1. CH160 General Chemistry II Lecture Presentation Solubility Equilibria
Chapter 17
11/16/2018
Chapter 17

2. Why Study Solubility Equilibria?
Many natural processes involve precipitation or dissolution of salts. A few examples:
Dissolving of underground limestone deposits (CaCO3) forms caves
Note: Limestone is water “insoluble” (How can this be?)
Precipitation of limestone (CaCO3) forms stalactites and stalagmites in underground caverns
Precipitation of insoluble Ca3(PO4)2 and/or CaC2O4 in the kidneys forms kidney stones
Dissolving of tooth enamel, Ca5(PO4)3OH, leads to tooth decay (ouch!)
Precipitation of sodium urate, Na2C5H2N4O2, in joints results in gouty arthritis.
11/16/2018
Chapter 17

3. Why Study Solubility Equilibria?
Many chemical and industrial processes involve precipitation or dissolution of salts. A few examples:
Production/synthesis of many inorganic compounds involves their precipitation reactions from aqueous solution
Separation of metals from their ores often involves dissolution
Qualitative analysis, i.e. identification of chemical species in solution, involves characteristic precipitation and dissolution reactions of salts
Water treatment/purification often involves precipitation of metals as insoluble inorganic salts
Toxic Pb2+, Hg2+, Cd2+ removed as their insoluble sulfide (S2-) salts
PO43- removed as insoluble calcium salts
Precipitation of gelatinous insoluble Al(OH)3 removes suspended matter in water
11/16/2018
Chapter 17

4. Why Study Solubility Equilibria?
To understand precipitation/dissolution processes in nature, and how to exploit precipitation/dissolution processes for useful purposes, we need to look at the quantitative aspects of solubility and solubility equilibria.
11/16/2018
Chapter 17

5. Solubility of Ionic Compounds
Solubility Rules
general rules for predicting the solubility of ionic compounds
strictly qualitative
11/16/2018
Chapter 17

6. Solubility of Ionic Compounds
Solubility Rule Examples
All alkali metal compounds are soluble
Most hydroxide compounds are insoluble. The exceptions are the alkali metals, Ba2+, and Ca2+
Most compounds containing chloride are soluble. The exceptions are those with Ag+, Pb2+, and Hg22+
All chromates are insoluble, except those of the alkali metals and the NH4+ ion
11/16/2018
Chapter 17

7. Solubility of Ionic Compounds
Fe(OH)3
Cr(OH)3
large excess added +
NaOH
Fe3+
Precipitation of both Cr3+ and Fe3+ occurs
Cr3+
11/16/2018
Chapter 17

8. Solubility of Ionic Compounds
small excess added slowly +
NaOH
Cr3+
Fe(OH)3
Fe3+
less soluble salt precipitates only
Cr3+
11/16/2018
Chapter 17

9. Solubility of Ionic Compounds
Solubility Rules
general rules for predicting the solubility of ionic compounds
strictly qualitative
Do not tell “how” soluble
Not quantitative
11/16/2018
Chapter 17

10. Solubility Equilibrium
My+
saturated solution
My+
xMy+
yAx-
Ax-
Ax-
solid
MxAy
11/16/2018
Chapter 17

11. Solubility of Ionic Compounds
Solubility Equilibrium
MxAy(s) <=> xMy+(aq) + yAx-(aq)
The equilibrium constant for this reaction is the solubility product, Ksp:
Ksp = [My+]x[Ax-]y
11/16/2018
Chapter 17

12. Solubility Product, Ksp
Ksp is related to molar solubility
11/16/2018
Chapter 17

13. Solubility Product, Ksp
Ksp is related to molar solubility
qualitative comparisons
11/16/2018
Chapter 17

14. Solubility Product, Ksp
Ksp used to compare relative solubilities
smaller Ksp = less soluble
larger Ksp= more soluble
11/16/2018
Chapter 17

