1. Chapter 16 Aqueous Ionic Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Copyright  2011 Pearson Education, Inc.

2. Buffers Buffers are solutions that resist changes in pH when an acid or base is added They act by neutralizing acid or base that is added to the buffered solution But just like everything else, there is a limit to what they can do, and eventually the pH changes Many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing the weak acid’s conjugate base anion blood has a mixture of H 2 CO 3 and HCO 3 − 2

3. Making an Acid Buffer 3

4. How Acid Buffers Work: Addition of Base HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) Buffers work by applying Le Châtelier’s Principle to weak acid equilibrium Buffer solutions contain significant amounts of the weak acid molecules, HA These molecules react with added base to neutralize the HA HA(aq) + OH − (aq) → A − (aq) + H 2 O(l)  you can also think of the H 3 O + combining with the OH − to make H 2 O; the H 3 O + is then replenished by the shifting equilibrium 4

5. H2OH2O HA How Buffers Work HA  + H3O+H3O+ A−A− Added HO − new A − A−A− 5

6. How Acid Buffers Work: Addition of Acid HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) The buffer solution also contains significant amounts of the conjugate base anion, A − These ions combine with added acid to make more HA H + (aq) + A − (aq) → HA(aq)  After the equilibrium shifts, the concentration of H 3 O + is kept constant 6

7. H2OH2O How Buffers Work HA  + H3O+H3O+ A−A− A−A− Added H 3 O + new HA 7

8. Common Ion Effect HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) Adding a salt NaA containing the A - anion, which is the conjugate base of the acid (the common ion), shifts the position of equilibrium to the left This causes the pH to be higher than the pH of the acid solution  lowering the H 3 O + ion concentration 8

9. Common Ion Effect 9

10. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 10 write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0 HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change equilibrium

11. [HA][A − ][H 3 O + ] initial 0.100 0 change equilibrium Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 11 represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x+x+x xx 0.100  x 0.100 + x x HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O +

12. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 12 determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and [A − ] eq = [A − ] init, then solve for x [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100 x 0.100  x 0.100 +x K a for HC 2 H 3 O 2 = 1.8 x 10 −5

13. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 13 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init K a for HC 2 H 3 O 2 = 1.8 x 10 −5 the approximation is valid x = 1.8 x 10 −5 [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100 x

14. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 14 substitute x into the equilibrium concentration definitions and solve x = 1.8 x 10 −5 [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100 1.8E-5 0.100 + x x 0.100  x

15. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 15 substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100 1.8E−5

16. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 16 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values match [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100 1.8E−5 K a for HC 2 H 3 O 2 = 1.8 x 10 −5

17. Practice − What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 17 write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0 HF + H 2 O  F  + H 3 O + [HA][A − ][H 3 O + ] initial 0.140.071≈ 0 change equilibrium

18. [HA][A − ][H 3 O + ] initial 0.140.0710 change equilibrium Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 18 represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x+x+x xx 0.14  x 0.071 + x x HF + H 2 O  F  + H 3 O +

19. pK a for HF = 3.15K a for HF = 7.0 x 10 −4 Practice – What is the pH of a buffer that is 0.14 M and 0.071 M KF? 19 determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and [A − ] eq = [A − ] init solve for x [HA][A − ][H 3 O + ] initial 0.140.071≈ 0 change −x−x+x+x+x+x equilibrium 0.012 0.100 x 0.14  x 0.071 +x

20. Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 20 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init K a for HF = 7.0 x 10 −4 the approximation is valid x = 1.4 x 10 −3 [HA][A − ][H 3 O + ] initial 0.140.071≈ 0 change −x−x+x+x+x+x equilibrium 0.140.071x

21. Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 21 substitute x into the equilibrium concentration definitions and solve x = 1.4 x 10 −3 [HA][A − ][H 3 O + ] initial 0.140.071≈ 0 change −x−x+x+x+x+x equilibrium 0.140.0721.4E-3 0.071 + x x 0.14  x

22. Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 22 substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial 0.140.071≈ 0 change −x−x+x+x+x+x equilibrium 0.140.0721.4E−3

23. Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 23 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values are close enough [HA][A − ][H 3 O + ] initial 0.140.071≈ 0 change −x−x+x+x+x+x equilibrium 0.140.0721.4E−3 K a for HF = 7.0 x 10 −4

24. Henderson-Hasselbalch Equation Calculating a buffer’s pH can be simplified using the Henderson-Hasselbalch Equation  the “x is small” approximation must be valid pH is calculated from the pK a and the initial concentrations of the weak acid and salt of the conjugate base 24

25. Deriving the Henderson-Hasselbalch Equation 25

26. Deriving the Henderson-Hasselbalch Equation 26

27. Example 16.2: What is the pH of a buffer that is 0.050 M HC 7 H 5 O 2 and 0.150 M NaC 7 H 5 O 2 ? 27 assume the [HA] and [A − ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch Equation check the “x is small” approximation HC 7 H 5 O 2 + H 2 O  C 7 H 5 O 2  + H 3 O + K a for HC 7 H 5 O 2 = 6.5 x 10 −5

28. Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 28 find the pK a from the given K a assume the [HA] and [A − ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch equation check the “x is small” approximation HF + H 2 O  F  + H 3 O +

29. Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation? Use the Henderson-Hasselbalch equation when “x is small” approximation is valid The “x is small” approximation will work when the initial concentrations of acid and salt are not very dilute, AND If the K a is fairly small For most problems, this means that the initial acid and salt concentrations should be over 100 to 1000x larger than the value of K a 29

30. How Much Does the pH of a Buffer Change When an Acid or Base Is Added? Though buffers do resist change in pH when acid or base is added to them, their pH does change Calculating the new pH after adding acid or base requires breaking the problem into two parts 1.a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A − to make more HA added base reacts with the HA to make more A − 2.an equilibrium calculation of [H 3 O + ] using the new initial values of [HA] and [A − ] 30

31. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 31 If the added chemical is a base, write a reaction for OH − with HA. If the added chemical is an acid, write a reaction for H 3 O + with A −. construct a stoichiometry table for the reaction HC 2 H 3 O 2 + OH −  C 2 H 3 O 2  + H 2 O HAOH − A−A− mols before 0.1000 mols added ─0.010─ mols after

32. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 32 fill in the table – tracking the changes in the number of moles for each component HC 2 H 3 O 2 + OH −  C 2 H 3 O 2  + H 2 O HAOH − A−A− mols before 0.100≈ 00.100 mols added ─0.010─ mols after 0.090≈ 00.110

33. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 33 If the added chemical is a base, write a reaction for OH − with HA. If the added chemical is an acid, write a reaction for it with A −. construct a stoichiometry table for the reaction enter the initial number of moles for each HC 2 H 3 O 2 + OH −  C 2 H 3 O 2  + H 2 O HAOH – A−A− mols before 0.1000.0100.100 mols change mols end new molarity

34. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 34 using the added chemical as the limiting reactant, determine how the moles of the other chemicals change add the change to the initial number of moles to find the moles after reaction divide by the liters of solution to find the new molarities HC 2 H 3 O 2 + OH −  C 2 H 3 O 2  + H 2 O HAOH − A−A− mols before 0.1000.0100.100 mols change mols end new molarity −0.010 +0.010 0.1100 0.090 00.110

35. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 35 write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0, and using the new molarities of the [HA] and [A − ] HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + [HA][A − ][H 3 O + ] initial 0.0900.110 ≈ 0≈ 0 change equilibrium

36. [HA][A − ][H 3 O + ] initial 0.0900.110≈ 0 change equilibrium Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 36 represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x+x+x xx 0.090  x 0.110 + x x HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O +

37. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 37 determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and [A − ] eq = [A − ] init solve for x [HA][A − ][H 3 O + ] initial 0.100 ≈ 0≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.110 x 0.090  x 0.110 +x K a for HC 2 H 3 O 2 = 1.8 x 10 −5

38. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 38 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init K a for HC 2 H 3 O 2 = 1.8 x 10 −5 the approximation is valid x = 1.47 x 10 −5 [HA][A − ][H 3 O + ] initial 0.0900.110≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.110 x

39. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 39 substitute x into the equilibrium concentration definitions and solve x = 1.47 x 10 −5 [HA][A - ][H 3 O + ] initial 0.0900.110 ≈ 0≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.1101.5E-5 0.110 + x x 0.090  x

40. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 40 substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial 0.0900.110 ≈ 0≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.1101.5E−5

41. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 41 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values match [HA][A − ][H 3 O + ] initial 0.0900.110 ≈ 0≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.1101.5E−5 K a for HC 2 H 3 O 2 = 1.8 x 10 −5

42. or, by using the Henderson- Hasselbalch Equation

43. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 43 write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0, and using the new molarities of the [HA] and [A − ] fll in the ICE table HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + [HA][A − ][H 3 O + ] initial 0.0900.110 ≈ 0≈ 0 change equilibrium +x+x+x+x xx 0.090  x 0.110 + x x

44. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 44 find the pK a from the given K a assume the [HA] and [A − ] equilibrium concentrations are the same as the initial HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + K a for HC 2 H 3 O 2 = 1.8 x 10 −5 [HA][A − ][H 3 O + ] initial 0.0900.110≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.110x

45. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 45 substitute into the Henderson- Hasselbalch equation check the “x is small” approximation HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + pK a for HC 2 H 3 O 2 = 4.745

46. Example 16.3: Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L to adding 0.010 mol NaOH to 1.00 L of pure water 46 HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + pK a for HC 2 H 3 O 2 = 4.745

47. What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? (pK a = 3.15) 47 If the added chemical is a base, write a reaction for OH − with HA. If the added chemical is an acid, write a reaction for H 3 O + with A −. construct a stoichiometry table for the reaction F − + H 3 O +  HF + H 2 O F−F− H3O+H3O+ HF mols before 0.07100.140 mols added– 0.020 – mols after (The “x is small” approximation is valid)

48. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? 48 fill in the table – tracking the changes in the number of moles for each component F − + H 3 O +  HF + H 2 O F−F− H3O+H3O+ HF mols before 0.07100.140 mols added– 0.020 – mols after 0.05100.160

49. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? 49 If the added chemical is a base, write a reaction for OH − with HA. If the added chemical is an acid, write a reaction for H 3 O + with A −. construct a stoichiometry table for the reaction enter the initial number of moles for each F − + H 3 O +  HF + H 2 O F−F− H3O+H3O+ HF mols before0.0710.0200.140 mols change mols after new molarity

50. F−F− H3O+H3O+ HF mols before0.0710.0200.140 mols change mols after new molarity Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? 50 using the added chemical as the limiting reactant, determine how the moles of the other chemicals change add the change to the initial number of moles to find the moles after reaction divide by the liters of solution to find the new molarities 0.16000.051 F − + H 3 O +  HF + H 2 O +0.020−0.020 0.16000.051

51. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? 51 write the reaction for the acid with water construct an ICE table. assume the [HA] and [A − ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch equation HF + H 2 O  F  + H 3 O + [HF][F − ][H 3 O + ] initial 0.1600.051≈ 0 change −x−x+x+x+x+x equilibrium 0.1600.051x

52. Basic Buffers B: (aq) + H 2 O (l)  H:B + (aq) + OH − (aq) Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B + Cl − NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH − (aq) 52

53. Henderson-Hasselbalch equation is written for a weak acid reactant and its conjugate base product For a basic buffer with a weak base as a reactant and its conjugate acid as a product B: + H 2 O  H:B + + OH − Therefore, the chemical equation of the basic buffer must be written as an acid reaction H:B + + H 2 O  B: + H 3 O + We can rewrite the Henderson-Hasselbalch equation for the chemical equation of the basic buffer in terms of pOH Henderson-Hasselbalch Equation for Basic Buffers 53

