1. Additional Aspects of Aqueous Equilibria
Chapter 17

2. Acid- Base Rxns – Write Molecular and Net ionic equations
Strong Acid and Strong Base
HCl(aq) + NaOH (aq) 
Strong Acid and Weak Base
HCl(aq) + NaC2H3O2(aq) 
Strong Base and Weak Acid
NaOH(aq) + HC2H3O2(aq) 

3. The Common-Ion Effect
The extent of ionization of a weak electrolyte is decreased by adding a strong electrolyte that has an ion in common with the weak electrolyte

4. The Common-Ion Effect
If you have a weak acid and a common ion, the [H+] will decrease, causing the pH to increase

5. The Common-Ion Effect
If you have a weak base and a common ion, the [OH-] will decrease, causing the pH to decrease

6. Sample Exercise 17.1 Calculating the pH When a Common Ion is Involved
What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? Ka = 1.8x10-5

7. Composition and Action of Buffered Solutions
Buffers are solns which contain a weak conj. acid-base pair and can resist drastic changes in pH upon the addition of small amounts of strong acid or strong base

8. Composition and Action of Buffered Solutions
A buffer resists changes in pH because it contains both an acidic species to neutralize OH- and a basic one to neutralize H+

9. Composition and Action of Buffered Solutions
Consider a buffer composed of weak acid HX and one of its salts MX (X-), where M is Na+, K+, or another cation.

10. Composition and Action of Buffered Solutions
If OH- ions are added, they react with the acid component of the buffer to produce water and base X- .
OH- + HX  H2O + X-
As long as the ratio of [HX]/[X-] doesn’t change too much, the change of pH is small.

11. Composition and Action of Buffered Solutions
If H+ ions are added, they react with the base component of the buffer
H+ + X-  HX
As long as the ratio of [HX]/[X-] doesn’t change too much, the change in pH is small.

13. Composition and Action of Buffered Solutions
Buffers
Made of conjugate acid-base pairs
most effectively resist change in pH when the concentration of weak acid and conjugate base are the same
Chosen so that the acid in the buffer has a pKa close to the desired pH

14. Sample Exercise 17.3 Calculating the pH of a Buffer
What is the pH of a buffer that is 0.12M
in lactic acid [HC3H5O3] (Ka = 1.4 × 10-4) and 0.10M in sodium lactate [NaC3H5O3] ?

15. Sample Exercise 17.4 Preparing a Buffer
How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? (Assume that the addition of NH4Cl does not change the volume of the solution.)

16. Addition of Strong Acids or Bases to Buffers
To calculate pH upon addition of a SA or SB
1. Consider the acid-base neutralization rxn and determine its effect on [HX] and [X-] (LR and stoichiometry)
2. Use Ka and the new concentrations of [HX] and [X-] to calculate[H+] (ICE or H-H)

17. Sample Exercise 17.5 Calculating pH Changes in Buffers
A buffer is made by adding mol CH3COOH and mol CH3COONa to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (Ka = 1.8 x 10-5)
(a) Calculate the pH of this solution after mol of NaOH is added.
(b) calculate the pH of the soln after the addition of mol HCl

18. Acid-Base Titrations Warm up:
How many mL of 0.105M HCl are needed to titrate 23.5mL of 0.117M KOH to the equivalence point?
Hint: Write a reaction and use stoichiometry

19. Acid-Base Titrations pH titration curve
1. graph of the pH as a function of the volume of added titrant
2. Shape helps determine equivalence pt, choose indicator, and determine Ka or Kb
3. The equivalence point is the point at which H+ moles = OH- moles
4. Any indicator beginning and ending its color change on the rapid-rise portion of the curve can be used

20. Strong Acid-Strong Base Titrations
Titration curve of 50.0mL of 0.100M HCl as 0.100M NaOH is added

21. Strong Acid-Strong Base Titrations
(1) The initial pH before any base is added can be determined from the initial concentration of the strong acid

22. Strong Acid-Strong Base Titrations
(2) Between the initial pH and the equivalence pt, the pH is determined by the concentration of the acid that has not yet been neutralized

23. Strong Acid-Strong Base Titrations
(3) At the equivalence point, the moles of H+ = moles of OH- and all that remains in solution is a salt. The pH is 7.00.

24. Strong Acid-Strong Base Titrations
(4) After the equivalence point, the pH can be determined by finding the excess[OH-], the pOH, and then pH.

25. Sample Exercise 17.6 Calculating pH for a Strong Acid-Strong Base Titration
Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution: (a) 49.0 mL, (b) 51.0 mL.

26. Weak Acid-Strong Base Titrations
Titration curve of 50.0mL of 0.10M HC2H3O2 with 0.100M NaOH

27. Weak Acid-Strong Base Titrations
(1) The initial pH can be determined by calculating the pH of weak acid

28. Weak Acid-Strong Base Titrations
(2) Between the initial pH and the equivalence pt, the pH can be determined using stoichiometry and the Ka
(Note: pH=pKa when [A-] = [HA] or [acid]=[base] which occurs halfway to equivalence)

29. Weak Acid-Strong Base Titrations
(3) At the equivalence pt, the pH must be calculated by finding pH of weak base formed

30. Weak Acid-Strong Base Titrations
(4) After the equivalence pt, the excess [OH-] from the addition of the strong base will account for the pH

31. Sample Exercise 17.8 Calculating pH for a Weak Acid-Strong Base Titration
Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH3COOH (Ka = 1.8 ×10-5).

