1. Acids and Bases 2009-20101 Acids and Bases

2. Acids and Bases 2009-20102 Acids Svante Arrhenius, a Swedish chemist, defines an acid as a substance that yields hydrogen ions (H + ) when dissolved in water. Formulas for acids contain one or more hydrogen atoms as well as an anion.

3. Acids and Bases 2009-20103 In some cases two different names seem to be assigned to the same chemical formula. HCl(g)hydrogen chloride HCl(l)hydrogen chloride HCl(aq)hydrochloric acid The name assigned to the compound depends on its physical state. In the gaseous or pure liquid state, HCl is a molecular compound called hydrogen chloride. When it is dissolved in water, the molecules break apart into H + and Cl - ions; in this state, the substance is called hydrochloric acid. Naming Acids

4. Acids and Bases 2009-20104 Binary acids (formed by hydrogen and one other element) are named with a “hydro-” prefix and an “-ic” ending on the anion root. Ex-HCl HBr Hydrochloric acid Hydrobromic acid

5. Acids and Bases 2009-20105 The formulas for oxoacids, (acids that contain hydrogen and an anion containing oxygen) are usually written with the H first, followed by the anion, as illustrated in the following examples: H 2 CO 3 HNO 3 HClO 2 Carbonic acidNitric acidChlorous acid If the anion ends in “-ate” then the acid ends in “-ic”, if the anion ends in “-ite”, then the acid ends in “-ous”. Remember, “ic” goes with the higher oxidation state, N has an oxidation state of ___ in HNO 3 (nitric acid) and ___ in HNO 2 (nitrous acid) +5 +3

6. Acids and Bases 2009-20106 Acids have a sour taste; for example, vinegar owes its sourness to acetic acid, and lemons and citrus fruits contain citric acid. Acids cause color changes in plant dyes; for example, they change the color of blue litmus red. Acids react with certain metals to produce hydrogen gas. Acids react with carbonates and bicarbonates to produce carbon dioxide gas. Aqueous acid solutions conduct electricity. Acids

7. Acids and Bases 2009-20107 Brønsted Acid Arrhenius’s definitions of acids are limited in that they apply only to aqueous solutions. Broader definitions were proposed by the Danish chemist Johannes Brønsted. A Br ø nsted acid is a proton donor. HCl(aq) H + (aq) + Cl - Remember, the H + ion is really just a proton (a hydrogen atom is one proton and one electron, you pull off the electron and all you are left with is…

8. Acids and Bases 2009-20108 The size of a proton is about 10 -15 m, compared to the diameter of 10 -10 m for an average atom or ion. Such an exceedingly small charged particle cannot exist as a separate entity in aqueous solution owing to its strong attraction for the negative region of the polar water molecule. Consequently, the proton exists in a hydrated form as H 3 O +, and is referred to as the hydronium ion H + + H 2 O H 3 O +

9. Acids and Bases 2009-20109 Since the acidic properties of the proton are unaffected by hydration, we will generally use H + (aq) to represent the hydrated proton. This notation is for convenience only, because H 3 O + is closer to reality. Keep in mind that both notations represent the same species in aqueous solution. H + (aq) = H 3 O + (aq)

10. Acids and Bases 2009-201010 Monoprotic acids each unit of acid yields one hydrogen upon ionization Diprotic acids each unit of an acid gives up two H + ions Triprotic acids yields three H + ions upon ionization HCl H 2 CO 3 H 3 PO 4

11. Acids and Bases 2009-201011 Diprotic acids give up their two H + ions in separate steps: H 2 SO 4 (aq) H + (aq) + HSO 4 - (aq) HSO 4 - (aq) H + (aq) + SO 4 -2 (aq) Triprotic acids give up their H + ions in three separate steps. Is HSO 4 - a strong or weak acid? Explain Weak, only partially ionizes

12. Acids and Bases 2009-201012 In another definition formulated by Svante Arrhenius, a base can be described as a substance that yields hydroxide ions (OH - ) when dissolved in water. Some examples are NaOH KOH Ba(OH) 2 Bases Sodium hydroxide Potassium hydroxide Barium hydroxide

13. Acids and Bases 2009-201013 Ammonia (NH 3 ) is also classified as a common base. At first glance this may seem to be an exception to the definition of a base. Note that as long as a substance yields hydroxide ions when dissolved in water, it need not contain hydroxide ions in its structure to be considered a base. In fact, when ammonia dissolves in water, the following reaction occurs: NH 3 + H 2 O NH 4 + + OH - Thus it is properly classified as a base.