54. Relationship between pK a and pK b The relationship between the K a of a weak acid and K b of its conjugate base is K a  K b = K w = 1.0 x 10 −14 There is also a relationship between the pK a of a weak acid and the pK b of its conjugate base −log(K a  K b ) = −log(K w ) = 14 −log(K a ) + −log(K b ) = 14 pK a + pK b = 14 54

55. Example 16.4: What is the pH of a buffer that is 0.50 M NH 3 (pK b = 4.75) and 0.20 M NH 4 Cl? 55 find the pK a of the conjugate acid (NH 4 + ) from the given K b assume the [B] and [HB + ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch equation check the “x is small” approximation NH 3 + H 2 O  NH 4 + + OH −

56. Example 16.4: What is the pH of a buffer that is 0.50 M NH 3 (pK b = 4.75) and 0.20 M NH 4 Cl? 56 find the pK b if given K b assume the [B] and [HB + ] equilibrium concentrations are the same as the initial substitute into the Henderson-Hasselbalch equation base form, find pOH check the “x is small” approximation calculate pH from pOH NH 3 + H 2 O  NH 4 + + OH −

57. Buffering Effectiveness Good buffers should be able to neutralize moderate amounts of added acid or base But, there is a limit to how much acid or base can be added before the pH changes significantly The buffering capacity is the amount of acid or base a buffer can neutralize The buffering range is the pH range the buffer is be effective The effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base 57

58. HAOH − A−A− mols before0.1800.020 mols added ─ 0.010 ─ mols after 0.17≈ 00.030 Effect of Relative Amounts of Acid and Conjugate Base Buffer 1 0.100 mol HA & 0.100 mol A − Initial pH = 5.00 Buffer 2 0.18 mol HA & 0.020 mol A − Initial pH = 4.05 pK a (HA) = 5.00 HA + OH −  A  + H 2 O HAOH − A−A− mols before0.1000 mols added ─ 0.010 ─ mols after 0.090≈ 00.110

59. HAOH − A−A− mols before0.1800.020 mols added ─ 0.010 ─ mols after 0.17≈ 00.030 Effect of Relative Amounts Acid & Conjugate Base after adding 0.010 mol NaOH pH = 5.09 HAOH − A−A− mols before0.1000 mols added ─ 0.010 ─ mols after 0.090≈ 00.110 after adding 0.010 mol NaOH pH = 4.25 A buffer is most effective with equal concentrations of acid and base

60. HAOH − A−A− mols before0.5000.500 mols added ─ 0.010 ─ mols after 0.49≈ 00.51 HAOH − A−A− mols before0.0500 mols added ─ 0.010 ─ mols after 0.040≈ 00.060 Effect of Absolute Concentrations of Acid and Conjugate Base Buffer 1 0.50 mol HA & 0.50 mol A − Initial pH = 5.00 Buffer 2 0.050 mol HA & 0.050 mol A − Initial pH = 5.00 pK a (HA) = 5.00 HA + OH −  A  + H 2 O

61. HAOH − A−A− mols before0.5000.500 mols added ─ 0.010 ─ mols after 0.49≈ 00.51 HAOH − A−A− mols before0.0500 mols added ─ 0.010 ─ mols after 0.040≈ 00.060 Effect of Absolute Concentrations of Acid and Conjugate Base after adding 0.010 mol NaOH pH = 5.02 after adding 0.010 mol NaOH pH = 5.18 A buffer is most effective when the concentrations of acid and base are largest

62. Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer 62

63. Effectiveness of Buffers A buffer will be most effective when there are equal concentrations of acid and base [base]:[acid] = 1 A buffer will be effective when 0.1< [base]:[acid] < 10 A buffer will be most effective when the [acid] and the [base] are large Buffers that need to work mainly with added acid generally have [base] > [acid] Buffers that need to work mainly with added base generally have [acid] > [base] 63

64. Buffering Range We have said that a buffer will be effective when 0.1 < [base]:[acid] < 10 Substituting into the Henderson-Hasselbalch equation we can calculate the minimum and maximum pH at which the buffer will be effective Lowest pHHighest pH Therefore, the effective pH range of a buffer is pK a ± 1 When choosing an acid to make a buffer, choose one whose is pK a closest to the pH of the buffer 64

65. Formic acid, HCHO 2 pK a = 3.74 Example 16.5a: Choose the best acid to combine with its sodium salt to make a buffer with pH 4.25? Chlorous acid, HClO 2 pK a = 1.95 Nitrous acid, HNO 2 pK a = 3.34 Hypochlorous acid, HClOpK a = 7.54 The pK a of HCHO 2 is closest to the desired pH of the buffer, so it would give the most effective buffering range 65

66. Example 16.5b: What ratio of NaCHO 2 : HCHO 2 would be required to make a buffer with pH 4.25? 66 Formic acid, HCHO 2, pK a = 3.74 To make a buffer with pH 4.25, you would use 3.24 times as much NaCHO 2 as HCHO 2

67. What ratio of NaC 7 H 5 O 2 : HC 7 H 5 O 2 would be required to make a buffer with pH 3.75? Benzoic acid, HC 7 H 5 O 2, pK a = 4.19 to make a buffer with pH 3.75, you would use 0.363 times as much NaC 7 H 5 O 2 as HC 7 H 5 O 2 67

68. Titration In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete when the neutralization reaction is complete, the endpoint of the titration is reached An indicator (a chemical that changes color when the pH changes) may be added to determine the endpoint When the moles of H 3 O + = moles of OH −, the titration has reached its equivalence point 68

69. Titration 69

70. The Titration Curve pH vs. amount of titrant added is plotted The inflection point of the curve is the equivalence point of the titration Prior to the equivalence point, the solution in the flask is in excess, so the pH is closer to its pH The pH of the equivalence point depends on the pH of the salt solution equivalence point of neutral salt, pH = 7 equivalence point of acidic salt, pH < 7 equivalence point of basic salt, pH > 7 Beyond the equivalence point, the solution from the burette is in excess, so the pH approaches its pH 70

71. Titration Curve: Unknown Strong Base Added to Known Strong Acid 71

72. Before Equivalence (excess acid) After Equivalence (excess base) Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH Equivalence Point equal moles of HCl and NaOH pH = 7.00  Because the solutions are equal concentration, and 1:1 stoichiometry, the equivalence point is at equal volumes 72

