1. Chapter 16 Aqueous Acidic Equilibrium

2. History of Acids and Bases In the early days of chemistry chemists were organizing physical and chemical properties of substances. They discovered that many substances could be placed in two different property categories: Substance A 1.Sour taste 2.Reacts with carbonates to make CO 2 3.Reacts with metals to produce H 2 4.Turns blue litmus pink 5.Reacts with B substances to make salt water Substance B 1.Bitter taste 2.Reacts with fats to make soaps 3.Do not react with metals 4.Turns red litmus blue 5.Reacts with A substances make salt and water Arrhenius was the first person to suggest a reason why substances are in A or B due to their ionization in water.

3. The Swedish chemist Svante Arrhenius proposed the first definition of acids and bases. (Substances A and B became known as acids and bases) According to the Arrhenius model: “acids are substances that dissociate in water to produce H + ions and bases are substances that dissociate in water to produce OH - ions” NaOH (aq)  Na + (aq) + OH - (aq) Base HCl (aq)  H + (aq) + Cl - (aq) Acid Arrhenius Acids

4. Unknown to Arrhenius free H + ions do not exist in water. They covalently react with water to produce hydronium ions, H 3 O +. or: H + (aq) + H 2 O (l)  H 3 O + (aq) This new bond is called a coordinate covalent bond since both new bonding electrons come from the same atom Hydronium ion

5. What is the difference between a strong acid and a weak acid? Acid Strength

6. What is the difference between a strong acid and a weak acid? Strong acids ionize 100% and weak ones do not! Acid Strength

7. What is the difference between a strong acid and a weak acid? Strong acids ionize 100% and weak ones do not! A single arrow is used to represent the ionization of a strong acid. Double arrows (Equilibrium) are used to represent weak acids. For example:HCl (g) H + (aq) + Cl - (aq) HF (g) H + (aq) + F - Acid Strength

8. What is the difference between a strong acid and a weak acid? Strong acids ionize 100% and weak ones do not! A single arrow is used to represent the ionization of a strong acid. Double arrows (Equilibrium) are used to represent weak acids. For example:HCl (g) H + (aq) + Cl - (aq) HF (g) H + (aq) + F - According to Arrhenius, is water an acid or base? HOH (l) H + (aq) + OH – (aq) Acid Strength

9. What is the difference between a strong acid and a weak acid? Strong acids ionize 100% and weak ones do not! A single arrow is used to represent the ionization of a strong acid. Double arrows (Equilibrium) are used to represent weak acids. For example:HCl (g) H + (aq) + Cl - (aq) HF (g) H + (aq) + F - According to Arrhenius, is water an acid or base? HOH (l) H + (aq) + OH – (aq) Neither, he called it Neutral (same amount of OH - and H + Acid Strength

10. How can we identify strong acids or bases? Acid Strength

11. How can we identify strong acids or bases? Easy memorize them! Acid Strength

12. How can we identify strong acids or bases? Easy, memorize them! Memorized Strong Acids 1.HClO 4 2.H 2 SO 4 3.HI 4.HBr 5.HCl 6.HNO 3 Memorized Strong Bases Hydroxides of group 1 and 2 metals, excluding Be and Mg Acid Strength

13. Johannes Brønsted and Thomas Lowry revised Arrhenius’s acid-base theory to include this behavior. They defined acids and bases as follows: “An acid is a hydrogen containing species that donates a proton. A base is any substance that accepts a proton” e.g. HCl (aq) + H 2 O (l)  Cl - (aq) + H 3 O + (aq) In the above example what is the Brønsted acid? What is the Brønsted base? Bronsted Lowry Acid Strength

14. HCl (aq) + H 2 O (l)  Cl - ( aq) + H 3 O + (aq) In reality, the reaction of HCl with H 2 O is an equilibrium and occurs in both directions, although in this case the equilibrium lies far to the right. For the reverse reaction Cl - behaves as a Brønsted base and H 3 O + behaves as a Brønsted acid. The Cl - is called the conjugate base of HCl. Brønsted acids and bases always exist as conjugate acid-base pairs. Bronsted Lowry Acids/Bases

15. In pure water (no solute) water molecules behave as both an acid and base!! e.g. H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) This is called the self-ionization (autoionizaion) of water. Although the equilibrium lies far to the left it is very important to take into consideration, especially for living systems. Auto Ionization of Water

