1. HOMEWORK Do in this order 51,52,53,55,61,65,67,23,25,27,29,31,33, 37a&e, 39,41,43a,45,47, and 49a&b

2. Acid-Base Equilibria

3. REMEMBER When working problems in this chapter, it will help if you always……..

5. Strong Acid/Strong Base Titration Excess Base Excess Acid moles of acid =moles of base

6. WHAT IS THE DIFFERENCE BETWEEN END POINT AND EQUIVALENCE POINT?

8. Strong Acid/Strong Base Titration Determine the pH if 50 ml of 0.25 M HCl is mixed with 30 ml of 0.25 M NaOH

9. Strong Acid/Strong Base Titration Determine the pH when 30.0 ml of 0.500 M HCl is added to 60 ml of 0.250 M NaOH.

10. Strong Acid/Strong Base Titration Determine the pH when 25.0 ml of 0.400 M HCl is added to 85.0 ml of 0.300 M NaOH.

11. Strong Acid/Strong Base Titration a.Determine the pH if enough 0.700 M NaOH is added to 30.00 ml of.500 M HCl to reach the equivalence point. b.Determine the volume of NaOH required to reach the equivalence point.

12. Weak Acid/Strong Base Titration All strong base is converted to CB Use Kb = X 2 moles base – x total volume

13. Weak Acid/Strong Base Titration Buffer Region Use [H + ] = Ka [ moles acid ] [ moles base] You are in in the buffer region when the moles of acid are greater than base. Excess base region Moles of base is greater than acid Determine the moles of strong base left over. Divide by total volume gives [OH - ] concentration. All strong base is converted to CB at the equivalence point Use Kb = X 2 moles base – x total volume

14. HENDERSON-HASSELBALCH EQUATION (Buffers) Derivative is H + = Ka [HB] [B - ] If you have a buffer, use the above equation. Makes the math easier.

15. HENDERSON-HASSELBALCH EQUATION (Buffers) Derivative is pKa = -log 10 Ka pH = pka when moles of acid = moles of base H + = Ka [HB] [B - ]

17. Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. 1. Before the addition of any base (initial pH). 2 After the addition of 8.00 mL of 0.500 M NaOH (buffer region). 3. After the addition of 10.00 mL of 0.500 M NaOH (half-neutralization point). 4. After the addition of 20.00 mL of 0.500 M NaOH (equivalence point). 5. After the addition of 21.00 mL of 0.500 M NaOH (beyond the equivalence point). Neutralization Reaction Weak Acid - Strong Base 5 Positions on the Titration Curve

18. Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. 1. Before the addition of any base. Neutralization Reaction Weak Acid - Strong Base 5 Positions on the Titration Curve

19. Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH 2. after the addition of 8.00 mL of 0.500 M NaOH Neutralization Reaction Weak Acid - Strong Base 5 Positions on the Titration Curve

20. Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH 3. after the addition of 10.00 mL of 0.500 M NaOH Neutralization Reaction Weak Acid - Strong Base 5 Positions on the Titration Curve

21. Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH 4. after the addition of 20.00 mL of 0.500 M NaOH Neutralization Reaction Weak Acid - Strong Base 5 Positions on the Titration Curve

22. Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH 5. After the addition of 21.00 mL of 0.500 M NaOH Neutralization Reaction Weak Acid - Strong Base 5 Positions on the Titration Curve

23. Determine the pH of a solution when 25.0 ml of 2.00 M HC 2 H 3 O 2 is titrated halfway to the equivalence point with NaOH Neutralization Reaction Weak Acid - Strong Base 5 Positions on the Titration Curve

24. Determine the pH of a solution when 25.0 ml of 2.00 M HC 2 H 3 O 2 is titrated with 1.50 M NaOH to the equivalence point, Neutralization Reaction Weak Acid - Strong Base 5 Positions on the Titration Curve

26. Buffer Region Excess Acid Region Use [H + ] = Ka [ moles acid ] [ moles base] You are in in the buffer region when the moles of weak is greater than strong All strong acid is converted to CA ate the equivalence point Use Ka = X 2 moles base – x total volume Moles of strong are greater than weak Determine the moles of acid left over then divide by total volume. Gives H + concentration WEAK BASE STRONG ACID TITRATION

27. Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia with 0.500 M HCl 1. Before the addition of any acid (initial pH). 2 After the addition of 8.00 mL of 0.500 M HCl (buffer region). 3. After the addition of 10.00 mL of 0.500 M HCl (half-neutralization point). 4. After the addition of 20.00 mL of 0.500 M HCl (equivalence point). 5. After the addition of 21.00 mL of 0.500 M HCl (beyond the equivalence point). Neutralization Reaction Weak Base - Strong Acid 5 Positions on the Titration Curve

28. Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia 1. Before the addition of any acid. Neutralization Reaction Weak Base - Strong Acid 5 Positions on the Titration Curve

29. Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia 2. After the addition of 8.00 mL of 0.500 M HCl Neutralization Reaction Weak Base - Strong Acid 5 Positions on the Titration Curve

30. Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia 3. After the addition of 10.00 mL of 0.500 M HCl Neutralization Reaction Weak Base - Strong Acid 5 Positions on the Titration Curve

31. 4. After the addition of 20.00 mL of 0.500 M HCl Neutralization Reaction Weak Base - Strong Acid 5 Positions on the Titration Curve Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia

32. 5. After the addition of 21.00 mL of 0.500 M HCl Neutralization Reaction Weak Base - Strong Acid 5 Positions on the Titration Curve Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M ammonia

33. Determine the pH of a solution when 25.0 ml of 2.00 M NH 3 is titrated halfway to the equivalence point with HCl Neutralization Reaction Weak Base - Strong Acid 5 Positions on the Titration Curve

34. Determine the pH of a solution when 25.0 ml of 2.00 M NH 3 is titrated to the equivalence point with 0.125 M HCl. Neutralization Reaction Weak Base - Strong Acid 5 Positions on the Titration Curve

35. Titration of a Polyprotic Weak Acid

36. Titration of a Polyprotic Strong Acid 2 nd equivalence point 1 st equivalence point

37. 37 pH Review pH = - log [H + ] H + is really a proton Range is from 0 - 14 If [H + ] is high, the solution is acidic; pH < 7 If [H + ] is low, the solution is basic or alkaline ; pH > 7

38. 38 The Body and pH Homeostasis of pH is tightly controlled Extracellular fluid = 7.4 Blood = 7.35 – 7.45 8.0 death occurs Acidosis (acidemia) below 7.35 Alkalosis (alkalemia) above 7.45

39. 39

40. 40 Small changes in pH can produce major disturbances Most enzymes function only with narrow pH ranges Acid-base balance can also affect electrolytes (Na +, K +, Cl - ) Can also affect hormones

41. 41 Acidosis Principal effect of acidosis is depression of the CNS through ↓ in synaptic transmission. Generalized weakness Deranged CNS function the greatest threat Severe acidosis causes – Disorientation – coma – death

42. 42 Alkalosis Alkalosis causes over excitability of the central and peripheral nervous systems. Numbness Lightheadedness It can cause : – Nervousness – muscle spasms or tetany – Convulsions – Loss of consciousness – Death

43. 43

44. The function of a buffer is to resist changes in the pH of a solution. Buffers are just a special case of the common ion effect. Buffer Composition Weak Acid+Conj. Base HOAc+OAc - H 2 PO 4 - +HPO 4 2- Weak Base+ Conj. Acid NH 3 + NH 4 + Buffer Solutions

45. ACID USES UP ADDED OH - CONJ. BASE USES UP ADDED H + OAc - + H 2 O HOAc + OH - LeChatelier's Principle LeChatelier's Principle NOTE: THE CONJUGATE ACID/BASE PAIRS ARE ON OPPOSITE SIDES OF THE EQUATION. NOTE: THE CONJUGATE ACID/BASE PAIRS ARE ON OPPOSITE SIDES OF THE EQUATION. Buffer Solutions

46. Preparing a Buffer A buffer can be prepared by 1.Using a weak acid and ½ the moles of a strong base or less 2.Using a weak base and ½ the moles of a strong acid or less 3.A weak acid and the salt of a weak base 4.A weak base and the salt of a weak acid

47. Preparing a Buffer Using a weak acid and ½ the moles of a strong base or less Using a weak acid and ½ the moles of a strong base or less Why does this work? 30.0 ml of 0.200 M acetic acid and 15.0 ml of 30.0 ml of 0.200 M acetic acid and 15.0 ml of 0.200 M sodium hydroxide 0.200 M sodium hydroxide C 2 H 3 O 2 - + OH - H C 2 H 3 O 2 + H 2 O C 2 H 3 O 2 - + OH - H C 2 H 3 O 2 + H 2 O I C E

48. Preparing a Buffer Using a weak base and ½ the moles of a strong base or less Using a weak base and ½ the moles of a strong base or less Why does this work? 30.0 ml of 0.200 M ammonia and 15.0 ml of 0.200 M HCl 30.0 ml of 0.200 M ammonia and 15.0 ml of 0.200 M HCl NH 3 + H 2 O NH 4 + + H 2 O NH 3 + H 2 O NH 4 + + H 2 O I C E