14. Acids and Bases 2009-201014 Bases Bases have a bitter taste Bases feel slippery; for example, soaps, which contain bases, exhibit this property Bases cause color changes in plant dyes; for example, they change the color of red litmus blue Aqueous base solutions conduct electricity

15. Acids and Bases 2009-201015 Brønsted Base A Brønsted base is defined by Johannes Brønsted as being a substance capable of accepting a proton. NH 3 is a Brønsted base because it accepts a proton to form NH 4 + NH 3 + H 2 O NH 4 + + OH -

16. Acids and Bases 2009-201016 Lewis Acids and Bases G.N. Lewis formulated a definition for what is now called a Lewis base – a substance that can donate a pair of electrons, and a Lewis acid, a substance that can accept a pair of electrons.

17. Acids and Bases 2009-201017 The significance of the Lewis concept is that it is much more general than other definitions. For example, the reaction between boron trifluoride and ammonia is a Lewis acid-base reaction.

18. Acids and Bases 2009-201018 Strength of Acids and Bases Strong acids are strong electrolytes, which, for practical purposes, are assumed to ionize completely in water. That means that at equilibrium, solutions of strong acids will not contain any nonionized acid molecules. Like strong acids, strong bases are all strong electrolytes that ionize completely in water.

19. Acids and Bases 2009-201019 The most common strong acids are HClO 4, HCl, HNO 3 and H 2 SO 4. Hydroxides of alkali metals and alkaline Earth metals are strong bases (like NaOH, KOH and Ba(OH) 2 ). Other strong acids and strong bases are listed on your Relative Strengths of Acids and Bases Reference Sheet.

20. Acids and Bases 2009-201020 The strength of an acid is measured by its tendency to ionize: HX  H + + X - The strength of the H-X bond influences the extent to which an acid undergoes ionization. The stronger the bond (the higher the bond dissociation energy in kJ/mol), the more difficult it is for the HX molecule to break up and hence the weaker the acid.

21. Acids and Bases 2009-201021 BondBond Dissociation Energy (kJ/mol) Acid Strength H-F568.2 H-Cl431.9 H-Br366.1 H-I298.3 Bond Dissociation Energies for Hydrogen Halides and Acid Strengths strong weak

22. Acids and Bases 2009-201022 The Strength of Oxoacids Oxoacids contain hydrogen, oxygen, and one other element Z, which occupies a central position. To compare oxoacid strength, it is convenient to separate the oxoacids into two groups.

23. Acids and Bases 2009-201023 Oxoacids having different central atoms that are from the same group of the periodic table and that have the same oxidation number. Within this group, acid strength increases with increasing electronegativity of the central atom. HClO 3 > HBrO 3 The Cl pulls more strongly on the electron pair shared with the O, making the O-H bond more polar, therefore making it easier to ionize (and the acid stronger)

24. Acids and Bases 2009-201024 Oxoacids having the same central atom but different numbers of attached groups. Within this group, acid strength increases as the oxidation number of the central atom increases. HClO 4 > HClO 3 > HClO 2 > HClO The greater the number of O atoms pulling on the Cl, the more that the electrons are pulled away from the O-H bond, making the O-H bond more polar, therefore making it easier to ionize (and the acid stronger)

25. Acids and Bases 2009-201025 You are going to remember these trends because… HCl is a strong acid and HF (with the higher bond dissociation energy) isn’t. HClO 4 is a strong acid and HClO 3 (where the Cl has a lower oxidation number because it has fewer oxygen atoms attached to it) isn’t.