16. The self-ionization of water is described by the equation: e.g. H 2 O (l) + H 2 O (l )  H 3 O + (aq) + OH - (aq) The equilibrium constant for this reaction is given by: K w = K[H 2 O] 2 = 10 -14 This equilibrium lies very much to the left i.e. mostly water. For pure water [OH - ] = [H + ] = 1 x 10 -7 M Auto Ionization of Water

17. As [OH - ] and [H + ] are so small the [H 2 O] is not affected by their formation. It is useful to define a new constant K w such that: K w is called the ion product of water. What is the value for the ion product of water? 1.00 g ml 18.0 g moleml 10 -3 L = 55.5M Auto Ionization of Water

18. As [OH - ] and [H + ] are so small the [H 2 O] is not affected by their formation. It is useful to define a new constant K w such that: K w is called the ion product of water. What is the value for the ion product of water? [H + ][OH - ] = 10 -14 1.00 g ml 18.0 g mole ml 10 -3 L = 55.5 M Auto Ionization of Water

19. Autoionization of Water An amphoteric substance is capable of behaving either as an acid or as a base. K w = [H + ][OH - ] = 1.00 x 10 -14

20. Hydronium Ion in Water

21. Acids in Water When a Brønsted-Lowry acid donates a H + ion, it forms its conjugate base. HF + H 2 O H 3 O + + F - Conjugate acid-base pairs differ only from each other only by the presence or absence of a proton. When a Brønsted-Lowry base accepts a proton, it becomes it conjugate acid. abcacb

22. We define an aqueous solution as being neutral when the [H + ] = [OH - ]. We define an aqueous solution as being acidic when [H + ] > [OH - ]. We define an aqueous solution as being basic when [H + ] H + ] < [OH - ]. However, in each case K w = 1 x 10 -14 M 2, since Acidic, Basic, or Neutral [H + ][OH-]=K w

23. We define an aqueous solution as being neutral when the [H + ] = [OH - ]. We define an aqueous solution as being acidic when [H + ] > [OH - ]. We define an aqueous solution as being basic when [H + ] H + ] < [OH - ]. However, in each case K w = 1 x 10 -14 M 2, since [H + ] = 0.0000001 = 10 -7 (how can this be abbreviated further?) Acidic, Basic, or Neutral [H + ][OH-]=K w

24. We define an aqueous solution as being neutral when the [H + ] = [OH - ]. We define an aqueous solution as being acidic when [H + ] > [OH - ]. We define an aqueous solution as being basic when [H + ] H + ] < [OH - ]. However, in each case K w = 1 x 10 -14 M 2, since [H + ] = 0.0000001 = 10 -7 (how can this be abbreviated further?) Acidic, Basic, or Neutral [H + ][OH-]=K w By just describing the power Called the power of H, or pH.

25. We define an aqueous solution as being neutral when the [H + ] = [OH - ]. We define an aqueous solution as being acidic when [H + ] > [OH - ]. We define an aqueous solution as being basic when [H + ] H + ] < [OH - ]. However, in each case K w = 1 x 10 -14 M 2, since [H + ] = 0.0000001 = 10 -7 (how can this be abbreviated further?) Acidic, Basic, or Neutral [H + ][OH-]=K w By just describing the power Called the power of H, or pH. Our math departments tells us that log means power too. pH = 7

26. The mathematical definition of pH using [H + ] for [H 3 O + ] is listed below: pH = -log [H + ], or [H + ]= 1x10 -pH (both are mathematically equivalent) How about the power for the OH -, what should this be called? pH and Logs

27. The mathematical definition of pH using [H + ] for [H 3 O + ] is listed below: pH = -log [H + ], or [H + ]= 1x10 -pH (both are mathematically equivalent) How about the power for the OH -, what should this be called? Would you believe pOH? pH and Logs

28. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2

29. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2 [OH - ] = 10 -12

30. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2 [OH - ] = 10 -12 [H+] > [OH -] acidic

31. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2 [OH - ] = 10 -12 [H + ] = 10 -7 [OH - ] = 10 -7 [H+] > [OH -] acidic

32. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2 [OH - ] = 10 -12 [H + ] = 10 -7 [OH - ] = 10 -7 [H+] = [OH -] neutral [H+] > [OH -] acidic

33. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2 [OH - ] = 10 -12 [H + ] = 10 -7 [OH - ] = 10 -7 [H + ] =10 -12 [OH - ] = 10 -2 [H+] = [OH -] neutral [H+] > [OH -] acidic

34. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2 [OH - ] = 10 -12 [H + ] = 10 -7 [OH - ] = 10 -7 [H + ] =10 -12 [OH - ] = 10 -2 [H + ] < [OH - ] basic [H+] = [OH -] neutral [H+] > [OH -] acidic

35. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2 [OH - ] = 10 -12 [H + ] = 10 -7 [OH - ] = 10 -7 [H + ] =10 -12 [OH - ] = 10 -2 [H + ] =10 -16 [OH - ] = [H + ] < [OH - ] basic [H+] = [OH -] neutral [H+] > [OH -] acidic

36. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2 [OH - ] = 10 -12 [H + ] = 10 -7 [OH - ] = 10 -7 [H + ] =10 -12 [OH - ] = 10 -2 [H + ] =10 -16 [OH - ] = 10 2 [H + ] < [OH - ] basic [H+] = [OH -] neutral [H+] > [OH -] acidic

37. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2 [OH - ] = 10 -12 [H + ] = 10 -7 [OH - ] = 10 -7 [H + ] =10 -12 [OH - ] = 10 -2 [H + ] =10 -1 [OH - ] = 10 2 [H+] < [OH - ] basic [H + ] < [OH - ] basic [H+] = [OH -] neutral [H+] > [OH -] acidic

38. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2 [OH - ] = 10 -12 [H + ] = 10 -7 [OH - ] = 10 -7 [H + ] =10 -12 [OH - ] = 10 -2 [H + ] =10 -16 [OH - ] = 10 2 [H+] < [OH - ] basic [H + ] < [OH - ] basic [H+] = [OH - ] neutral [H + ] > [OH - ] acidic

39. A pH Number line Number lines have been used in history and math classes, so to keep up we use them in chemistry classes. pH = 16 pH = 12 pH = 7 pH = 2 [H + ] = 10 -2 [OH - ] = 10 -12 [H + ] = 10 -7 [OH - ] = 10 -7 [H + ] =10 -12 [OH - ] = 10 -2 [H + ] =10 -16 [OH - ] = 10 2 [H+] < [OH - ] basic [H + ] < [OH - ] basic [H+] = [OH -] neutral [H+] > [OH -] acidic basic

40. Relative Strengths of Acids and Bases The leveling effect of water refers to the observation that strong acids all have the same strength in water and are completely converted into a solution of hydronium ions.

41. The pH Scale pH = -log[H + ] pOH = -log[OH - ] pK w = pH + pOH

42. Comparison of Strong and Weak Acids

43. Weak Acids HNO 2 (aq) + H 2 O(l) H 3 O + (aq) + NO 2 - (aq) Degree of ionization is the ratio of the quantity of a substance that is ionized to the concentration of the substance before ionization. When expressed as a percentage, it is called percent ionization.

44. Practice Calculate and compare [H + ] for a 0.100 M solution of HClO 4 and a 0.100 M solution of HClO (K a = 2.9 x 10 -8 for HClO).

45. Practice Calculate and compare [H + ] for a 0.100 M solution of HClO 4 and a 0.100 M solution of HClO (K a = 2.9 x 10 -8 for HClO). HClO 4 H + + ClO 4 - 0.10000 initial 00.1000.100 final [H + ]=0.100 = 10 -1 pH = 1 HClO H + + ClO - 0.10000 initial 0.100 - x xxequilibrium

46. Continued K a = [H + ][ClO - ] [HClO] = X2X2 0.100 - x

47. Continued K a = [H + ][ClO - ] [HClO] = X2X2 0.100 - x = 2.9 x 10 -8

48. Continued K a = [H + ][ClO - ] [HClO] = X2X2 0.100 - x = 2.9 x 10 -8 Assume X small X 2 = (0.100)(2.9X10 -8 ) X = [H + ] = 5.4 X 10 -5 pH = 4.27

49. Continued K a = [H + ][ClO - ] [HClO] = X2X2 0.100 - x = 2.9 x 10 -8 Assume X small X 2 = (0.100)(2.9X10 -8 ) X = [H + ] = 5.4 X 10 -5 pH = 4.27 % ionization = [H+] final [HClO] final X 100 = 5.4 X 10 -2 %

50. Practice Calculate the degree of ionization for 1.28 M HNO 2 and 0.015 M HNO 2 solutions (K a = 4.0 x 10 -4 ). HNO 2 NO 2 - + H + 0.0150 0 Initial 0.0150 - x X x equilibrium