73. 5.0 x 10 −4 mole NaOH added Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 mol Before equivalence point added 5.0 mL NaOH HClNaOHNaCl mols before0.0025000 mols change mols end molarity, new HClNaOHNaCl mols before0.0025000 mols change −.00050+.00050 mols end molarity, new HClNaOHNaCl mols before0.0025000 mols change −.00050+.00050 mols end 0.0020000.00050 molarity, new HClNaOHNaCl mols before0.0025000 mols change −.00050+.00050 mols end 0.0020000.00050 molarity, new 0.066700.017

74. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (aq) To reach equivalence, the added moles NaOH = initial moles of HCl = 2.50 x 10 −3 moles At equivalence, we have 0.00 mol HCl and 0.00 mol NaOH left over Because the NaCl is a neutral salt, the pH at equivalence = 7.00 74

75. 0.00250 mole NaOH added Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 At equivalence point added 25.0 mL NaOH 75 HClNaClNaOH mols before0.0025000 mols change mols end molarity, new HClNaClNaOH mols before0.0025000 mols change −.00250+.00250+.0025 mols end molarity, new HClNaClNaOH mols before0.0025000 mols change −.00250+.00250+.0025 mols end 00.002500 molarity, new HClNaClNaOH mols before0.0025000 mols change −.00250+.00250+.0025 mols end 00.002500 molarity, new 00.0500

76. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 After equivalence point added 30.0 mL NaOH 0.0030 mole NaOH added HClNaClNaOH mols before0.0025000 mols change mols end molarity, new HClNaClNaOH mols before0.0025000 mols change −.00250+.00250+.00300 mols end 00.002500.00050 molarity, new HClNaClNaOH mols before0.0025000 mols change −.00250+.00250+.00300 mols end 00.002500.00050 molarity, new 00.0450.0091

77. added 5.0 mL NaOH 0.00200 mol HCl pH = 1.18 added 10.0 mL NaOH 0.00150 mol HCl pH = 1.37 added 25.0 mL NaOH equivalence point pH = 7.00 added 30.0 mL NaOH 0.00050 mol NaOH pH = 11.96 added 40.0 mL NaOH 0.00150 mol NaOH pH = 12.36 added 50.0 mL NaOH 0.00250 mol NaOH pH = 12.52 added 35.0 mL NaOH 0.00100 mol NaOH pH = 12.22 added 15.0 mL NaOH 0.00100 mol HCl pH = 1.60 added 20.0 mL NaOH 0.00050 mol HCl pH = 1.95 Adding 0.100 M NaOH to 0.100 M HCl 77 25.0 mL 0.100 M HCl 0.00250 mol HCl pH = 1.00

78. HNO 3 NaNO 3 NaOH mols before0.012500 mols change −.0015+.0015 mols end 0.0110.00150 molarity, new Calculate the pH of the solution that results when 10.0 mL of 0.15M NaOH is added to 50.0 mL of 0.25M HNO 3 HNO 3 (aq) + NaOH(aq)  NaNO 3 (aq) + H 2 O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/L=1.25 x 10 −2 Before equivalence point added 10.0 mL NaOH 78 HNO 3 NaNO 3 NaOH mols before0.012500 mols change −.0015+.0015 mols end 0.0110.00150 molarity, new 0.0180.0250 = 0.0015 mole NaOH

79. Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M HNO 3 to reach equivalence HNO 3 (aq) + NaOH(aq)  NaNO 3 (aq) + H 2 O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/L=1.25 x 10 −2 At equivalence point: moles of NaOH = 1.25 x 10 −2 79

80. Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO 3 HNO 3 (aq) + NaOH(aq)  NaNO 3 (aq) + H 2 O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/L=1.25 x 10 −2 After equivalence point added 100.0 mL NaOH = 0.015 moles NaOH HNO 3 NaNO 3 NaOH mols before0.012500 mols change −0.0125+0.0125+0.015 mols end 00.01250.0025 molarity, new HNO 3 NaNO 3 NaOH mols before0.012500 mols change −0.0125+0.0125+0.015 mols end 00.01250.0025 molarity, new 00.08330.017

81. Titration of a Strong Base with a Strong Acid If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one where the base in the burette and the acid is in the flask 81

82. Titration of a Weak Acid with a Strong Base Titrating a weak acid with a strong base results in differences in the titration curve at the equivalence point and excess acid region The initial pH is determined using the K a of the weak acid The pH in the excess acid region is determined as you would determine the pH of a buffer The pH at the equivalence point is determined using the K b of the conjugate base of the weak acid The pH after equivalence is dominated by the excess strong base the basicity from the conjugate base anion is negligible 82

83. Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH HCHO 2(aq) + NaOH (aq)  NaCHO 2(aq) + H 2 O (aq) Initial pH [HCHO 2 ][CHO 2 − ][H 3 O + ] initial 0.1000.000≈ 0 change −x−x+x+x+x+x equilibrium 0.100 − xxx K a = 1.8 x 10 −4 83

84. Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 Before equivalence added 5.0 mL NaOH HAA−A− OH − mols before0.0025000 mols added ––.00050 mols after 0.00200.00050≈ 0 84

85. HAA−A− OH − mols before0.0025000 mols added ––0.00250 mols after 00.00250≈ 0 [HCHO 2 ][CHO 2 − ][OH − ] initial 00.0500≈ 0 change +x+x−x−x+x+x equilibrium x5.00E-2-xx Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH added 25.0 mL NaOH CHO 2 − (aq) + H 2 O (l)  HCHO 2(aq) + OH − (aq) K b = 5.6 x 10 −11 [OH − ] = 1.7 x 10 −6 M HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 At equivalence

