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Solubility Equilibria

Solubility Equilibria. Why Study Solubility Equilibria?. Many natural processes involve precipitation or dissolution of salts. A few examples: Dissolving of underground limestone deposits (CaCO 3 ) forms caves Note: Limestone is water “insoluble” (How can this be?)

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Solubility Equilibria

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  1. Solubility Equilibria

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

  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

  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.

  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. 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

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

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

  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

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

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

  12. Solubility Product, Ksp • Ksp is related to molar solubility

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

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

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

  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

  17. 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)

  18. 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)

  19. Example 3 • The Ksp for CaF2 (FW = 78 g/mol) is 4.0 x 10-11. 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, 0.017 g/L)

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

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

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

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

  24. Precipitation • 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.

  25. 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)

  26. Predicting Precipitation • To determine if solution is supersaturated: • Compare ion product (Q or IP) to Ksp • ForMxAy(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

  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

  28. Example 4 • Will a precipitate form if (a) 500.0 mL of 0.0030 M lead nitrate, Pb(NO3)2, and 800.0 mL of 0.0040 M sodium fluoride, NaF, are mixed, and (b) 500.0 mL of 0.0030 M Pb(NO3)2 and 800.0 mL of 0.040 M NaF are mixed? (ans.: (a) No, Q = 7.5 x 10-9; (b) Yes, Q = 7.5 x 10-7)

  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)

  30. Example 5 • A student carefully adds solid silver nitrate, AgNO3, to a 0.0030 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.: 0.068 M)

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

  32. 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)

  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) 2 sources of Cl- Cl- is common ion

  34. Example 6 • What is the molar solubility of AgCl (Ksp = 1.8 x 10-10) in a 0.020 M NaCl solution? What is the molar solubility of AgCl in pure water? (ans.: 8.5 x 10-9, 1.3 x 10-5)

  35. 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 + 0.020 M NaCl Molar solubility AgCl in 0.020 M NaCl Molar solubility 8.5 x 10-9 M

  36. 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-

  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- AgCl(s) <=> Ag+ + Cl-

  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- Common-Ion Effect

  39. pH and Solubility • How can pH influence solubility? • Solubility of “insoluble” salts will be affected by pH changes if the anion of the salt isat least moderately basic • Solubility increases as pH decreases • Solubility decreases as pH increases

  40. 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

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

  42. pH and Solubility • Example: • Fe(OH)2-Add acid Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq)

  43. pH and Solubility • Example: • Fe(OH)2-Add acid Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq) 2H3O+(aq) + 2OH-(aq)  4H2O

  44. 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

  45. 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-]

  46. 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

  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 decrease pH solubility increases increase pH solubility decreases

  48. 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)

  49. 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)

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