49. Derivative is H + = Ka [HB] [B - ] If you have a buffer, use the above equation. Makes the math easier. You have a buffer if have a conjugate acid base pair or the moles of weak (acid/base) is greater than the strong (acid/base) Buffer Solutions

50. Which of the following mixtures would result in a buffered solution when 1.0 L of each of the two solutions are mixed? a.0.2 M HNO 3 and 0.4 M NaNO 3 b.0.2 M HNO 3 and 0.4 M HF c.0.2 M HNO 3 and 0.4 M NaF d.0.2 M HNO 3 and 0.4 M NaOH

51. H + = Ka [HB] [B - ] Buffer Solutions How does adding water to a buffer system affect the pH?

52. H + = Ka [HB] [B - ] Buffer Solutions How does adding water to a buffer system affect the pH? CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each. The volumes cancel. You can ignore the volume and use H + = Ka HB moles B - moles

53. Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? K a = 1.8 x 10 -5

54. Adding an Acid to a Buffer Problem: What is the pH when 1.00 mL of 1.00 M HCl is added to 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M

55. Adding an Base to a Buffer What is the pH when 1.00 mL of 1.00 M NaOH is added to 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M. What is the pH when 1.00 mL of 1.00 M NaOH is added to 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M.

56. Buffer Solutions How many moles of HCl must be added to 1.0 L of 1.0 M sodium acetate to produce a solution buffered at a.pH= pKa b.pH =4.20

57. Buffer Solutions Consider a solution that contains both ammonia and ammonium chloride. Calculate the ratio of [NH 3 ]/[NH 4 Cl ] at a pH of 5.23.

58. Preparing a Buffer You want to buffer a solution at pH = 4.30. This means [H 3 O + ] = 10 -pH = 5.0 x 10 -5 M It is best to choose an acid such that [H 3 O + ] is about equal to K a (or pH ­ pK a ). —then you get the exact [H 3 O + ] by adjusting the ratio of acid to conjugate base.

59. Preparing a Buffer Solution Buffer prepared from HCO 3 - weak acid CO 3 2- conjugate base HCO 3 - + H 2 O H 3 O + + CO 3 2-

60. Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M POSSIBLE ACIDS K a POSSIBLE ACIDS K a HSO 4 - / SO 4 2- 1.2 x 10 -2 HSO 4 - / SO 4 2- 1.2 x 10 -2 HOAc / OAc - 1.8 x 10 -5 HOAc / OAc - 1.8 x 10 -5 HCN / CN - 4.0 x 10 -10 HCN / CN - 4.0 x 10 -10 Best choice is acetic acid / acetate. Remember pKa = -log 10 Ka pH = pKa when moles acid = moles of base pH = pKa when moles acid = moles of base

61. You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M Preparing a Buffer

62. You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M Preparing a Buffer

63. You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M Solve for [HOAc]/[OAc - ] ratio Solve for [HOAc]/[OAc - ] ratio = 2.78/ 1 = 2.78/ 1 Therefore, if you use 0.100 mol of NaOAc and 0.278 mol of HOAc, you will have pH = 4.30. Preparing a Buffer

64. Buffer capacity  The buffer capacity is the measure of the amount of acid or base that the solution can absorb without significant change in pH  Depend on how many moles of the conjugate acid-base pair are present - for solutions having the same concentration, the greater the volume  greater buffer capacity

65. Buffer capacity  To increase the buffering capacity of a buffer system, increase the moles of acid and moles of base CONCENTRATION of the acid and conjugate base are CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES not important. It is the RATIO OF THE NUMBER OF MOLES of each. of each.

66. Buffer Capacity And Buffer Range There is a limit to the capacity of a buffer solution to neutralize added acid or base, and this limit is reached before all of one of the buffer components has been consumed. In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize. As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal ( pH=pKa).

67. Acid-Base Indicators An acid-base indicator is a weak acid having one color and the conjugate base of the acid having a different color. One of the “colors” may be colorless. HIn + H 2 O º H 3 O + + In - color 1 color 2 Acid-base indicators are often used for applications in which a precise pH reading isn’t necessary. A common indicator used in introductory chemistry laboratories is litmus.

68. 17-3 Acid-Base Indicators Color of some substances depends on the pH. HIn + H 2 O In - + H 3 O + In the acid form the color appears to be the acid color. In the base form the color appears to be the base color. Intermediate color is seen in between these two states. The complete color change occurs over about 2 pH units. Remember….LeChatelier’s Principle

69. Slide 69 of 45 Indicator Colors and Ranges Slide 27 of 45General Chemistry: Chapter 17Prentice-Hall © 2007

70. Titration Curve For Strong Acid - Strong Base

71. Acid-Base Titrations Indicators should change color close to the equivalence point.