1. REACH will present significant challenges to all of us
REACH will present significant challenges to all of us. There’s still much water to flow under the bridge before the detailed requirements of REACH are set in tablets of stone.
However, there is one thing we can anticipate and that is that REACH will demand much greater supply chain co-operation on the use of chemicals.
From the chemical manufacturer, through the formulator, to the product manufacturer to the final retailer we will all have to work together far more, understand each others challenges, languages and contribution to a mutual value chain.
Today we want to share with you some ideas about how such partnerships can work. We don’t offer a panacea, nor do we imagine that our relationship is necessarily representative of those that you in this room may have, but we feel much of the learning from our experience can be interpreted constructively in your supply chains.

2. Solubility Equilibria

3. 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
Precipitation of sodium urate, Na2C5H2N4O2, in joints results in gouty arthritis.

4. 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

5. 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.

6. 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.

7. 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

8. Solubility of Ionic Compounds
Fe(OH)3
Cr(OH)3
large excess added +
NaOH
Fe3+
Precipitation of both Cr3+ and Fe3+ occurs
Cr3+

9. Solubility of Ionic Compounds
small excess added slowly +
NaOH
Cr3+
Fe(OH)3
Fe3+
less soluble salt precipitates only
Cr3+

10. 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. Solubility Equilibrium
My+
saturated solution
My+
xMy+
yAx-
Ax-
Ax-
solid
MxAy

12. 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

13. Solubility Product, Ksp
Ksp is related to molar solubility

14. Solubility Product, Ksp
Ksp is related to molar solubility
qualitative comparisons

15. Solubility Product, Ksp
Ksp used to compare relative solubilities
smaller Ksp = less soluble
larger Ksp= more soluble

16. Solubility Product, Ksp
Ksp is related to molar solubility
qualitative comparisons
quantitative calculations

17. 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

18. Example 1
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)

19. Example 2
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)

20. Example 3
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)

21. Precipitation Precipitation reaction exchange reaction
one product is insoluble
Example
Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)

22. 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

23. Precipitation Precipitation reaction exchange reaction Example
one product is insoluble
Example
Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)
Net Ionic: Ca2+(aq) + CO32-(aq) <=> CaCO3(s)

24. Precipitation saturated solution
Compare precipitation to solubility equilibrium
Ca2+(aq) + CO32-(aq) <=> CaCO3(s) prec. vs
CaCO3(s) <=> Ca2+(aq) + CO32-(aq) sol. Equil.
saturated solution

25. Precipitation saturated solution
Compare precipitation to solubility equilibrium:
Ca2+(aq) + CO32-(aq) <=> CaCO3(s) vs
CaCO3(s) <=> Ca2+(aq) + CO32-(aq)
saturated solution
Precipitation occurs until solubility equilibrium is established.

26. Precipitation 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)

27. 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

28. 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

29. Example 4
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)

30. 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)

31. Example 5
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)

32. Problem Solving Strategy
Precipitation does not occur until Q exceeds Ksp. (Q > Ksp)
We need to add enough Ag+ to make the solution supersaturated
Use the saturated solution (Q = Ksp) as a reference point
Calculate the [Ag+] needed to give a saturated solution.
Add more Ag+ than this to give a precipitate

33. Factors that Affect Solubility
Common Ion Effect pH
Complex-Ion Formation

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)

35. 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

36. Example 6
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)

37. 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

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-

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-

40. 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-
Increase = stress

41. 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-
Increase = stress
Stress relief = remove some [Cl-]

42. 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-
reacts w/some Cl-
Reverse reaction removes some excess

43. 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-
Shifts towards reactants
Equilibrium reestablished
More AgCl present = less dissolved = lower solubility

44. 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

45. 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

46. 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

47. Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)
pH and Solubility
Example:
Fe(OH)2
Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)

48. 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)

49. 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

50. 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

51. 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-]

52. 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

53. 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

54. 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)

55. 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)

56. pH, Solubility, and Tooth Decay
metabolism
+ food organic acids
(H3O+)
bacteria in mouth

57. 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)

58. 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

59. Tooth Decay

60. 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

61. 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!

62. 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

63. Complex Ion Formation and Solubility
Metals act as Lewis acids
Example
Fe3+(aq) + 6H2O(l)  Fe(H2O)63+(aq)
Lewis acid
Lewis base

64. Complex Ion Formation and Solubility
Metals act as Lewis acids
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

65. Complex Ion Formation and Solubility
Metals act as Lewis acids
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.

66. Complex Ion Formation and Solubility
Metals act as Lewis acids
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
Complex ion
Lewis acid
Lewis base

67. 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.

68. Complex-Ion Formation and Solubility
Example
AgCl
AgCl(s)  Ag+(aq) + Cl-(aq)

69. Complex-Ion Formation and Solubility
Example
AgCl Add NH3
AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)

70. 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?

71. 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

72. 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

73. 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

74. تا جلسه بعد ...