51. Practice Calculate the degree of ionization for 1.28 M HNO 2 and 0.015 M HNO 2 solutions (K a = 4.0 x 10 -4 ). HNO 2 NO 2 - + H + 0.0150 0 Initial 0.0150 - x X x K a = [NO 2 - ][ H + ] [HNO 2 ]

52. Practice Calculate the degree of ionization for 1.28 M HNO 2 and 0.015 M HNO 2 solutions (K a = 4.0 x 10 -4 ). HNO 2 NO 2 - + H + 0.0150 0 Initial 0.0150 - x X x K a = [NO 2 - ][ H + ] [HNO 2 ] = (X)(X) 0.015-X

53. Practice Calculate the degree of ionization for 1.28 M HNO 2 and 0.015 M HNO 2 solutions (K a = 4.0 x 10 -4 ). HNO 2 NO 2 - + H + 0.0150 0 Initial 0.0150 - x X x K a = [NO 2 - ][ H + ] [HNO 2 ] = (X)(X) 0.015-X = 4.0 x 10 -4 small

54. Practice Calculate the degree of ionization for 1.28 M HNO 2 and 0.015 M HNO 2 solutions (K a = 4.0 x 10 -4 ). HNO 2 NO 2 - + H + 0.0150 0 Initial 0.0150 - x X x K a = [NO 2 - ][ H + ] [HNO 2 ] = (X)(X) 0.015-X = 4.0 x 10 -4 small X 2 = (0.0150)(4.0 x 10 -4 ) = 6.00 X 10 -6 X = 2.45 x 10 -3

55. Practice Calculate the degree of ionization for 1.28 M HNO 2 and 0.015 M HNO 2 solutions (K a = 4.0 x 10 -4 ). HNO 2 NO 2 - + H + 0.0150 0 Initial 0.0150 - x X x K a = [NO 2 - ][ H + ] [HNO 2 ] = (X)(X) 0.015-X = 4.0 x 10 -4 small X 2 = (0.0150)(4.0 x 10 -4 ) = 6.00 X 10 -6 X = 2.45 x 10 -3 % ionization = [H + ] [HNO 2 ] Iinitial X 100

56. Practice Calculate the degree of ionization for 1.28 M HNO 2 and 0.015 M HNO 2 solutions (K a = 4.0 x 10 -4 ). HNO 2 NO 2 - + H + 0.0150 0 Initial 0.0150 - x X x K a = [NO 2 - ][ H + ] [HNO 2 ] = (X)(X) 0.015-X = 4.0 x 10 -4 small X 2 = (0.0150)(4.0 x 10 -4 ) = 6.00 X 10 -6 X = 2.45 x 10 -3 % ionization = [H + ] [HNO 2 ] Iinitial X 100 = 2.45 x 10 -3 0.015 X 100 = 16 %

57. Practice Calculate the pH for a 0.0500 M HCl solution.

58. Practice Calculate the pH of a 1.22 M HNO 2 solution.

59. Strong Bases Strong bases completely dissociate in water producing hydroxide ions. All soluble metal hydroxides are strong bases.

60. Weak Bases Weak bases accept hydrogen ions from water to form hydroxide ions.

61. Weak Base Equilibria NH 3 + HOH NH 4 + + OH -

62. Practice Calculate and contrast [OH - ] in 0.200 M LiOH and 0.200 M NH 3 (K b = 1.76 x 10 -5 ).

63. Calculation of pH for Bases pOH = -log[OH - ] K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 -log(K w ) = -log[OH - ] - log[H 3 O + ] = 14.00 pK w = pOH + pH = 14.00

64. Practice Calculate the pH of a 0.200 M KOH solution.

65. Polyprotic Acids Monoprotic acids have one ionizable hydrogen atom per molecule, whereas polyprotic have two or more. Carbonic Acid H 2 CO 3 (aq) H + (aq) + HCO 3 - (aq) HCO 3 - (aq) H + (aq) + CO 3 2- (aq)

66. Practice Calculate the pH of 0.100 M carbonic acid solution (K a1 = 4.3 x 10 -7 and K a2 = 4.7 x 10 -11 ). H 2 CO 3 H + + HCO 3 - 0.10000 initial xx0.100 - x K a1 = [H + ][HCO 3 ] Equilibrium [H 2 CO 3 ]

67. Practice Calculate the pH of 0.100 M carbonic acid solution (K a1 = 4.3 x 10 -7 and K a2 = 4.7 x 10 -11 ). H 2 CO 3 H + + HCO 3 - 0.10000 initial xx0.100 - x K a1 = [H + ][HCO 3 ] Equilibrium [H 2 CO 3 ] = x2x2 0.100-x small