86. HAA−A− NaOH mols before0.0025000.0030 mols change −.00250+.00250−.00250 mols end 00.002500.00050 molarity, new Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH added 30.0 mL NaOH =0.00330 mole NaOH added HCHO 2(aq) + NaOH (aq)  NaCHO 2(aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 After equivalence HAA−A− NaOH mols before0.0025000.0030 mols change mols end molarity, new HAA−A− NaOH mols before0.0025000.0030 mols change −.00250+.00250−.00250 mols end 00.002500.00050 molarity, new 00.0450.0091

87. added 35.0 mL NaOH 0.00100 mol NaOH xs pH = 12.22 initial HCHO 2 solution 0.00250 mol HCHO 2 pH = 2.37 added 5.0 mL NaOH 0.00200 mol HCHO 2 pH = 3.14 added 10.0 mL NaOH 0.00150 mol HCHO 2 pH = 3.56 added 25.0 mL NaOH equivalence point 0.00250 mol CHO 2 − [CHO 2 − ] init = 0.0500 M [OH − ] eq = 1.7 x 10 −6 pH = 8.23 added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 20.0 mL NaOH 0.00050 mol HCHO 2 pH = 4.34 added 15.0 mL NaOH 0.00100 mol HCHO 2 pH = 3.92 added 12.5 mL NaOH 0.00125 mol HCHO 2 pH = 3.74 = pK a half-neutralization Adding NaOH to HCHO 2 added 40.0 mL NaOH 0.00150 mol NaOH xs pH = 12.36 added 50.0 mL NaOH 0.00250 mol NaOH xs pH = 12.52 87-stop

88. Titrating Weak Acid with a Strong Base The initial pH is that of the weak acid solution calculate pH by weak acid equilibrium problem Prior to equivalence point, the solution becomes a buffer calculate mol HA init and mol A − init using reaction stoichiometry calculate pH with Henderson-Hasselbalch equation using mol HA init and mol A − init Half-neutralization point is where pH = pK a 88

89. Titrating Weak Acid with a Strong Base At the equivalence point: mol A ¯ = original starting moles HA before equilibrium is established solution only has conjugate base anion in it o calculate the volume of added base o [A − ] init = mol A − /total liters  calculate like a weak base equilibrium problem Beyond equivalence point, the OH is in excess [OH − ] = xs moles MOH/total liters Kw = [H 3 O + ][OH − ]=1 x 10 −14 89

90. Example 16.7a: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the volume of KOH at the equivalence point. 90 write an equation for the reaction for B with HA use stoichiometry to determine the volume of added B HNO 2 + KOH  NO 2  + H 2 O

91. Example 16.7b: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH. 91 write an equation for the reaction for B with HA determine the moles of HA before & moles of added B make a stoichiometry table and determine the moles of HA in excess and moles A  made HNO 2 + KOH  NO 2  + H 2 O HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00100 mols after ≈ 0≈ 0 0.00300 0.00100

92. Example 16.7b: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH. 92 write an equation for the reaction of HA with H 2 O determine K a and pK a for HA use Henderson- Hasselbalch equation to determine pH HNO 2 + H 2 O  NO 2  + H 3 O + HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00100 mols after 0.003000.00100 ≈ 0≈ 0 Table 15.5 K a = 4.6 x 10 −4

93. Example 16.7b: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. 93 write an equation for the reaction for B with HA determine the moles of HA before & moles of added B make a stoichiometry table and determine the moles of HA in excess and moles A  made HNO 2 + KOH  NO 2  + H 2 O HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00200 mols after ≈ 0≈ 0 0.00200 at half-equivalence, moles KOH = ½ mole HNO 2

94. Example 16.7b: A 40.0mL sample of 0.100M HNO 2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. 94 write an equation for the reaction of HA with H 2 O determine K a and pK a for HA use the Henderson- Hasselbalch equation to determine the pH HNO 2 + H 2 O  NO 2  + H 3 O + HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00200 mols after 0.00200 ≈ 0≈ 0 Table 15.5 K a = 4.6 x 10 -4

95. Titration Curve of a Weak Base with a Strong Acid 95

96. Titration of 25.0 mL of 0.10 M NH 3 with 0.10 M HCl. Calculate the initial pH of the NH 3 (aq) 96 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial: NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH − (aq) [HCl][NH 4 + ][NH 3 ] initial 000.10 change +x+x+x+x−x−x equilibrium xx0.10−x pK b = 4.75 K b = 10 −4.75 = 1.8 x 10 −5

97. Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 Before equivalence: after adding 5.0 mL of HCl NH 3 NH 4 ClHCl mols before0.0025000.00050 mols change −.00050 mols end 0.002000.000500 molarity, new 0.06670.0170 97 NH 4 + (aq) + H 2 O (l)  NH 4 + (aq) + H 2 O (l) pK b = 4.75 pK a = 14.00 − 4.75 = 9.25

98. Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. 98 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 At equivalence mol NH 3 = mol HCl = 2.50 x 10 −3 added 25.0 mL HCl NH 3 NH 4 ClHCl mols before0.002500 mols change −.00250+.00250−.00250 mols end 00.002500 molarity, new 00.0500

99. Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. 99 NH 3(aq) +HCl (aq)  NH 4 Cl (aq) at equivalence [NH 4 Cl] = 0.050M [NH 3 ][NH 4 + ][H 3 O + ] initial 00.050≈ 0 change +x+x−x−x+x+x equilibrium x0.050−xx NH 4 + (aq) + H 2 O (l)  NH 3(aq) + H 3 O + (aq)

100. Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 30.0 mL of HCl. 100 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 After equivalence: after adding 30.0 mL HCl NH 3 NH 4 ClHCl mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new 00.0450.0091 when you mix a strong acid, HCl, with a weak acid, NH 4 +, you only need to consider the strong acid

101. Titration of a Polyprotic Acid If K a1 >> K a2, there will be two equivalence points in the titration the closer the K a ’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of 0.100 M H 2 SO 3 with 0.100 M NaOH 101

102. Monitoring pH During a Titration Monitoring pH during the titration is done by measuring solution conductivity due to the [H 3 O + ] using a probe that specifically measures just H 3 O + The endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve If the amount of titrant added to reach the endpoint is needed, the titration is monitored using an indicator 102