15. Solubility Product, Ksp
Ksp is related to molar solubility
qualitative comparisons
quantitative calculations
11/16/2018
Chapter 17

16. Calculations with Ksp
Basic steps for solving solubility equilibrium problems
Write the balanced chemical equation for the solubility equilibrium and the expression for Ksp
Derive the mathematical relationship between Ksp and molar solubility (x)
Make an ICE table
Substitute equilibrium concentrations of ions into Ksp expression
Using Ksp, solve for x or visa versa, depending on what is wanted and the information provided
11/16/2018
Chapter 17

17. Example 1 (1 on Example Problems Handout)
Calculate the Ksp for MgF2 if the molar solubility of this salt is 2.7 x 10-3 M.
(ans.: 7.9 x 10-8)
11/16/2018
Chapter 17

18. Example 2 (2 on Example Problems Handout)
Calculate the Ksp for Ca3(PO4)2 (FW = 310.2) if the solubility of this salt is 8.1 x 10-4 g/L.
(ans.: 1.3 x 10-26)
11/16/2018
Chapter 17

19. Example 3 (4 on Example Problems Handout)
The Ksp for CaF2 (FW = 78 g/mol) is 4.0 x What is the molar solubility of CaF2 in water? What is the solubility of CaF2 in water in g/L?
(ans.: 2.2 x 10-4 M, g/L)
11/16/2018
Chapter 17

20. Precipitation Precipitation reaction Example exchange reaction
one product is insoluble
Example
Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)
11/16/2018
Chapter 17

21. Precipitation Precipitation reaction Example exchange reaction
one product is insoluble
Example
Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)
Na+ and Ca2+ “exchange” anions
11/16/2018
Chapter 17

22. Precipitation Precipitation reaction Example exchange reaction
one product is insoluble
Example
Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)
Net Ionic: Ca2+(aq) + CO32-(aq) <=> CaCO3(s)
11/16/2018
Chapter 17

23. Precipitation Compare precipitation to solubility equilibrium vs
Ca2+(aq) + CO32-(aq) <=> CaCO3(s) prec. vs
CaCO3(s) <=> Ca2+(aq) + CO32-(aq) sol. Equil.
saturated solution
11/16/2018
Chapter 17

24. Precipitation Compare precipitation to solubility equilibrium: vs
Ca2+(aq) + CO32-(aq) <=> CaCO3(s) vs
CaCO3(s) <=> Ca2+(aq) + CO32-(aq)
saturated solution
Precipitation occurs until solubility equilibrium is established.
11/16/2018
Chapter 17

25. Precipitation saturated solution
Ca2+(aq) + CO32-(aq) <=> CaCO3(s) vs
CaCO3(s) <=> Ca2+(aq) + CO32-(aq)
saturated solution
Key to forming ionic precipitates: Mix ions so concentrations exceed those in saturated solution (supersaturated solution)
11/16/2018
Chapter 17

26. Predicting Precipitation
To determine if solution is supersaturated:
Compare ion product (Q or IP) to Ksp
For MxAy(s) <=> xMy+(aq) + yAx-(aq)
Q = [My+]x[Ax-]y
Q calculated for initial conditions
Q > Ksp  supersaturated solution, precipitation occurs, solubility equilibrium established (Q = Ksp)
Q = Ksp  saturated solution, no precipitation
Q < Ksp  unsaturated solution, no precipitation
11/16/2018
Chapter 17

27. Predicting Precipitation
Basic Steps for Predicting Precipitation
Consult solubility rules (if necessary) to determine what ionic compound might precipitate
Write the solubility equilibrium for this substance
Pay close attention to the stoichiometry
Calculate the moles of each ion involved before mixing
moles = M x L or moles = mass/FW
Calculate the concentration of each ion involved after mixing assuming no reaction
Calculate Q and compare to Ksp
11/16/2018
Chapter 17