26. Acids and Bases 2009-201026 Note: H 3 O + is the strongest acid that can exist in aqueous solutions. Acids stronger than H 3 O + react with water to produce H 3 O + and their conjugate bases. Thus, HCl, which is a stronger acid than H 3 O +, reacts with water completely to form H 3 O + and Cl -. HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq)

27. Acids and Bases 2009-201027 The OH - ion is the strongest base that can exist in aqueous solution. Bases stronger than OH - react with water to produce OH - and their conjugate acids. For example, the oxide ion, (O -2 ) is a stronger base than OH -, so it reacts with water completely as follows: O -2 (aq) + H 2 O(l)  2OH - (aq) For this reason, the oxide ion does not exist in aqueous solutions.

28. Acids and Bases 2009-201028 Amphoteric Compounds NH 3 (g) + H 2 O(l)  OH - (aq) + NH 4 + (aq) H 2 SO 4 (aq) + H 2 O(l)  H 3 O + (aq) + HSO 4 - (aq) As you could see from the previous two examples, water will act as either an acid or a base, depending on the strength of the acid or base with which it is reacting. Any species that can react as either an acid or a base is described as amphoteric. Proton acceptor (base) Proton donor (acid)

29. Conjugate base Remains when one proton has been removed from the acid Conjugate acid Results from the addition of a proton to a Bronsted base Base Proton (H + ) acceptor Acid Proton (H + ) donor An extension of the Brønsted definition of acids and bases is the concept of the conjugate acid-base pair CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) A conjugate acid-base pair is defined as an acid and its conjugate base (what’s left after the H+ was removed from the acid) or a base and its conjugate acid (substance formed by the addition of the H+ to the base). ** Because the acid and base are always stronger than the conjugate acid and conjugate base, the direction of the reaction proceeds from acid/base  conjugate acid/conjugate base.

30. Acids and Bases 2009-201030 Identify the acid, base, conjugate acid and conjugate base in the following reaction (**Reaction proceeds from stronger to weaker…) NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) Base Proton (H + ) acceptor (stronger base) Acid Proton donor (Stronger acid) Conjugate Acid Accepted proton (H + ) (weaker acid) Conjugate Base What’s left after H + was donated by acid (weaker base)

31. Acids and Bases 2009-201031 The Acid-Base Properties of Water Water is a very weak electrolyte and therefore a poor conductor of electricity, but it does undergo ionization to a small extent: H 2 O(l) H + (aq) + OH - (aq) This reaction is sometimes called the autoionization of water.

32. Acids and Bases 2009-201032 In the study of acid-base reactions in aqueous solutions, the hydrogen ion concentration is the key, because it indicates the acidity or alkalinity of the solution. Expressing the hydrogen ion as H +, we can write the equilibrium constant for the autoionization of water as k w = [H + ][OH - ] [H 2 O] Remember, pure liquids and solids are not listed in the ionization equation, therefore…

33. Acids and Bases 2009-201033 k w = [H + ][OH - ] k w is called the ion-product constant, and is the product of the molar concentrations of H + and OH - ions at a particular temperature.

34. Acids and Bases 2009-201034 In pure water at 25 o C, the concentrations of H + and OH - ions are equal and found to be [H + ] = 1.0 x 10 -7 M and [OH - ] = 1.0 x 10 -7 M. Thus, k w = [H + ][OH - ] k w = (1.0 x 10 -7 )(1.0 x 10 -7 ) k w = 1.0 x 10 -14

35. Acids and Bases 2009-201035 Whether we have pure water or a solution of dissolved species, the following relation ALWAYS holds at 25 o C k w = [H + ][OH - ] = 1.0 x 10 -14

36. 36 Because HCl is a strong acid… HCl  H + + Cl - Calculate the concentration of OH - ions in an HCl solution whose hydrogen ion concentration is 1.3 M. k w = [H + ][OH - ] 1.0 x 10 -14 = (1.3)[OH - ] [OH - ] = 7.7 x 10 -15 M [HCl] [H + ] [Cl - ] I (initial) C (change) E (end) 1.3 0.0 0.0 -1.3 +1.3 +1.3 0.0 1.3 1.3