68. Practice Calculate the pH of 0.100 M carbonic acid solution (K a1 = 4.3 x 10 -7 and K a2 = 4.7 x 10 -11 ). H 2 CO 3 H + + HCO 3 - 0.100 0 0 initial xx0.100 - x K a1 = [H + ][HCO 3 ] Equilibrium [H 2 CO 3 ] = x2x2 0.100-x = 4.3 x 10 -7 small X=2.07x10 -4

69. Practice Calculate the pH of 0.100 M carbonic acid solution (K a1 = 4.3 x 10 -7 and K a2 = 4.7 x 10 -11 ). H 2 CO 3 H + + HCO 3 - 0.100 0 0 initial xx0.100 - x K a1 = [H + ][HCO 3 ] Equilibrium [H 2 CO 3 ] = x2x2 0.100-x = 4.3 x 10 -7 X=2.07x10 -4 HCO 3 - H + + CO 3 - 2.07x10 -4 0 initial 2.07x10 -4 + y y 2.07x10 -4 - y Equilibrium

70. Practice Calculate the pH of 0.100 M carbonic acid solution (K a1 = 4.3 x 10 -7 and K a2 = 4.7 x 10 -11 ). H 2 CO 3 H + + HCO 3 - 0.10000 initial xx0.100 - x K a1 = [H + ][HCO 3 ] Equilibrium [H 2 CO 3 ] = x2x2 0.100-x = 4.3 x 10 -7 X=2.07x10 -4 HCO 3 - H + + CO 3 - 2.07x10 -4 0 initial 2.07x10 -4 + y y 2.07x10 -4 - y K a2 = [H + ][CO 3 2- ] [ HCO 3 - ] = 2.07x10 -4 - y [2.07x10 -4 + y][y] = small 4.7 x 10 -11

71. Summary of Results [H + ] = 2.07 x 10 -4 [HCO 3 - ] = 2.07 x 10 -4 [CO 3 2- ] = 4.7 x10 -11 pH = -log [ 2.07 x 10 -4 ] = 3.684 pOH = 10.32

72. Acid Strength and Structure The greater the ratio of O to H the stronger the acid, since the electronegative nature of the oxygen withdraws elections away from the hydrogen weakening the H-O bond.

73. Structure and Acid Strength

74. Periodic Trends and Acid Strengths

75. Oxy Acid Strength Oxo acid is an acid where the proton is directly attached to the oxygen. Examples: sulfuric, perchloric, nitric Since the oxygen is attached to another atom (Z) then that atom defines the strength The strength depends on the electronegativity of Z, which draws electrons away from the H-O bond The greater the electronegativity, the weaker the bond and the greater the strength of the acid, thus HOCl is stronger than HOBr

76. Oxy Acid Strength Oxygen atoms attached to Z is important, as well. The more the better, thus removing more electrons and weakening the bond, thus HClO 4 should be the strongest acid. Oxygen proton ratio; the greater the stronger 1.oxygen to Proton ratio is most important 2.The larger oxygen to proton ratio, the stronger the acid.

77. Binary Acid Strength Molecular structure and acid strength (Bond polarity and strength) Binary acids In group VIIA (The halogens) 1.Since acids loose protons, then bond strength ought to be a factor in proton loss 2.Generally speaking the stronger the H-X bond the weaker the acid 3.Bond strength is related to bond length 4.The shorter the bond the stronger it is

78. Binary Acid Strength Why is HF a weak acid and all the rest strong?

79. Binary Acid Strength Why is HF a weak acid and all the rest strong? Probably due to the fact that the H 3 O + and F - are held together by hydrogen bonds thus keeping the hydronium ion concentration low

80. Binary Acid Strength Polarity of the HA bond is important within a period For example period number 2 acids Thus HA strength increases as the electronegativity of A increases from left to right on the periodic chart. HCH 3 <HNH 3 <HOH<HF With in a period since A- relatively the same size. For the binary acids we see that within a group bond length is the deciding factor, but with in a period bond polarity is a factor, since size is not changing as much as it does down a group.

