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.

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 Fe(OH)3 Cr(OH)3 large excess added + NaOH Fe3+ Precipitation of both Cr3+ and Fe3+ occurs Cr3+

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

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 My+ saturated solution My+ xMy+ yAx- Ax- Ax- solid MxAy

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

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

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.

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)

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

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 Cl- Cl- 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? 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

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

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-]

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

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

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

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

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

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)

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

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-]

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

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

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) strong base weak base Ca10(PO4)6(OH)2  10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)

pH, Solubility, and Tooth Decay metabolism + food 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 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

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) 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 OH- more 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) 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

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) Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq) More AgCl dissolves in response Solubility increases

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

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

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