103. Monitoring pH During a Titration 103

104. Indicators many dyes change color based on the solution’s pH The dyes are weak acids an equilibrium is established with H 2 O & H 3 O + in the solution HInd (aq) + H 2 O (l)  Ind  (aq) + H 3 O + (aq) The color of the solution depends on the relative concentrations of Ind  : HInd when the ratio ≈ 1, the color will be mix of the two when the ratio > 10, the color will be that of Ind  when the ratio < 0.1, the color will be that of HInd 104

105. Phenolphthalein 105

106. Methyl Red 106

107. Monitoring a Titration with an Indicator the titration curve usually shows a very large change in pH for very small additions of titrant near the equivalence point indicators can therefore be used to determine the endpoint of the titration if the indicator changes color within the same range as the equivalence point pK a of HInd ≈ pH at equivalence point 107

108. Acid-Base Indicators 108

109. Solubility Equilibria All ionic compounds dissolve in water to some degree many compounds have such low solubility in water that we classify them as insoluble the concept of equilibrium can be applied to salts dissolving the equilibrium constant for the process can measure relative solubility in water 109

110. Solubility Product The equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp For an ionic solid M n X m, the dissociation reaction is: M n X m (s)  nM m+ (aq) + mX n− (aq) K sp = [M m+ ] n [X n− ] m The dissociation reaction for PbCl 2 is: PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) K sp = [Pb 2+ ][Cl − ] 2 110

111. 111

112. Molar Solubility Solubility is the mass of solute dissolved in a given amount of solution at a set temperature molar solubility is the moles of solute dissolved in a liter of solution the molarity of the dissolved solute in a saturated solution for M n X m (s)  nM m+ (aq) + mX n− (aq) 112 S =

113. Example 16.8: Calculate the molar solubility of PbCl 2 in pure water at 25  C 113 write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Pb 2+ ][Cl − ] Initial00 Change+S+2S EquilibriumS2S PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) K sp = [Pb 2+ ][Cl − ] 2

114. Example 16.8: Calculate the molar solubility of PbCl 2 in pure water at 25  C 114 substitute into the K sp expression find the value of K sp from Table 16.2, plug into the equation, and solve for S [Pb 2+ ][Cl − ] Initial00 Change+S+2S EquilibriumS2S K sp = [Pb 2+ ][Cl − ] 2 K sp = (S)(2S) 2

115. Practice – Determine the K sp of PbBr 2 if its molar solubility in water at 25  C is 1.05 x 10 −2 M 115 write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Pb 2+ ][Br − ] initial00 change+(1.05 x 10 −2 )+2(1.05 x 10 −2 ) equilibrium(1.05 x 10 −2 )(2.10 x 10 −2 ) PbBr 2 (s)  Pb 2+ (aq) + 2 Br − (aq) K sp = [Pb 2+ ][Br − ] 2

116. Practice – Determine the K sp of PbBr 2 if its molar solubility in water at 25  C is 1.05 x 10 −2 M 116 substitute into the K sp expression plug into the equation and solve K sp = [Pb 2+ ][Br − ] 2 K sp = (1.05 x 10 −2 )(2.10 x 10 −2 ) 2 [Pb 2+ ][Br − ] initial00 change+(1.05 x 10 −2 )+2(1.05 x 10 −2 ) equilibrium(1.05 x 10 −2 )(2.10 x 10 −2 )

117. K sp and Relative Solubility Molar solubility is related to K sp but you can only compare solubilities of compounds using K sp if the compounds have the same dissociation stoichiometry 117

118. The Effect of Common Ion on Solubility Addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt For example, addition of NaCl to the solubility equilibrium of solid PbCl 2 decreases the solubility of PbCl 2 PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) the addition of Cl − shifts the equilibrium to the left 118

119. Example 16.10: Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25  C 119 write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Ca 2+ ][F − ] initial00.100 change+S+2S equilibriumS0.100 + 2S CaF 2 (s)  Ca 2+ (aq) + 2 F − (aq) K sp = [Ca 2+ ][F − ] 2

120. Example 16.10: Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25  C 120 substitute into the K sp expression, assume S is small find the value of K sp from Table 16.2, plug into the equation, and solve for S [Ca 2+ ][F − ] initial00.100 change+S+2S equilibriumS0.100 + 2S K sp = [Ca 2+ ][F − ] 2 K sp = (S)(0.100 + 2S) 2 K sp = (S)(0.100) 2

121. Determine the concentration of Ag + ions in seawater that has a [Cl − ] of 0.55 M K sp of AgCl = 1.77 x 10 −10 write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Ag + ][Cl − ] initial00.55 change+S+S+S+S equilibriumS0.55 + S AgCl(s)  Ag + (aq) + Cl − (aq) K sp = [Ag + ][Cl − ] 121

122. Practice – Determine the concentration of Ag + ions in seawater that has a [Cl − ] of 0.55 M 122 substitute into the K sp expression, assume S is small find the value of K sp from Table 16.2, plug into the equation, and solve for S [Ag + ][Cl − ] Initial00.55 Change+S+S+S+S EquilibriumS0.55 + S K sp = [Ag + ][Cl − ] K sp = (S)(0.55 + S) K sp = (S)(0.55)

123. The Effect of pH on Solubility For insoluble ionic hydroxides, solubility increases as pH decreases M(OH) n (s)  M n+ (aq) + nOH − (aq) For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M 2 (CO 3 ) n (s)  2 M n+ (aq) + nCO 3 2− (aq) H 3 O + (aq) + CO 3 2− (aq)  HCO 3 − (aq) + H 2 O(l) 123

124. Precipitation Precipitation occurs when the ion concentration exceeds the solubility of the ionic compound By comparing the reaction quotient, Q, to K sp, we can determine if precipitation will occur Q = K sp, solution is saturated, no precipitation Q < K sp, solution is unsaturated, no precipitation Q > K sp, solution exceeds saturation point, salt precipitates Some solutions with Q > K sp will not precipitate unless disturbed – these are called supersaturated solutions 124

125. precipitation occurs if Q > K sp a supersaturated solution will precipitate if a seed crystal is added 125-stop