81. Acid-Base Properties of Salts

82. K a /K b Relationship HA (aq) + H 2 O (l ) H 3 O + ( aq) + A - (aq) A - (aq) + H 2 O (l) OH - (aq) + HA (aq) +2H 2 O (l) Ka=Ka= Kb= [H 3 O + ][A - ] [HA] [OH - ][HA] [A - ] HA(aq)H 3 O + (aq)+A - (aq)+OH - (aq)+HA (aq) [H 3 O + ][A - ][OH-][HA] [HA] A - (aq) [A - ] K= [OH - ][H 3 O + ]Kw=Kw= K a xKb=Kb= [H 3 O + ][A - ][OH - ][HA] [A - ] [HA] K w =K a xKb=Kb= [H + ][OH - ]

83. Practice Calculate the pH of a 0.575 M of sodium acetate.

84. Practice Calculate the pH of a 0.880 M of ammonium nitrate solution.

85. Common-Ion Effect The common-ion effect is the shift in the position of an equilibrium caused by the addition of an ion taking part in the reaction. An example would be combining a solution of Na 2 CO 3 with a solution of H 2 CO 3. The presence of Na 2 CO 3 would cause stress of the H 2 CO 3 equilibrium so that less would ionize.

86. Henderson-Hasselbalch Equation This equation is used to calculate an approximate pH of a common-ion effect or buffer solution. The pH of this solution does not change when it is diluted with deionized water.

87. pH Buffers Buffer: a solution of a weak acid or base and its conjugate partner (HA and A-). Buffers resist changes in pH. The weak acid reacts with any added OH - HA + OH - H 2 O + A - The weak base reacts with any added H 3 O + A - + H 3 O + HA + H 2 O

88. pH Change in a Buffer as a Function of Buffer Concentration Addition of strong base Addition of strong acid

89. Buffer Capacity Buffer capacity is the quantity of acid or base that can be added to a pH buffer without significantly changing the pH of the buffer. Buffer capacity is a function of:  concentrations of buffer components  pH of the buffer

90. Practice Calculate the pH of a solution containing 0.22 M acetic acid and 0.13 M sodium acetate.

91. Practice Determine the number of moles of sodium acetate that must be added to 250.0 mL of 0.16 M acetic acid in order to prepare a pH 4.70 buffer (K a = 1.76 x 10 -5 ).

92. Titration A method for determining concentrations of unknown solutions Titrant- Substance in burette Analate-Unknown solution Standard solution-A stable precisely know solution End point-When the indicator changes color Equivalence point-No excess reactants present Indicator-Weak organic acid used to indicate end point Inflection point-When slope of titration curve changes sign Titration curve-Graph pH vs. volume titrant added End point vs Equivalence point-hopefully the same

93. Titration Titrating a SA with a SB  Equivalence point should be around pH=7  Choose an indicator that changes around 7, phenolphthalein is typically the indicator of choice, but it changes at 8.5 is this OK?  Go thru the pH changes vs ml NaOH added when titrating 10.0 ml of 0.10M HCl with 0.10 M NaOH

94. Titration

95. Titration WA with a SB (NaOH and acetic acid)  The titration produces a salt of a conjugate base, which is relatively strong since it came from a WA  The initial pH is higher than in the titration of a strong acid (Weak only partially ionizes.  There is an initial rather sharp increase in pH at the start of the titration, because a common ion (acetate) shifts the equilibrium back to the left, thus raising the pH  After the sharp increase and before the equivalence point is a relatively flat region called the buffer region

96. Titration  [H+] = [A-] at the half point of the equivalence point and the pH = pKa according to the Henderson/Heselbalch equation  At the equivalence point the pH>7 due to the hydrolysis of acetate ion producing more OH-  The steep portion of the curve is between pH7- 10, less than a drop of titrant to get there  More limitation in indicators since steep part starts at pH 7, i.e. cannot use a indicator that changes color less than pH7  The salt will react with water (hydrolysis) to produce OH- thus a basic equivalence point

97. Acid/Base Indicators Figure 16.13

98. Choosing an Indicator

99. Titration Curves for NH 3 and NaOH

100. Titration Curves for Acetic and Hydrochloric Acids

101. Titration Curve for Carbonate Ion A titration curve is a graph of pH of a solution as titrant is added.

102. Practice A 40.0 mL solution of 0.100 M sodium hypochlorite is titrated with a 0.100 M hydrochloric acid solution. Calculate the pH after 20.0 mL and 30.0 mL additions of the acid.

103. THE END

104. ChemTour: Acid Rain Click to launch animation PCPC | MacMac This ChemTour explores the effects of fossil fuel burning on the pH of rainwater, as well as the resulting environmental and industrial consequences.

105. ChemTour: Acid-Base Ionization Click to launch animation PCPC | MacMac This ChemTour explores the differences among Brønsted– Lowry acids, Brønsted–Lowry bases, Lewis acids, and Lewis bases.

