Ch. 15: Applications of Aqueous Equilibria

1. Ch. 15: Applications of Aqueous Equilibria 15.4 Titrations and pH curves

2. Titration • used to find the concentration of a solution using a solution of known concentration • _____________ • solution with unknown concentration • _____________ • solution with known concentration • equivalence point is usually marked by the end point of an indicator

3. pH curve • also called a titration curve • plotting of pH of the analyte as a function of the amount of titrant added • millimole (mmol) • 1/1000th of a mole • = molarity

4. Strong Acid and Strong Base • Titrate 50 mL of 0.2 HNO3 with 0.1 M NaOH • no NaOH- 50.0 mL HNO3 • 10.0 mLNaOH- 50.0 mL HNO3

5. Strong Acid and Strong Base • 20 mLNaOH and 50 mL HNO3 • 50 mLNaOH and 50 mL HNO3

6. Strong Acid and Strong Base • 100 mLNaOH and 50 mL HNO3 • 150 mLNaOH and 50 mL HNO3

7. Strong Acid and Strong Base very gradual changes in pH until close to the equivalence pt.

8. Weak Acid and Strong Base • Titrate 50 mL of 0.10M HC2H3O2 with 0.1 M NaOH • no NaOH- 50.0 mLHC2H3O2

9. Weak Acid and Strong Base • 10mL NaOH and 50 mL HC2H3O2

10. Weak Acid and Strong Base • 50mL NaOH and 50 mL HC2H3O2

11. Weak Acid and Strong Base • 75mL NaOH and 50 mL HC2H3O2 there are two bases to consider: which will be most important? Can just use the OH- to determine the pH

12. Comparing Shapes

13. Weak Base and Strong Acid

14. Weak Base and Strong Acid

15. Ch. 15: Applications of Aqueous Equilibria 15.5 Indicators

16. phenolphthalein Acid-Base Indicators • used to mark end of titration (____________________) • only works if you choose one whose end point is the same as eq. point • usually are complex molecules that act as weak acids • change colors when H+ is removed

17. Indicators HIn  H+ + In- yellow blue • If more H+ (acid) is added to the solution, the reaction shifts to the left to make it yellow • If more OH- (base) is added to the solution, the reaction shifts to the right to make it blue Bromthymol Blue

18. Calculating the pH at end point • Assume that color change is visible to the eye when ratio of [In-]/[HIn] is 1/10 • that means that one tenth of the indicator must have changed forms • Using the H-H equation and the one tenth rule, the pH of end point or color change is:

19. Only shows where the color change occurs

20. Choosing an Indicator The weaker the acid being titrated, the smaller the vertical area around eq. pt, that means less flexibility in choosing an indicator.

21. Example • Estimate the pH of a solution in which bromcresol green is blue and thymol blue is yellow. • A 0.100 M HCl solution is made. 2 drops of methyl orange are added. What color is the solution?

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

23. 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 ________________ equals the rate of _____________________ CaF2(s) ↔ Ca2+(aq) + 2F-(aq) • ______________ solution • When no more solid can dissolve at equilibrium

24. Solubility Product • Solubility product: Ksp CaF2(s) ↔ Ca2+(aq) + 2F-(aq) • Ksp= • 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

25. Ksp Values

26. 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=

27. 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

28. Comparing Solubilities • You can only compare solubilities using Ksp values for compounds containing the same number of ions CaSO4 > CuCO3 > AgI Ksp values: 6.1x10-5 > 2.5x10-10 > 1.5x10-6 • Why can we use Ksp values to judge solubility? • Can only compare using actual solubility values (x) when compounds have different numbers of ions

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

30. 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

31. 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 = _______ [OH-]  solubility ____________ • Ex: Ag3PO4(s) ↔ 3Ag+(aq) + PO43-(aq) • What would happen if H+ is added? • H+ uses up _____ to make phosphoric acid • Eq. shifts to _______ - Solubility __________

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

33. 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 ________________ • Q > K: more reactant will form, ____________ until equilibrium reached • Q < K: more product will form, ____________

34. 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

35. 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

36. 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

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

38. Example 2

39. 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 = (2.0x10-4)[I-]2 • [I-]= : [I-]> than that to cause PbI2 to ppt • 5.3x10-12=[Cu+][I-] = (1.0x10-4)[I-] • [I-]= : [I-]> than that to cause CuI to ppt • Takes a much lower conc to cause CuI to ppt so it will happen first

40. Qualitative Analysis

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

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

43. 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+

44. 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

45. Example 3