32. Sample Exercise 17.9 Calculating the pH at the Equivalence Point
Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH3COOH with M NaOH.

33. Titrations
WA/SB Titrations
(Note: pH =pKa halfway to equivalence)

34. Titrations An acid acting as the titrant in an acid-base titration
Red SA/SB
Blue SA/WB

35. Titrations and choosing indicators

36. Titrations and choosing indicators

37. Titrations of Polyprotic Acids
Phosphorous acid is neutralized in a series of steps

38. Titration Simulation

39. Solubility and Salts http://phet. colorado

40. Solubility Equilibria
the amount of a substance that dissolves in a given amount of solvent at a certain T to form a saturated soln

41. Solubility Equilibria
Unsaturated Solutions – more solute can be added to solvent. It will continue to dissolve solute into solvent.
Saturated solutions – no more solute can be added to the solvent. It can no longer dissolve any more solute.

42. Solubility Equilibria
A saturated solution of a slightly soluble salt is one in which the solution is in equilibrium with undissolved solute
The solubility product constant Ksp tells us how soluble a solid is in water

43. Solubility Equilibria
Predict which solution would have a greater solubility
AgCl Ksp = 1.8x10-10
AgBr Ksp = 5.0 x 10-13
AgI Ksp = 8.3 x 10-17

44. Sample Exercise 17.9 Writing Solubility-Product (Ksp) Expressions
Write the expression for the solubility product constant for CaF2, and look up the corresponding Ksp value in Appendix D.

45. Sample Exercise 17.10 Calculating Ksp from Solubility
Solid silver chromate is added to pure water at 25 ºC. Some of the solid remains undissolved at the bottom of the flask. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 × 10-4 M. Calculate Ksp for this compound.

46. Sample Exercise 17.11 Calculating Solubility from Ksp
The Ksp for CaF2 is 3.9 ×10-11 at 25 ºC.
Calculate the solubility of CaF2 in grams per liter.

47. Factors that affect solubility
Solubility of a substance is affected by T

48. Factors that affect solubility
Solubility of a slightly soluble salt is decreased by the presence of a second solute that has a common ion

49. Sample Exercise 17.12 Calculating the Effect of a Common Ion on Solubility
Calculate the molar solubility of CaF2 at 25 °C in a solution that is (a) M in Ca(NO3)2, (b) M in NaF.

50. Factors that affect solubility
The solubility of any substance whose anion is basic will be affected to some extent by the pH of the soln

51. Factors that affect solubility
A saturated solution of Mg(OH)2 has a pH of and contains 1.7 x 10-4 M Mg2+. The pH is then changed to 9.00, what is [Mg2+]?

52. Precipitation and Separation of ions
If Q > Ksp, precipitation occurs until Q=Ksp
If Q = Ksp, equilibrium exists (saturated solution)
If Q < Ksp, solid dissolves until Q=Ksp (unsaturated)

53. Sample Exercise 17.15 Predicting Whether a Precipitate Will Form
Will a precipitate form when 0.10 L of 8.0 x 10-3 M Pb(NO3)2 is added to 0.40 L of 5.0 x 10-3 M Na2SO4?

54. Precipitation and Separation of ions
Selective precipitation - Ions can be separated from each other based upon the solubilities of their salts

55. Sample Exercise 17.16 Calculating Ion Concentrations for Precipitation
A solution contains 1.0 × 10-2 M Ag+ and 2.0 ×10-2 M Pb2+. When Cl– is added to the solution, both AgCl
(Ksp = 1.8× 10-10) and PbCl2 (Ksp = 1.7×10-5) precipitate from the solution. What concentration of Cl– is necessary to begin the precipitation of each salt? Which salt precipitates first?

56. Practice Exercise 17.16 Calculating Ion Concentrations for Precipitation
A solution consists of M Mg2+ and M Cu2+. Which ion will precipitate first as OH– is added to the solution? What concentration of OH– is necessary to begin the precipitation of each cation?
[Ksp = 1.8 × for Mg(OH)2,
and Ksp = 4.8 ×10-20 for Cu(OH)2.]

57. Practice Exercise 17.7
(a) Calculate the pH in the solution formed by adding 10.0 mL of M NaOH to 40.0 mL of M benzoic acid (C6H5COOH, Ka = 6.3 × 10-5) .

58. Practice Exercise 17.7
(b) Calculate the pH in the solution formed by adding 10.0 mL of M HCl to 20.0 mL of M NH3.(Kb=1.8x10-5)

59. Practice Exercise 17.8
Calculate the pH at the equivalence point when (a) 40.0 mL of M benzoic acid (C6H5COOH, Ka = 6.3 × 10-5 ) is titrated with M NaOH