106. ChemTour: Self-Ionization of Water Click to launch animation PCPC | MacMac This ChemTour illustrates the process by which water molecules act as both a proton acceptor (base) and a proton donor (acid), and explores the equilibrium constant (Kw) for the self-ionization of water.

107. ChemTour: pH Scale Click to launch animation PCPC | MacMac This ChemTour introduces the pH scale and uses interactive graphs to explain the relationship between pH, pOH [H3O+], and [OH–].

108. ChemTour: Acid Strength and Molecular Structure Click to launch animation PCPC | MacMac Students learn to determine relative acid strength based on the molecular and electronic structure of the acid.

109. ChemTour: Buffers Click to launch animation PCPC | MacMac Students learn to use the Henderson–Hasselbalch equation to predict the pH of a buffer.

110. ChemTour: Strong Acid and Strong Base Titration Click to launch animation PCPC | MacMac In this interactive virtual titration lab, students are introduced to the titration apparatus and learn to determine the concentration of an unknown acid from the volume of basic solution added.

111. ChemTour: Titrations of Weak Acids Click to launch animation PCPC | MacMac Students learn to read and understand the different stages of a titration curve for a weak acid or polyprotic acid, and understand what is happening at a molecular level.

112. Adding a Drop of HCl to a AgNO 3 Solution If a single drop of water containing 0.02 mmol of HCl is added to 1.0 L of a 10 ‑ 5 M solution of AgNO 3, will a precipitate form? AgCl(s) Ag + (aq) + Cl ‑ (aq) K sp = 1.8×10 -10 A) Yes B) No C) Can’t tell

113. Adding a Drop of HCl to a AgNO 3 Solution If a single drop of water containing 0.02 mmol of HCl is added to 1.0 L of a 10 ‑ 5 M solution of AgNO 3, will a precipitate form? AgCl(s) Ag + (aq) + Cl ‑ (aq) K sp = 1.8×10 -10 A) Yes B) No C) Can’t tell 0.02 mmole Cl - 1.0 L 10 3 mmole mole = 2.0 x 10 -5 M Cl -

114. Adding a Drop of HCl to a AgNO 3 Solution If a single drop of water containing 0.02 mmol of HCl is added to 1.0 L of a 10 ‑ 5 M solution of AgNO 3, will a precipitate form? AgCl(s) Ag + (aq) + Cl ‑ (aq) K sp = 1.8×10 -10 A) Yes B) No C) Can’t tell 0.02 mmole Cl - 1.0 L 10 3 mmole mole = 2.0 x 10 -5 [Cl - ] = 10 -5

115. Adding a Drop of HCl to a AgNO 3 Solution If a single drop of water containing 0.02 mmol of HCl is added to 1.0 L of a 10 ‑ 5 M solution of AgNO 3, will a precipitate form? AgCl(s) Ag + (aq) + Cl ‑ (aq) K sp = 1.8×10 -10 A) Yes B) No C) Can’t tell 0.02 mmole Cl - 1.0 L 10 3 mmole mole = 2.0 x 10 -5 [Ag + ] = 10 -5 Q = [Ag + ][Cl - ] = (2.0X10 -5 )(10 -5 ) = 2.0 X 10 -10

116. Adding a Drop of HCl to a AgNO 3 Solution If a single drop of water containing 0.02 mmol of HCl is added to 1.0 L of a 10 ‑ 5 M solution of AgNO 3, will a precipitate form? AgCl(s) Ag + (aq) + Cl ‑ (aq) K sp = 1.8×10 -10 A) Yes B) No C) Can’t tell 0.02 mmole Cl - 1.0 L 10 3 mmole mole = 2.0 x 10 -5 [Cl - ] = 10 -5 Q = [Ag + ][Cl - ] = (2.0X10 -5 )(10 -5 ) = 2.0 X 10 -10 Q > K sp Yes

117. Concentration of Ag + In Ionic Solutions Given that Ksp(AgCl) > K sp (AgBr), which of the following salts, when added in excess to an aqueous 0.1 M AgNO 3 solution, will result in the lowest concentration of Ag + (aq)? A) AgNO 3 B) NaCl C) AgBr

118. Concentration of Ag + In Ionic Solutions Consider the following arguments for each answer and vote again: A.Because of the common-ion effect, the addition of AgNO 3 (s) will cause a net decrease in the concentration of Ag + (aq). B.Adding NaCl will induce the precipitation of AgCl(s) from the solution, thus lowering the Ag + (aq) concentration. C.AgBr(s) is less soluble than AgCl(s), and so its addition will cause the greatest decrease in the Ag + (aq) concentration.