28. Example 4 (7 and 8 on Example Problems Handout)
Will a precipitate form if (a) mL of M lead nitrate, Pb(NO3)2, and mL of M sodium fluoride, NaF, are mixed, and (b) mL of M Pb(NO3)2 and mL of M NaF are mixed?
(ans.: (a) No, Q = 7.5 x 10-9; (b) Yes, Q = 7.5 x 10-7)
11/16/2018
Chapter 17

29. Solubility of Ionic Compounds
Solubility Rules
All alkali metal compounds are soluble
The nitrates of all metals are soluble in water.
Most compounds containing chloride are soluble. The exceptions are those with Ag+, Pb2+, and Hg22+
Most compounds containing fluoride are soluble. The exceptions are those with Mg2+, Ca2+, Sr2+, Ba2+, and Pb2+
Ex. 4: Possible precipitate = PbF2 (Ksp = 4.1 x 10-8)
11/16/2018
Chapter 17

30. Example 5 (10 on Example Problem Handout)
A student carefully adds solid silver nitrate, AgNO3, to a M solution of sodium sulfate, Na2SO4. What [Ag+] in the solution is needed to just initiate precipitation of silver sulfate, Ag2SO4 (Ksp = 1.4 x 10-5)?
(ans.: M)
11/16/2018
Chapter 17

31. Factors that Affect Solubility
Common Ion Effect pH
Complex-Ion Formation
11/16/2018
Chapter 17

32. Factors that Affect Solubility
Common Ion Effect pH
Complex-Ion Formation
These sure sound familiar. Where have I seen them before?
11/16/2018
Chapter 17

33. Common Ion Effect and Solubility
Consider the solubility equilibrium of AgCl.
AgCl(s) <=> Ag+(aq) + Cl-(aq)
How does adding excess NaCl affect the solubility equilibrium?
NaCl(s)  Na+(aq) + Cl-(aq)
11/16/2018
Chapter 17

34. Common Ion Effect and Solubility
Consider the solubility equilibrium of AgCl.
AgCl(s) <=> Ag+(aq) + Cl-(aq)
How does adding excess NaCl affect the solubility equilibrium?
NaCl(s)  Na+(aq) + Cl-(aq)
2 sources of Cl-
Cl- is common ion
11/16/2018
Chapter 17

35. Example 6 (11 on Example Problem Handout)
What is the molar solubility of AgCl (Ksp = 1.8 x 10-10) in a M NaCl solution? What is the molar solubility of AgCl in pure water?
(ans.: 8.5 x 10-9, 1.3 x 10-5)
11/16/2018
Chapter 17

36. Common Ion Effect and Solubility
How does adding excess NaCl affect the solubility equilibrium of AgCl?
AgCl in H2O
1.3 x 10-5 M
M NaCl
Molar solubility
AgCl in M NaCl
Molar solubility
8.5 x 10-9 M
11/16/2018
Chapter 17

37. Common Ion Effect and Solubility
Why does the molar solubility of AgCl decrease after adding NaCl?
Understood in terms of LeChatelier’s principle:
NaCl(s) --> Na+ + Cl-
11/16/2018
Chapter 17

38. Common Ion Effect and Solubility
Why does the molar solubility of AgCl decrease after adding NaCl?
Understood in terms of LeChatelier’s principle:
NaCl(s) --> Na+ + Cl-
AgCl(s) <=> Ag+ + Cl-
11/16/2018
Chapter 17

39. Common Ion Effect and Solubility
Why does the molar solubility of AgCl decrease after adding NaCl?
Understood in terms of LeChatelier’s principle:
NaCl(s) --> Na+ + Cl-
AgCl(s) <=> Ag+ + Cl-
Common-Ion Effect
11/16/2018
Chapter 17

40. pH and Solubility How can pH influence solubility?
Solubility of “insoluble” salts will be affected by pH changes if the anion of the salt is at least moderately basic
Solubility increases as pH decreases
Solubility decreases as pH increases
11/16/2018
Chapter 17

