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Ch. 15: Applications of Aqueous Equilibria

Ch. 15: Applications of Aqueous Equilibria. 15.6: Solubility Equilibria and Solubility Products. Solubility. As a salt dissolves in water and ions are released, they can collide and re-from the solid Equilibrium is reached when the rate of dissolution equals the rate of recrystallization

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Ch. 15: Applications of Aqueous Equilibria

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

  2. Solubility • As a salt dissolves in water and ions are released, they can collide and re-from the solid • Equilibrium is reached when the rate of dissolution equals the rate of recrystallization CaF2(s) ↔ Ca2+(aq) + 2F-(aq) • Saturated solution • When no more solid can dissolve at equilibrium

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

  4. Ksp Values

  5. Example 1 • CuBr has a solubility of 2.0x10-4 mol/L at 25°C. Find the Ksp value. • The solubility tells us the amount of solute that can dissolves in 1 L of water • Use ICE chart: solubility tells you x value • Ksp=[Cu+][Br-]= (2.0x10-4 M)2 = 4.0x10-8

  6. Example 2 • The Ksp value for Cu(IO3)2 is 1.4x10-7 at 25°C. Calculate its solubility. • Solve for solubility = x value using ICE chart • Ksp=[Cu2+][IO3-]2= (x)(2x)2 = 1.4x10-7 = 4x3 • X = (3.5x10-8)1/3 = 3.3x10-3

  7. Comparing Solubilities • You can only compare solubilities using Ksp values for compounds containing the same number of ions CaSO4 > AgI > CuCO3 Ksp 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

  8. Common Ion Effect • Solubility of a solid is lowered when a solution already contains one of the ions it contains • Why?

  9. Example 3 • Find 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 10-11 • x = 6.4x10-8

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

  11. Ch. 15: Applications of Aqueous Equilibria 15.7: Precipitation and Qualitative Analysis

  12. Precipitation • Opposite of dissolution • Can predict whether precipitation or dissolution will occur • Use Q: ion product • Equals Ksp but doesn’t have to be at equilibrium • Q > K: more reactant will form, precipitation until equilibrium reached • Q < K: more product will form, dissolution

  13. Example 1 • A 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

  14. Example 2 • A 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 first • Find Q and compare to K

  15. Example 2 • Q > K so shift to left, precipitation occurs • Will all of it precipitate out?? • No- • we need to figure out how much is created using stoichiometry • then how much ion is left over using ICE chart • Like doing acid/base problem

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

  17. Example 2

  18. Qualitative Analysis • Process used to separate a solution containing different ions using solubilities • A 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 ppt • 5.3x10-12=[Cu+][I-] = (1.0x10-4)[I-] • [I-]=5.3x10-8 M : [I-]> than that to cause CuI to ppt • Takes a much lower conc to cause CuI to ppt so it will happen first

  19. Qualitative Analysis

  20. Ch. 15: Applications of Aqueous Equilibria 15.8: Complex Ion Equilibria

  21. Complex Ion Equilibria • Complex ion • Charged species containing metal ion surrounded by ligands • Ligands • Lewis bases donating electron pair to empty orbitals on metal ion • Ex: H2O, NH3, Cl-, CN-, OH- • Coordination number • Number of ligands attached

  22. Complex Ion Equilibria • Usually, the conc of the ligand is very high compared to conc of metal ion in the solution • Ligands attach in stepwise fashion • Ag+ + NH3 Ag(NH3)+ • Ag(NH3)+ + NH3  Ag(NH3)2+

  23. Example 3 • Find 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=7.4x108 • Ag(S2O3)- + S2O32- Ag(S2O3)23- K2=3.9x104 • Because of the difference in conc between ligand and metal ion, the reactions can be assumed to go to completion

  24. Example 3

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