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Solubility Equilibria 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. 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 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. 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. 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 Solubility of Ionic Compounds large excess added + NaOH Fe(OH)3 Cr(OH)3 Fe3+ Cr3+ Precipitation of both Cr3+ and Fe3+ occurs Solubility of Ionic Compounds small excess added slowly + NaOH Cr3+ Fe(OH)3 Fe3+ Cr3+ less soluble salt precipitates only Solubility of Ionic Compounds • Solubility Rules – general rules for predicting the solubility of ionic compounds – strictly qualitative • Do not tell “how” soluble • Not quantitative Solubility Equilibrium saturated solution solid My+ xMy+ yAx- My+ Ax- AxMxAy 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 Solubility Product, Ksp • Ksp is related to molar solubility Solubility Product, Ksp • Ksp is related to molar solubility –qualitative comparisons Solubility Product, Ksp • Ksp used to compare relative solubilities –smaller Ksp = less soluble –larger Ksp= more soluble Solubility Product, Ksp • Ksp is related to molar solubility –qualitative comparisons –quantitative calculations 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 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) 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) 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) Precipitation • Precipitation reaction – exchange reaction • one product is insoluble • Example Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq) Precipitation • Precipitation reaction –exchange reaction • one product is insoluble • Example Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq) Na+ and Ca2+ “exchange” anions 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) Precipitation • Compare precipitation to solubility equilibrium Ca2+(aq) + CO32-(aq) <=> CaCO3(s) prec. vs CaCO3(s) <=> Ca2+(aq) + CO32-(aq) sol. Equil. saturated solution 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. 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) 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 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 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) 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) 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) Factors that Affect Solubility • Common Ion Effect • pH • Complex-Ion Formation 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) 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 ClCl- is common ion 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) Common Ion Effect and Solubility • How does adding excess NaCl affect the solubility equilibrium of AgCl? 1.3 x 10-5 M + 0.020 M NaCl Molar solubility Molar solubility AgCl in H2O AgCl in 0.020 M NaCl 8.5 x 10-9 M 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- 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+ + ClAgCl(s) <=> Ag+ + Cl- 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+ + ClAgCl(s) <=> Ag+ + ClCommon-Ion Effect 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 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 pH and Solubility • Example: –Fe(OH)2 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) pH and Solubility • Example: –Fe(OH)2-Add acid Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq) 2H3O+(aq) + 2OH-(aq) 4H2O 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 pH and Solubility • Example: –Fe(OH)2-Add acid More Fe(OH)2 dissolves in response Solubility increases Decrease = stress Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq) Stress relief = increase [OH-] 2H3O+(aq) + 2OH-(aq) 4H2O pH and Solubility • Example: –Fe(OH)2 overall Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq) 2H3O+(aq) + 2OH-(aq) 4H2O(l) Fe(OH)2(s) + 2H3O+(aq) <=> Fe2+(aq) + 4H2O(l) pH and Solubility • Example: – Fe(OH)2 Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq) 2H3O+(aq) + 2OH-(aq) 4H2O(l) overall Fe(OH)2(s) + 2H3O+(aq) <=> Fe2+(aq) + 4H2O(l) decrease pH solubility increases increase pH solubility decreases 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) pH, Solubility, and Tooth Decay Enamel (hydroxyapatite) = Ca10(PO4)6(OH)2 (insoluble ionic compound) weak base strong base Ca10(PO4)6(OH)2 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq) pH, Solubility, and Tooth Decay + food metabolism organic acids (H3O+) bacteria in mouth 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) pH, Solubility, and Tooth Decay Solubility increases Leads to tooth decay OH-(aq) + H3O+(aq) 2H2O(l) PO43-(aq) + H3O+(aq) HPO43-(aq) + H2O(l) Decrease = stress More Ca10(PO4)6(OH)2 dissolves in response Decrease = stress Ca10(PO4)6(OH)2(s) 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq) Tooth Decay pH, Solubility, and Tooth Decay • Why fluoridation? –F- replaces OH- in enamel Ca10(PO4)6(F)2(s) 10Ca2+(aq) + 6PO43-(aq) (aq) + 2F fluorapatite 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 OHmore resistant to acid attack Factors together fight tooth decay! 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 • 0.1 - 0.15 % w/w Complex Ion Formation and Solubility • Metals act as Lewis acids • Example Fe3+(aq) + 6H2O(l) Fe(H2O)63+(aq) Lewis acid Lewis base 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 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. 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) Lewis acid Lewis base Complex ion Ni2+(aq) + 6NH3(aq) Ni(NH3)62+(aq) Lewis acid Lewis base Complex ion Ag+(aq) + 2S2O32-(aq) Ag(S2O3)23-(aq) Lewis acid Lewis base Complex ion 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. Complex-Ion Formation and Solubility • Example –AgCl AgCl(s) Ag+(aq) + Cl-(aq) Complex-Ion Formation and Solubility • Example –AgCl Add NH3 AgCl(s) Ag+(aq) + Cl-(aq) Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) 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? Complex-Ion Formation and Solubility • Example – AgCl-Add NH3 AgCl(s) Ag+(aq) + Cl-(aq) More AgCl dissolves in response Solubility increases Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) Complex-Ion Formation and Solubility • Example –AgCl overall AgCl(s) Ag+(aq) + Cl-(aq) Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl-(aq) Addition of ligand solubility increases Summary: Factors that Influence Solubility • Common Ion Effect – Decreases solubility • pH – pH decreases • Increases solubility – pH increases • Decreases solubility – Salt must have basic anion • Complex-Ion Formation – Increases solubility تا جلسه بعد ...