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 CaF 2 (s) ↔ Ca 2+ (aq) + 2F - (aq)  Saturated solution When no more solid can dissolve at equilibrium When no more solid can dissolve at equilibrium

3. Solubility Product  Solubility product: K sp CaF 2 (s) ↔ Ca 2+ (aq) + 2F - (aq)  Ksp=[Ca 2+ ][F - ] 2  Why do we leave out the CaF 2 ?  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. K sp Values

5. Example 1  CuBr has a solubility of 2.0x10 -4 mol/L at 25°C. Find the K sp value. The solubility tells us the amount of solute that can dissolves in 1 L of water The solubility tells us the amount of solute that can dissolves in 1 L of water Use ICE chart: solubility tells you x value Use ICE chart: solubility tells you x value Ksp=[Cu + ][Br - ]= (2.0x10 -4 M) 2 = 4.0x10 -8 Ksp=[Cu + ][Br - ]= (2.0x10 -4 M) 2 = 4.0x10 -8 CuBr(s) ↔ Cu + (aq) + Br - (aq) CuBr(s) ↔ Cu + (aq) + Br - (aq) I Not imp. 00 C -2.0x10 -4 M +2.0x10 -4 M E Not imp 2.0x10 -4 M

6. Example 2  The K sp value for Cu(IO 3 ) 2 is 1.4x10 -7 at 25°C. Calculate its solubility. Solve for solubility = x value using ICE chart Solve for solubility = x value using ICE chart K sp =[Cu 2+ ][IO 3 - ] 2 = (x)(2x) 2 = 1.4x10 -7 = 4x 3 K sp =[Cu 2+ ][IO 3 - ] 2 = (x)(2x) 2 = 1.4x10 -7 = 4x 3 X = (3.5x10 -8 ) 1/3 = 3.3x10 -3 X = (3.5x10 -8 ) 1/3 = 3.3x10 -3 Cu(IO 3 ) 2 (s) ↔ Cu 2+ (aq) + 2IO 3 - (aq) Cu(IO 3 ) 2 (s) ↔ Cu 2+ (aq) + 2IO 3 - (aq) I Not imp. 00 C-x+x+2x E Not imp x2x

7. Comparing Solubilities  You can only compare solubilities using K sp values for compounds containing the same number of ions CaSO 4 > AgI > CuCO 3 CaSO 4 > AgI > CuCO 3 K sp values: 6.1x10 -5 > 1.5x10 -6 > 2.5x10 -10  Why can we use K sp 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 CaF 2 (s) if the K sp is 4.0 x 10 -11 and it is in a 0.025 M NaF solution. K sp =[Ca 2+ ][F - ] 2 = (x)(2x+0.025) 2 = 4.0 x 10 -11 K sp =[Ca 2+ ][F - ] 2 = (x)(2x+0.025) 2 = 4.0 x 10 -11 (x)(0.025) 2 ≈ 4.0 x 10 -11 (x)(0.025) 2 ≈ 4.0 x 10 -11 x = 6.4x10 -8 x = 6.4x10 -8 CaF 2 (s) ↔ Ca 2+ (aq) + 2F - (aq) CaF 2 (s) ↔ Ca 2+ (aq) + 2F - (aq) I Not imp. 00.025 C-x+x+2x E Not imp x.025+2x

10. pH and solubility  pH can effect solubility because of the common ion effect  Ex: Mg(OH) 2 (s) ↔ Mg 2+ (aq) + 2OH - (aq) How would a high pH effect solubility? How would a high pH effect solubility? High pH = high [OH-]  decrease solubility High pH = high [OH-]  decrease solubility  Ex: Ag 3 PO 4 (s) ↔ 3Ag + (aq) + PO 4 3- (aq) What would happen if H+ is added? What would happen if H+ is added? H + uses up PO 4 3- to make phosphoric acid H + uses up PO 4 3- to make phosphoric acid Eq. shifts to right - Solubility increases 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 K sp but doesn’t have to be at equilibrium Equals K sp 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 -3 M Ce(NO 3 ) 3 and 300.0 mL 2.00x10 -2 M KIO 3. Will Ce(IO 3 ) 3 precipitate out? Calculate Q value and compare to K (on chart) Calculate Q value and compare to K (on chart) Ce(IO 3 ) 3 (s) ↔ Ce 3+ (aq) + 3IO 3 - (aq) Ce(IO 3 ) 3 (s) ↔ Ce 3+ (aq) + 3IO 3 - (aq) Q=[Ce 3+ ][IO 3 - ] 3 Q=[Ce 3+ ][IO 3 - ] 3 K=1.9x10 -10 < Q so YES 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(NO 3 ) 2 and 250.0 mL of 1.00x10 -1 M NaF. Find concentration of Mg 2+ and F - at equilibrium with solid MgF 2 (K sp =6.4x10 -9 ) MgF 2 (s)  Mg 2+ (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 we need to figure out how much is created using stoichiometry then how much ion is left over using ICE chart then how much ion is left over using ICE chart  Like doing acid/base problem