41. pH and Solubility Salts contain either basic or neutral anions:
basic anions
Strong bases: OH-, O2-
Weak bases (conjugate bases of weak molecular acids): F-, S2-, CH3COO-, CO32-, PO43-, C2O42-, CrO42-, etc.
Solubility affected by pH changes
neutral anions (conjugate bases of strong monoprotic acids)
Cl-, Br-, I-, NO3-, ClO4-
Solubility not affected by pH changes
11/16/2018
Chapter 17

42. Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
pH and Solubility
Example:
Fe(OH)2
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
11/16/2018
Chapter 17

43. Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
pH and Solubility
Example:
Fe(OH)2-Add acid
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
11/16/2018
Chapter 17

44. Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
pH and Solubility
Example:
Fe(OH)2-Add acid
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
2H3O+(aq) + 2OH-(aq)  4H2O
11/16/2018
Chapter 17

45. pH and Solubility Example: Fe(OH)2-Add acid
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
2H3O+(aq) + 2OH-(aq)  4H2O
Which way does this reaction shift the solubility equilibrium? Why?
Understood in terms of LeChatlier’s principle
11/16/2018
Chapter 17

46. More Fe(OH)2 dissolves in response Stress relief = increase [OH-]
pH and Solubility
Example:
Fe(OH)2-Add acid
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
2H3O+(aq) + 2OH-(aq)  4H2O
More Fe(OH)2 dissolves in response
Solubility increases
Decrease = stress
Stress relief = increase [OH-]
11/16/2018
Chapter 17

47. pH and Solubility Example: Fe(OH)2
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
2H3O+(aq) + 2OH-(aq)  4H2O(l)
Fe(OH)2(s) + 2H3O+(aq) <=> Fe2+(aq) + 4H2O(l)
overall
11/16/2018
Chapter 17

48. pH and Solubility Example: Fe(OH)2
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
2H3O+(aq) + 2OH-(aq)  4H2O(l)
Fe(OH)2(s) + 2H3O+(aq) <=> Fe2+(aq) + 4H2O(l)
overall
decrease pH
solubility increases
increase pH
solubility decreases
11/16/2018
Chapter 17

49. pH, Solubility, and Tooth Decay
Enamel (hydroxyapatite) = Ca10(PO4)6(OH)2
(insoluble ionic compound)
Ca10(PO4)6(OH)2  10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
11/16/2018
Chapter 17

50. pH, Solubility, and Tooth Decay
Enamel (hydroxyapatite) = Ca10(PO4)6(OH)2
(insoluble ionic compound)
strong base
weak base
Ca10(PO4)6(OH)2  10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
11/16/2018
Chapter 17

51. pH, Solubility, and Tooth Decay
metabolism
+ food organic acids
(Yummy)
(H3O+)
bacteria in mouth
11/16/2018
Chapter 17

52. pH, Solubility, and Tooth Decay
Ca10(PO4)6(OH)2(s)  10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
OH-(aq) + H3O+(aq)  2H2O(l)
PO43-(aq) + H3O+(aq)  HPO43-(aq) + H2O(l)
11/16/2018
Chapter 17

53. pH, Solubility, and Tooth Decay
Ca10(PO4)6(OH)2(s)  10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
OH-(aq) + H3O+(aq)  2H2O(l)
PO43-(aq) + H3O+(aq)  HPO43-(aq) + H2O(l)
More Ca10(PO4)6(OH)2 dissolves in response
Solubility increases
Leads to tooth decay
Decrease = stress
Decrease = stress
11/16/2018
Chapter 17

54. Tooth Decay
11/16/2018
Chapter 17

55. pH, Solubility, and Tooth Decay
Why fluoridation?
F- replaces OH- in enamel
Ca10(PO4)6(F)2(s)  10Ca2+(aq) + 6PO43-(aq) + 2F-(aq)
fluorapatite
11/16/2018
Chapter 17

