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CH160 General Chemistry II Lecture Presentation Solubility Equilibria

CH160 General Chemistry II Lecture Presentation Solubility Equilibria. Chapter 17. 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

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CH160 General Chemistry II Lecture Presentation Solubility Equilibria

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  1. CH160 General Chemistry IILecture PresentationSolubility Equilibria Chapter 17 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. 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 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. Chapter 17

  5. Solubility of Ionic Compounds • Solubility Rules • general rules for predicting the solubility of ionic compounds • strictly qualitative 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 Chapter 17

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

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

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

  12. Solubility Product, Ksp • Ksp is related to molar solubility Chapter 17

  13. Solubility Product, Ksp • Ksp is related to molar solubility • qualitative comparisons Chapter 17

  14. Solubility Product, Ksp • Ksp used to compare relative solubilities • smaller Ksp = less soluble • larger Ksp= more soluble Chapter 17

  15. Solubility Product, Ksp • Ksp is related to molar solubility • qualitative comparisons • quantitative calculations 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 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) 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) Chapter 17

  19. Example 3(4 on Example Problems Handout) • 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) Chapter 17

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

  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 Chapter 17

  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) Chapter 17

  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 Chapter 17

  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. Chapter 17

  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) 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 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 Chapter 17

  28. Example 4(7 and 8 on Example Problems Handout) • 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) 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) Chapter 17

  30. Example 5(10 on Example Problem Handout) • 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) Chapter 17

  31. Factors that Affect Solubility • Common Ion Effect • pH • Complex-Ion Formation Chapter 17

  32. Factors that Affect Solubility • Common Ion Effect • pH • Complex-Ion Formation These sure sound familiar. Where have I seen them before? 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) 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 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 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) 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 + 0.020 M NaCl Molar solubility AgCl in 0.020 M NaCl Molar solubility 8.5 x 10-9 M 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- 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- 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 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 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 Chapter 17

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

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

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

  46. 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-] 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 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 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) 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) Chapter 17

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