16. Example 2 Mg 2+ (aq) + 2F - (aq) ↔ MgF 2 (s) Mg 2+ (aq) + 2F - (aq) ↔ MgF 2 (s) I 1.50 mmol 25.0 mmol Not imp. C-1.50-1.50+2(1.50) E0 23.5 mmol Not imp. MgF 2 (s) ↔ Mg 2+ (aq) + 2F - (aq) MgF 2 (s) ↔ Mg 2+ (aq) + 2F - (aq)I Not imp. 023.5mmol/400mL C-x+x+2x E Not imp x 5.5x10 -2 +2x 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 Pb 2+. If I - is gradually added, which will precipitate out first, CuI or PbI 2 ? 1.4x10 -8 =[Pb 2+ ][I - ] 2 = (2.0x10 -3 )[I - ] 2 1.4x10 -8 =[Pb 2+ ][I - ] 2 = (2.0x10 -3 )[I - ] 2 [I - ]=2.6x10 -3 M : [I - ]> than that to cause PbI 2 to ppt [I - ]=2.6x10 -3 M : [I - ]> than that to cause PbI 2 to ppt 5.3x10 -12 =[Cu + ][I - ] = (1.0x10 -4 )[I - ] 5.3x10 -12 =[Cu + ][I - ] = (1.0x10 -4 )[I - ] [I - ]=5.3x10 -8 M : [I - ]> than that to cause CuI to ppt [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 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 Charged species containing metal ion surrounded by ligands  Ligands Lewis bases donating electron pair to empty orbitals on metal ion Lewis bases donating electron pair to empty orbitals on metal ion Ex: H 2 O, NH 3, Cl-, CN-, OH- Ex: H 2 O, NH 3, Cl-, CN-, OH-  Coordination number Number of ligands attached 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 + + NH 3  Ag(NH 3 ) + Ag + + NH 3  Ag(NH 3 ) + Ag(NH 3 ) + + NH 3  Ag(NH 3 ) 2+ Ag(NH 3 ) + + NH 3  Ag(NH 3 ) 2+

23. Example 3  Find the [Ag + ], [Ag(S 2 O 3 ) - ], and [Ag(S 2 O 3 ) 2 3- ] in solution made with 150.0 mL of 1.00x10 -3 M AgNO 3 with 200.0 mL of 5.00 M Na 2 S 2 O 3. Ag + + S 2 O 3 2-  Ag(S 2 O 3 ) - K 1 =7.4x10 8 Ag + + S 2 O 3 2-  Ag(S 2 O 3 ) - K 1 =7.4x10 8 Ag(S 2 O 3 ) - + S 2 O 3 2-  Ag(S 2 O 3 ) 2 3- K 2 =3.9x10 4 Ag(S 2 O 3 ) - + S 2 O 3 2-  Ag(S 2 O 3 ) 2 3- K 2 =3.9x10 4  Because of the difference in conc between ligand and metal ion, the reactions can be assumed to go to completion

24. Example 3 Ag + + 2S 2 O 3 2-  Ag(S 2 O 3 ) 2 3- Ag + + 2S 2 O 3 2-  Ag(S 2 O 3 ) 2 3- I (150.0)(1.0x10 -3 ) = 0.150mmol (200.0)(5.00) = 1.00x10 3 mmol 0 C -0.150 mmol +0.150 mmol E0 1.00x10 3 mmol 0.150 mmol