1 / 25

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

lauradclark
Download Presentation

Ch. 15: Applications of Aqueous Equilibria

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related