119. Dissolution of a Speck of BaSO 4 in H 2 O Suppose water is slowly added to a vessel containing a speck of the sparingly soluble salt BaSO 4 (s). Which of the following plots shows the equilibrium concentration of Ba 2+ (aq) in the resulting solution versus the amount of water added? A)B) C)

120. Dissolution of a Speck of BaSO 4 in H 2 O Consider the following arguments for each answer and vote again: A.As water is added and more BaSO 4 (s) is dissolved, the concentration of Ba 2+ (aq) will increase until the solution becomes saturated. B.The concentration of Ba 2+ (aq) will increase until all the BaSO 4 (s) has dissolved, after which additional water will decrease the Ba 2+ (aq) concentration. C.Until the BaSO 4 (s) has completely dissolved, the concentration of Ba 2+ (aq) will remain constant.

121. Conductivity of a NaCl + AgNO 3 Solution The conductivity of an aqueous solution is directly proportional to the concentration of the ions present. Given this fact, which of the following plots shows the conductivity of a NaCl solution as a function of the amount of AgNO 3 (s) added? A) B) C)

122. Conductivity of a NaCl + AgNO 3 Solution Consider the following arguments for each answer and vote again: A.Adding AgNO 3 (s) will increase the total ion concentration, so the conductivity will increase until the solution is saturated. B.As AgNO 3 (s) is added, the conductivity of the solution will decrease because of the precipitation of AgCl(s) until all of the Cl - (aq) is consumed. C.Although AgCl(s) will precipitate as AgNO 3 (s) is added, the total concentration of ions will remain constant until the Cl - (aq) is depleted.

123. Dependence of pH on Temperature To the left is a plot of the autoionization constant, K w, versus temperature. What is the pH of hot water? A) 7

124. Dependence of pH on Temperature Consider the following arguments for each answer and vote again: A.At higher temperatures, the concentrations of H 3 O + and OH - increase. Therefore, the pH of hot water is less than 7. B.Regardless of temperature, the concentrations of H 3 O + and OH - remain equal, so the pH remains 7, which is neutral. C.At higher temperatures, H + ions acquire enough kinetic energy to escape the solution, leaving a predominance of OH - ions.

125. NH 3 Buffer Solution Which of the following, when added to an NH 3 (aq) solution, will form a basic buffer? A) NaOH B) HCl C) NaCl

126. NH 3 Buffer Solution Consider the following arguments for each answer and vote again: A.NH 3, a weak base, is normally an acidic buffer, so to create a basic buffer, one must add NaOH. B.By adding HCl to the NH 3 solution to form some NH 4 +, the solution will become a basic buffer. C.NH 3 is already a weak base, so to create a basic buffer solution, one need only add a neutral buffering salt like NaCl.

127. Titration of a Diluted Weak Base To the left is a plot that shows the pH of a weak acid as it is titrated with 0.01 M NaOH. Which of the following plots would correspond to the same titration if the same weak acid were diluted with water and then titrated with 0.01 M NaOH? A)B) C)

128. Titration of a Diluted Weak Base Consider the following arguments for each answer and vote again: A.Diluting a weak acid with water will increase the initial pH of the solution and decrease the final pH of the solution. B.The dilution would have little effect on the initial pH of the weak acid, especially in the buffer region. However, the pH after the equivalence point will be lower. C.If the weak acid is diluted, the titration will reach the equivalence point sooner, since the concentration of the acid will be lower.

129. Conductivity of a H 2 SO 4 /Ba(OH) 2 Solution Given that the conductivity of an aqueous solution depends on the concentration of the ions present, which of the following graphs shows conductivity (y-axis) plotted against the acid added (x-axis) for the titration of the strong base Ba(OH) 2 with the strong acid H 2 SO 4 ? A)B) C)

130. Conductivity of a H 2 SO 4 /Ba(OH) 2 Solution Consider the following arguments for each answer and vote again: A.This is a titration of a strong base with a strong acid, so the conductivity will track the pH of the solution. B.Although BaSO 4 (s) will precipitate as H 2 SO 4 is added, the total concentration of ions will remain constant until the Ba 2 + (aq) is depleted. C.The conductivity will decrease as BaSO 4 (s) and H 2 O(λ) are formed, after which excess H 2 SO 4 will increase the conductivity.