126. Selective Precipitation Solutions containing 2 or more cations, can be separated by addition of a reagent that selectively forms an insoluble salt with one of the ions, but not the others A successful reagent can precipitate with more than one cation, as long as their K sp values are significantly different 126

127. Example 16.12: Will a precipitate form when we mix Pb(NO 3 ) 2(aq) with NaBr (aq) if the concentrations after mixing are 0.0150 M and 0.0350 M respectively? write the equation for the reaction determine the ion concentrations of the original salts determine the K sp for any “insoluble” product write the dissociation reaction for the insoluble product calculate Q, using the ion concentrations compare Q to K sp. If Q > K sp, precipitation Pb(NO 3 ) 2(aq) + 2 NaBr (aq) → PbBr 2(s) + 2 NaNO 3(aq) K sp of PbBr 2 = 4.67 x 10 –6 PbBr 2(s)  Pb 2+ (aq) + 2 Br − (aq) Pb(NO 3 ) 2 = 0.0150 M  Pb 2+ = 0.0150 M, NO 3 − = 2(0.0150 M) NaBr = 0.0350 M  Na + = 0.0350 M, Br − = 0.0350 M Q = 1.84x10 -7 < K sp, so no ppt occurs 127

128. Will a precipitate form when we mix Ca(NO 3 ) 2(aq) with NaOH (aq) if the concentrations after mixing are both 0.0175 M? K sp of Ca(OH) 2 = 4.68 x 10 −6 write the equation for the reaction determine the ion concentrations of the original salts determine the K sp for any “insoluble” product write the dissociation reaction for the insoluble product calculate Q, using the ion concentrations compare Q to K sp. If Q > K sp, precipitation Ca(NO 3 ) 2(aq) + 2 NaOH (aq) → Ca(OH) 2(s) + 2 NaNO 3(aq) K sp of Ca(OH) 2 = 4.68 x 10 –6 Ca(OH) 2(s)  Ca 2+ (aq) + 2 OH − (aq) Ca(NO 3 ) 2 = 0.0175 M  Ca 2+ = 0.0175 M, NO 3 − = 2(0.0175 M) NaOH = 0.0175 M  Na + = 0.0175 M, OH − = 0.0175 M Q = 5.36x10 -6 > K sp, so ppt occurs 128

129. 129 Example 16.13: What is the minimum [OH − ] necessary to just begin to precipitate Mg 2+ (with [0.059]) from seawater? ppt starts when Q = K sp Mg(OH) 2(s)  Mg 2+ (aq) + 2 OH − (aq)

130. 130 What is the minimum concentration of Ca(NO 3 ) 2(aq) that will precipitate Ca(OH) 2 from 0.0175 M NaOH (aq) ? K sp of Ca(OH) 2 = 4.68 x 10 −6 ppt starts to occur when Q = K sp [Ca(NO 3 ) 2 ] = [Ca 2+ ] = 0.0153 M Ca(NO 3 ) 2(aq) + 2 NaOH (aq) → Ca(OH) 2(s) + 2 NaNO 3(aq) Ca(OH) 2(s)  Ca 2+ (aq) + 2 OH − (aq)

131. 131 Example 16.14: What is the [Mg 2+ ] when Ca 2+ (Ca 2+ = [0.011]) just begins to precipitate from seawater? ppt starts when Q = K sp Ca(OH) 2(s)  Ca 2+ (aq) + 2 OH − (aq)

132. 132 precipitating Mg 2+ begins when [OH − ] = 1.9 x 10 −6 M precipitating Ca 2+ begins when [OH − ] = 2.06 x 10 −2 M when Ca 2+ just begins to precipitate out, the [Mg 2+ ] has dropped from 0.059 M to 4.8 x 10 −10 M Mg(OH) 2(s)  Mg 2+ (aq) + 2 OH − (aq) Example 16.14: What is the [Mg 2+ ] when Ca 2+ (Ca 2+ = [0.011]) just begins to precipitate from seawater?

133. A solution is made by mixing Pb(NO 3 ) 2(aq) with AgNO 3(aq) so both compounds have a concentration of 0.0010 M. NaCl (s) is added to precipitate out both AgCl (s) and PbCl 2(aq). What is the [Ag + ] concentration when the Pb 2+ just begins to precipitate? 133

134. 134 What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate? AgCl ppt starts when Q = K sp AgNO 3(aq) + NaCl (aq) → AgCl (s) + NaNO 3(aq) AgCl (s)  Ag + (aq) + Cl − (aq)

135. 135 What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate? PbCl 2 ppt occurs when Q = K sp Pb(NO 3 ) 2(aq) + 2 NaCl (aq) → PbCl 2(s) + 2 NaNO 3(aq) PbCl 2(s)  Pb 2+ (aq) + 2 Cl − (aq)

136. 136 What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate precipitating Ag + begins when [Cl − ] = 1.77 x 10 −7 M precipitating Pb 2+ begins when [Cl − ] = 1.08 x 10 −1 M when Pb 2+ just begins to precipitate out, the [Ag + ] has dropped from 0.0010 M to 1.6 x 10 −9 M AgCl (s)  Ag + (aq) + Cl − (aq)

137. 137 Qualitative Analysis Analytical schemes that utilize selective precipitation to identify ions present is called a qualitative analysis scheme A sample containing several ions is subjected to the addition of several precipitating agents Addition of each reagent causes one of the ions present to precipitate out

138. 138 Qualitative Analysis

139. 139

140. 140 Group 1 cations: Ag +, Pb 2+ and Hg 2 2+ Group 1 cations form insoluble compounds with Cl − in water as long as the concentration is large enough PbCl 2 may be borderline  molar solubility of PbCl 2 = 1.43 x 10 −2 M Group 1 cations are precipitated by the addition of HCl

141. 141 Group 2 cations: Cd 2+, Cu 2+, Bi 3+, Sn 4+, As 3+, Pb 2+, Sb 3+, Hg 2+ Group 2 cations form insoluble compounds with HS − and S 2− in water at low pH Group 2 cations are precipitated by the addition of H 2 S in HCl

