with a Common Ion
What is the molar solubility of CaF2 in M Ca(NO3)2?
Follow the same format as before:
CaF2(s) ⇌ Ca2+(aq) + 2 F–(aq)
2) Ksp = [Ca2+][F–]2 = 3.9 × 10–11 3)
CaF2(s)
[Ca2+](M)
[F–](M)
Initial concentration
(M)
---
0.010
Change in concentration +s
+2s
Equilibrium concentration s 2s

16. (We assume that s << 0.010, so that 0.010 + s = 0.010!)
Example (completed)
Solve: Substitute the equilibrium concentration values from the table into the solubility-product equation:
3.9 × 10–11 = ( s)(2s)2
(We assume that s << 0.010, so that s = 0.010!)
3.9 × 10–11 = (0.010)(2s)2
s = 3.1 × 10–5 M

17. 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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18. Factors Affecting Solubility
pH: If a substance has a basic anion, it will be more soluble in an acidic solution.

19. 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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20. 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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21. 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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22. AP Learning Objectives, Margin Notes and References
LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.
LO 6.23 The student can interpret data regarding the relative solubility of salts in terms of factors (common ions, pH) that influence the solubility.

23. Precipitation (Mixing Two Solutions of Ions)
Will a Precipitate Form?
To decide, we calculate the reaction quotient, Q, and compare it to the solubility product constant, Ksp.
If Q = Ksp, the system is at equilibrium and the solution is saturated.
If Q < Ksp, more solid can dissolve, so no precipitate forms.
If Q > Ksp, a precipitate will form.
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24. Predicting Whether a Precipitate Forms
Does a precipitate form when 0.10 L of 8.0 × 10–3 M Pb(NO3)2 is added to 0.40 L of 5.0 × 10–3 M Na2SO4? (Ksp of 6.3 × 10–7 )
Because Q > Ksp, PbSO4 precipitates.

25. What are the equilibrium concentrations of the remaining ions once precipitation has occurred?
[Mg2+] = 2.1 x [F-] = 5.50 x10-2
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26. 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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27. Selective Precipitation
A solution contains 1.0 × 10–2 M Ag+ and 2.0 × 10–2 M Pb2+. When Cl– is added, both AgCl(Ksp = 1.8 × 10–10) and PbCl2(Ksp = 1.7 × 10–5) can precipitate. What concentration of Cl– is necessary to begin the precipitation of each salt? Which salt precipitates first?
AgCl [Cl-]= 1.8x PbCl2 [Cl-] = 2.9x10-2

28. 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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29. Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
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30. Separating the Common Cations by Selective Precipitation
Ksp values can be compare directly for ions that produce the same number of ions in solution.
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31. potassium.
sodium
lithium

32. AP Learning Objectives, Margin Notes and References
LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

33. Complex Ion Equilibria
Charged species consisting of a metal ion surrounded by ligands.
Ligand: Lewis base (a molecule with a lone pair that can be donated to an empty orbital on the metal form a covalent bond)
The number of ligands attached to the metal is called the coordination number. (commonly 6,4,or 2)
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34. Complex Ion Equilibria
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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35. 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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37. How Complex Ion Formation Affects Solubility
Silver chloride is insoluble. It has a Ksp of 1.6 × 10–10.
In the presence of NH3, the solubility greatly increases because Ag+ will form complex ions with NH3.

38. 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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39. Amphoterism and Solubility
Amphoteric oxides and hydroxides are soluble in strong acids or base, because they can act either as acids or bases.
Examples are oxides and hydroxides of Al3+, Cr3+, Zn2+, and Sn2+.

40. 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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