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Ch. 15: Applications of Aqueous Equilibria 15.6: Solubility Equilibria and Solubility Products

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Text of Ch. 15: Applications of Aqueous Equilibria 15.6: Solubility Equilibria and Solubility Products

  • Ch. 15: Applications of Aqueous Equilibria15.6: Solubility Equilibria and Solubility Products

  • SolubilityAs a salt dissolves in water and ions are released, they can collide and re-from the solidEquilibrium is reached when the rate of dissolution equals the rate of recrystallizationCaF2(s) Ca2+(aq) + 2F-(aq)Saturated solutionWhen no more solid can dissolve at equilibrium

  • Solubility ProductSolubility product: KspCaF2(s) Ca2+(aq) + 2F-(aq)Ksp=[Ca2+][F-]2Why do we leave out the CaF2?Adding more solid will not effect the amount of solid that can dissolve at a certain temperatureIt would increase both reverse and forward reaction rates b/c there is a greater amount

  • Ksp Values

  • Example 1CuBr has a solubility of 2.0x10-4 mol/L at 25C. Find the Ksp value.The solubility tells us the amount of solute that can dissolves in 1 L of waterUse ICE chart: solubility tells you x value

    Ksp=[Cu+][Br-]= (2.0x10-4 M)2 = 4.0x10-8

    CuBr(s) Cu+(aq) + Br-(aq) INot imp.00C-2.0x10-4 M+2.0x10-4 M+2.0x10-4 MENot imp2.0x10-4 M2.0x10-4 M

  • Example 2The Ksp value for Cu(IO3)2 is 1.4x10-7 at 25C. Calculate its solubility.Solve for solubility = x value using ICE chart

    Ksp=[Cu2+][IO3-]2= (x)(2x)2 = 1.4x10-7 = 4x3X = (3.5x10-8)1/3 = 3.3x10-3

    Cu(IO3)2(s) Cu2+(aq) + 2IO3-(aq) INot imp.00C-x+x+2xENot impx2x

  • Comparing Solubilities You can only compare solubilities using Ksp values for compounds containing the same number of ions CaSO4 > AgI > CuCO3Ksp values: 6.1x10-5 > 1.5x10-6 > 2.5x10-10 Why can we use Ksp values to judge solubility?Can only compare using actual solubility values (x) when compounds have different numbers of ions

  • Common Ion EffectSolubility of a solid is lowered when a solution already contains one of the ions it containsWhy?

  • Example 3Find the solubility of CaF2 (s) if the Ksp is 4.0 x 10-11 and it is in a 0.025 M NaF solution.

    Ksp=[Ca2+][F-]2= (x)(2x+0.025)2 = 4.0 x 10-11(x)(0.025)2 4.0 x 10-11x = 6.4x10-8

    CaF2(s) Ca2+(aq) + 2F-(aq)INot imp.00.025C-x+x+2xENot impx.025+2x

  • pH and solubilitypH can effect solubility because of the common ion effectEx: Mg(OH)2(s) Mg2+(aq) + 2OH-(aq)How would a high pH effect solubility?High pH = high [OH-] decrease solubilityEx: Ag3PO4(s) 3Ag+(aq) + PO43-(aq)What would happen if H+ is added?H+ uses up PO43- to make phosphoric acidEq. shifts to right - Solubility increases

  • Ch. 15: Applications of Aqueous Equilibria15.7: Precipitation and Qualitative Analysis

  • PrecipitationOpposite of dissolutionCan predict whether precipitation or dissolution will occurUse Q: ion productEquals Ksp but doesnt have to be at equilibriumQ > K: more reactant will form, precipitation until equilibrium reachedQ < K: more product will form, dissolution

  • Example 1A solution is prepared by mixing 750.0 mL of 4.00x10-3M Ce(NO3)3 and 300.0 mL 2.00x10-2M KIO3. Will Ce(IO3)3 precipitate out?Calculate Q value and compare to K (on chart)Ce(IO3)3(s) Ce3+(aq) + 3IO3-(aq)Q=[Ce3+][IO3-]3

    K=1.9x10-10 < Q so YES

  • Example 2A solution is made by mixing 150.0 mL of 1.00x10-2 M Mg(NO3)2 and 250.0 mL of 1.00x10-1 M NaF. Find concentration of Mg2+ and F- at equilibrium with solid MgF2 (Ksp=6.4x10-9)MgF2(s) Mg2+(aq) + 2F-(aq)Need to figure out whether the concentrations of the ions are high enough to cause precipitation firstFind Q and compare to K

  • Example 2Q > K so shift to left, precipitation occursWill all of it precipitate out??No- we need to figure out how much is created using stoichiometrythen how much ion is left over using ICE chartLike doing acid/base problem

  • Example 2How much will be use if goes to completion?Some of it dissolved- how much are left in solution?

    Mg2+(aq) + 2F-(aq) MgF2(s)I1.50 mmol25.0 mmolNot imp.C-1.50-1.50+2(1.50)E023.5 mmolNot imp.

    MgF2(s) Mg2+(aq) + 2F-(aq)INot imp.023.5mmol/400mLC-x+x+2xENot impx5.5x10-2+2x

  • Example 2

  • Qualitative AnalysisProcess used to separate a solution containing different ions using solubilitiesA solution of 1.0x10-4 M Cu+ and 2.0x10-3 M Pb2+. If I- is gradually added, which will precipitate out first, CuI or PbI2?1.4x10-8=[Pb2+][I-]2 = (2.0x10-3)[I-]2[I-]=2.6x10-3 M : [I-]> than that to cause PbI2 to ppt5.3x10-12=[Cu+][I-] = (1.0x10-4)[I-][I-]=5.3x10-8 M : [I-]> than that to cause CuI to pptTakes a much lower conc to cause CuI to ppt so it will happen first

  • Qualitative Analysis

  • Ch. 15: Applications of Aqueous Equilibria15.8: Complex Ion Equilibria

  • Complex Ion EquilibriaComplex ionCharged species containing metal ion surrounded by ligandsLigandsLewis bases donating electron pair to empty orbitals on metal ionEx: H2O, NH3, Cl-, CN-, OH-Coordination numberNumber of ligands attached

  • Complex Ion EquilibriaUsually, the conc of the ligand is very high compared to conc of metal ion in the solutionLigands attach in stepwise fashionAg+ + NH3 Ag(NH3)+Ag(NH3)+ + NH3 Ag(NH3)2+

  • Example 3Find the [Ag+], [Ag(S2O3)-], and [Ag(S2O3)23-] in solution made with 150.0 mL of 1.00x10-3 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3.Ag+ + S2O32- Ag(S2O3)-K1=7.4x108Ag(S2O3)- + S2O32- Ag(S2O3)23-K2=3.9x104Because of the difference in conc between ligand and metal ion, the reactions can be assumed to go to completion

  • Example 3

    Ag+ + 2S2O32- Ag(S2O3)23-I(150.0)(1.0x10-3) = 0.150mmol(200.0)(5.00) = 1.00x103mmol0C-0.150 mmol-0.150 mmol+0.150 mmolE01.00x103mmol0.150 mmol

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