142. 142 Group 3 Cations: Fe 2+, Co 2+, Zn 2+, Mn 2+, Ni 2+ Group 3 cations are precipitated as sulfides Group 3 cations form insoluble compounds with S 2− in water at high pH Group 3 cations Cr 3+, Fe 3+, and Al 3+ are precipitated as hydroxides by the addition of H 2 S in NaOH Group 3 Cations: Cr 3+, Fe 3+, Al 3+

143. 143 Group 4 cations: Mg 2+, Ca 2+, Ba 2+ Group 4 cations form insoluble compounds with PO 4 3− in water at high pH Precipitated by the addition of (NH 4 ) 2 HPO 4

144. 144 Group 5cations: Na +, K +, NH 4 + Are all soluble in water Identified by the color of their flame

145. Complex Ion Formation Transition metals tend to be good Lewis acids often bond to one or more H 2 O to form a hydrated ion H 2 O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag + (aq) + 2 H 2 O(l)  Ag(H 2 O) 2 + (aq) Ions that form by combining a cation with several anions or neutral molecules are called complex ions e.g., Ag(H 2 O) 2 + The attached ions or molecules are called ligands 145

146. Complex Ion Equilibria If a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand Ag(H 2 O) 2 + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) + 2 H 2 O (l) generally H 2 O is not included, because its complex ion is always present in aqueous solution Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) 146

147. Formation Constant The reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) The equilibrium constant for the formation reaction is called the formation constant, K f 147

148. Formation Constants 148

149. Example 16.15: 200.0 mL of 1.5 x 10 −3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Write the formation reaction and K f expression. Look up K f value determine the concentration of ions in the diluted solutions Cu 2+ (aq) + 4 NH 3 (aq)  Cu(NH 3 ) 4 2+ (aq) 149

150. Example 16.15: 200.0 mL of 1.5 x 10 −3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Create an ICE table. Because K f is large, assume all the Cu 2+ is converted into complex ion, then the system returns to equilibrium. [Cu 2+ ][NH 3 ][Cu(NH 3 ) 2 2+ ] initial6.7E-40.110 change≈−6.7E-4≈−4(6.7E-4)≈+6.7E-4 equilibriumx0.116.7E-4 Cu 2+ (aq) + 4 NH 3 (aq)  Cu(NH 3 ) 4 2+ (aq) 150

151. Example 16.15: 200.0 mL of 1.5 x 10 -3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Cu 2+ (aq) + 4 NH 3 (aq)  Cu(NH 3 ) 4 2+ (aq) substitute in and solve for x confirm the “x is small” approxima- tion [Cu 2+ ][NH 3 ][Cu(NH 3 ) 2 2+ ] initial6.7E-40.110 change≈−6.7E-4≈−4(6.7E-4)≈+6.7E-4 equilibriumx0.116.7E-4 2.7 x 10 −13 << 6.7 x 10 −4, so the approximation is valid 151

152. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) 152

153. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) Write the formation reaction and K f expression. Look up K f value determine the concentration of ions in the diluted solutions Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq) 153

154. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) Create an ICE table. Because K f is large, assume all the lim. rgt. is converted into complex ion, then the system returns to equilibrium. [Hg 2+ ][I − ][HgI 4 2− ] initial3.75E-46.25E-40 change ≈¼ ( −6.25E-4)≈−(6.25E-4)≈¼ ( +6.25E-4) equilibrium2.19E-4x1.56E-4 Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq) I − is the limiting reagent 154

155. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) substitute in and solve for x confirm the “x is small” approximation 2 x 10 −8 << 1.6 x 10 −4, so the approximation is valid Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq) [HgI 4 2− ] = 1.6 x 10 −4 [Hg 2+ ][I − ][HgI 4 2− ] initial3.75E-46.25E-40 change ≈¼ ( −6.25E-4)≈−(6.25E-4)≈¼ ( +6.25E-4) equilibrium2.19E-4x1.56E-4 155

156. 156 The Effect of Complex Ion Formation on Solubility The solubility of an ionic compound containing a metal cation that forms a complex ion, increases in the presence of aqueous ligands AgCl (s)  Ag + (aq) + Cl − (aq) K sp = 1.77 x 10 −10 Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7 Adding NH 3 to a solution in equilibrium with AgCl (s) increases the solubility of Ag +

157. 157

158. 158 Solubility of Amphoteric Metal Hydroxides Many metal hydroxides are insoluble Metal hydroxides are more soluble in acidic solution  equilibrium shifts to the right by removing OH − Some metal hydroxides also become more soluble in basic solution  acting as a Lewis base forming a complex ion Substances that behave as both an acid and base are said to be amphoteric Some cations that form amphoteric hydroxides include Al 3+, Cr 3+, Zn 2+, Pb 2+, and Sb 2+

159. 159 Al 3+ Aqueous Al 3+ is hydrated, forming acidic solutions Al(H 2 O) 6 3+ (aq) + H 2 O (l)  Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) Addition of OH − drives the equilibrium to the right and continues to remove H from the molecules Al(H 2 O) 5 (OH) 2+ (aq) + OH − (aq)  Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O (l) Al(H 2 O) 4 (OH) 2 + (aq) + OH − (aq)  Al(H 2 O) 3 (OH) 3(s) + H 2 O (l)

160. 160

161. 161

162. The Danger of Antifreeze Each year, thousands of pets and wildlife species die from consuming antifreeze Most brands of antifreeze contain ethylene glycol (aka 1,2-ethandiol) sweet taste initial effect = drunkenness Metabolized in the liver to glycolic acid (  - hydroxyethanoic acid) HOCH 2 COOH 162

163. Why Is Glycolic Acid Toxic? If present in high enough concentration in the bloodstream, glycolic acid overwhelms the buffering ability of the HCO 3 − in the blood, causing the blood pH to lower When the blood pH is low, its ability to carry O 2 is compromised (acidosis) HbH + (aq) + O 2 (g)  HbO 2 (aq) + H + (aq) One treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol 163