56. pH, Solubility, and Tooth Decay
Why fluoridation?
F- replaces OH- in enamel
Ca10(PO4)6(F)2(s)  10Ca2+(aq) + 6PO43-(aq) + 2F-(aq)
Less soluble (has lower Ksp) than Ca10(PO4)6(OH)2
weaker base than OH- more resistant to acid attack
Factors together fight tooth decay!
11/16/2018
Chapter 17

57. pH, Solubility, and Tooth Decay
Why fluoridation?
F- replaces OH- in enamel
Ca10(PO4)6(F)2(s)  10Ca2+(aq) + 6PO43-(aq) + 2F-(aq)
F- added to drinking water as NaF or Na2SiF6
1 ppm = 1 mg/L
F- added to toothpastes as SnF2, NaF, or Na2PO3F
% w/w
11/16/2018
Chapter 17

58. Complex Ion Formation and Solubility
Metals act as Lewis acids (see Chapter 15)
Example
Fe3+(aq) + 6H2O(l)  Fe(H2O)63+(aq)
Lewis acid
Lewis base
11/16/2018
Chapter 17

59. Complex Ion Formation and Solubility
Metals act as Lewis acids (see Chapter 15)
Example
Fe3+(aq) + 6H2O(l)  Fe(H2O)63+(aq)
Complex ion
Complex ion/complex contains central metal ion bonded to one or more molecules or anions called ligands
Lewis acid = metal
Lewis base = ligand
11/16/2018
Chapter 17

60. Complex Ion Formation and Solubility
Metals act as Lewis acids (see Chapter 15)
Example
Fe3+(aq) + 6H2O(l)  Fe(H2O)63+(aq)
Complex ion
Complex ions are often water soluble
Ligands often bond strongly with metals
Kf >> 1: Equilibrium lies very far to right.
11/16/2018
Chapter 17

61. Complex Ion Formation and Solubility
Metals act as Lewis acids (see Chapter 15)
Other Lewis bases react with metals also
Examples
Fe3+(aq) + 6CN-(aq)  Fe(CN)63-(aq)
Ni2+(aq) + 6NH3(aq)  Ni(NH3)62+(aq)
Ag+(aq) + 2S2O32-(aq)  Ag(S2O3)23-(aq)
Lewis acid
Lewis base
Complex ion
Lewis acid
Lewis base
Complex ion
Lewis acid
Lewis base
Complex ion
11/16/2018
Chapter 17

62. Complex-Ion Formation and Solubility
How does complex ion formation influence solubility?
Solubility of “insoluble” salts increases with addition of Lewis bases if the metal ion forms a complex with the base.
11/16/2018
Chapter 17

63. Complex-Ion Formation and Solubility
Example
AgCl
AgCl(s)  Ag+(aq) + Cl-(aq)
11/16/2018
Chapter 17

64. Complex-Ion Formation and Solubility
Example
AgCl-Add NH3
AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
11/16/2018
Chapter 17

65. Complex-Ion Formation and Solubility
Example
AgCl-Add NH3
AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
Which way does this reaction shift the solubility equilibrium? Why?
11/16/2018
Chapter 17

66. Complex-Ion Formation and Solubility
Example
AgCl-Add NH3
AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
More AgCl dissolves in response
Solubility increases
Decrease = stress
11/16/2018
Chapter 17

67. Complex-Ion Formation and Solubility
Example
AgCl
AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
AgCl(s) + 2NH3(aq)  Ag(NH3)2+(aq) + Cl-(aq)
overall
Addition of ligand
solubility increases
11/16/2018
Chapter 17

68. Summary: Factors that Influence Solubility
Common Ion Effect
Decreases solubility pH
pH decreases
Increases solubility
pH increases
Salt must have basic anion
Complex-Ion Formation
11/16/2018
Chapter 17

69. End of Presentation
11/16/2018
Chapter 17