1. Additional Aspects of Aqueous Equilibria Chapter 17

2. 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

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

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

5. 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? K a = 1.8x10 -5

6. 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

7. 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 +

8. 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.

9. 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  H 2 O + X - As long as the ratio of [HX]/[X - ] doesn’t change too much, the change of pH is small.

10. 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.

12. 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

13. Sample Exercise 17.3 Calculating the pH of a Buffer What is the pH of a buffer that is 0.12 M in lactic acid [CH 3 CH(OH)COOH, or HC 3 H 5 O 3 ] and 0.10 M in sodium lactate [CH 3 CH(OH)COONa or NaC 3 H 5 O 3 ]? For lactic acid, K a = 1.4 × 10 -4.

14. Sample Exercise 17.4 Preparing a Buffer How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH is 9.00? (Assume that the addition of NH 4 Cl does not change the volume of the solution.)

15. Addition of Strong Acid 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 K a and the new concentrations of [HX] and [X - ] to calculate[H + ] (ICE or H-H)

16. Sample Exercise 17.5 Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol CH 3 COOH and 0.300 mol CH 3 COONa to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the pH of this solution after 0.020 mol of NaOH is added. (b) calculate the pH of the soln after the addition of 0.020 mol HCl

17. 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

18. 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 K a or K b 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

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

20. 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

21. 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

22. 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.

23. 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.

24. Sample Exercise 17.6 Calculating pH for a Strong Acid-Strong Base Titration Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution: (a) 49.0 mL, (b) 51.0 mL.

25. Weak Acid-Strong Base Titrations Titration curve of 50.0mL of HC 2 H 3 O 2 with 0.100M NaOH

26. Weak Acid-Strong Base Titrations (1) The initial pH can be determined by calculating the pH of acetic acid using an equilibrium expression

27. Weak Acid-Strong Base Titrations (2) Between the initial pH and the equivalence pt, the pH can be determined using stoichiometry and the Ka expression

28. Weak Acid-Strong Base Titrations (3) At the equivalence pt, the pH must be calculated using a base equilibrium expression.

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

30. Sample Exercise 17.7 Calculating pH for a Weak Acid-Strong Base Titration Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH (K a = 1.8 ×10 -5 ).

31. Sample Exercise 17.8 Calculating the pH at the Equivalence Point Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH.

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

33. Titration Simulation http://group.chem.iastate.edu/Greenbo we/sections/projectfolder/flashfiles/stoic hiometry/acid_base.html http://chem- ilp.net/labTechniques/AcidBaseIdicatorS imulation.htm

34. Solubility and Salts http://phet.colorado.edu/en/si mulation/soluble-salts

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

36. 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.

37. 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 K sp tells us how soluble a solid is in water

38. Solubility Equilibria Predict which solution would have a greater solubility AgCl K sp = 1.8x10 -10 AgBr K sp = 5.0 x 10 -13 AgIK sp = 8.3 x 10 -17

39. Sample Exercise 17.9 Writing Solubility-Product (Ksp) Expressions Write the expression for the solubility product constant for CaF 2, and look up the corresponding K sp value in Appendix D.

40. 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 Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 × 10 -4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving the Ag + or CrO 4 2– ions in the solution, calculate Ksp for this compound.

41. Sample Exercise 17.11 Calculating Solubility from Ksp The Ksp for CaF 2 is 3.9 ×10 -11 at 25 ºC. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter.

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

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

44. Sample Exercise 17.12 Calculating the Effect of a Common Ion on Solubility Calculate the molar solubility of CaF 2 at 25 °C in a solution that is (a) 0.010 M in Ca(NO 3 ) 2, (b) 0.010 M in NaF.

45. 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

46. Factors that affect solubility Example: A saturated solution of Mg(OH) 2 has a pH of 10.52 and contains 1.7 x 10 -4 M Mg 2+. The pH is then changed to 9.00, what is [Mg 2+ ]?

48. Precipitation and Separation of ions The use of a reaction quotient Q can be used to predict what is going on in the solution

49. 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

50. 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(NO 3 ) 2 is added to 0.40 L of 5.0 x 10 -3 M Na 2 SO 4?

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

52. Sample Exercise 17.16 Calculating Ion Concentrations for Precipitation A solution contains 1.0 × 10 -2 M Ag + and 2.0 ×10 -2 M Pb 2+. When Cl – is added to the solution, both AgCl (Ksp = 1.8× 10 -10 ) and PbCl 2 (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?

53. Practice Exercise 17.16 Calculating Ion Concentrations for Precipitation A solution consists of 0.050 M Mg 2+ and 0.020 M Cu 2+. 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? [K sp = 1.8 × 10 -11 for Mg(OH) 2, and K sp = 4.8 ×10 -20 for Cu(OH) 2.]

54. Practice Exercise 17.7 (a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (C 6 H 5 COOH, Ka = 6.3 × 10 -5 ).

55. Practice Exercise 17.7 (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH 3.(Kb=1.8x10 -5 )

56. Practice Exercise 17.8 Calculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid (C 6 H 5 COOH, K a = 6.3 × 10 -5 ) is titrated with 0.050 M NaOH