37. Acids and Bases 2009-201037 pH – A Measure of Acidity Because the concentrations of H + and OH - ions in aqueous solutions are frequently very small numbers and therefore inconvenient to work with, Soren Sorensen in 1909 proposed a more practical measure called pH. The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration (in mol/L) pH = -log [H + ] pH is a dimensionless quantity (it will not have a label)

38. Acids and Bases 2009-201038 Since pH is simply a way to express hydrogen ion concentration, acidic and basic solutions at 25 o C can be distinguished by their pH values, as follows: Acidic solutions:[H + ] > 1.0 x 10 -7 M, pH < 7.00 In an acidic solution there is an excess of H + ions; [H+] > [OH-] Basic solutions:[H + ] 7.00 In a basic solution there is an excess of OH - ions; [OH - ] > [H + ] Neutral solutions[H + ] = 1.0 x 10 -7 M, pH = 7.00 Whenever [H+] = [OH-], the aqueous solution is said to be neutral. **Note – when concentration has two significant figures, pH will have two numbers TO THE RIGHT OF THE DECIMAL!

39. Acids and Bases 2009-201039 Calculate the pH of a 1.0 x 10 -3 M HCl solution. Remember, these are molarities HCl(aq)H + (aq)Cl - (aq) I 1.0 x 10 -3 0.0 C -1.0 x 10 -3 +1.0 x 10 -3 E 0.01.0 x 10 -3 Thus, [H + ] = 1.0 x 10 -3 M pH = -log(1.0 x 10 -3 ) pH = 3.00 Since HCl is a strong acid, it is completely ionized in solution: HCl(aq)  H + (aq) + Cl - (aq)

40. 40 The concentration of H + ions in a bottle of table wine was 3.2 x 10 -4 M right after the cork was removed. Only half of the wine was consumed. The other half, after it had been standing open to the air for a month, was found to have a hydrogen ion concentration equal to 1.0 x 10 -3 M. Calculate the pH of the wine on these two occasions. Some of the ethanol converted to acetic acid, a reaction that takes place in the presence of O 2. Why did the acidity increase? When the wine was first opened pH = -log [H + ] pH = -log (3.2 x 10 -4 ) = 3.49 After the wine sat for a month pH = -log [H + ] pH = -log (1.0 x 10 -3 ) = 3.00

41. Acids and Bases 2009-201041 Given the pH of a solution, you can figure out the [H + ] concentration by using the simple formula [H + ] = 10 -pH What is the hydrogen ion concentration of an acid with a pH of 3.00? [H + ] = 10 -pH [H + ] = 10 -3.00 [H + ] = 1.0 x 10 -3

42. Acids and Bases 2009-201042 A pH meter is commonly used in the laboratory to determine the pH of a solution. Although many pH meters have scales marked with values from 1 to 14, pH values can, in fact, be less than 1 and greater than 14.

43. Acids and Bases 2009-201043 A pOH scale analogous to the pH scale can be devised using the negative logarithm of the hydroxide ion concentration of a solution. Thus we define pOH as pOH = -log[OH - ]

44. Acids and Bases 2009-201044 Now consider again the ion-product constant for water: k w = [H+][OH - ] = 1.0 x 10 -14 Taking the negative logarithm of both sides we obtain -log[H + ] + -log[OH - ] = -log(1.0 x 10 -14 ) -log[H+] + -log[OH-] = 14.00 pHpOH pH + pOH = 14.00 “Logs make adders multiply”

45. Acids and Bases 2009-201045 In a NaOH solution [OH - ] is 2.9 x 10 -4 M. Calculate the pH of the solution. First, figure out the pOH… Then use the pOH to figure out the pH… pOH = -log [OH - ] pOH = -log (2.9 x 10 -4 ) pOH = 3.54 pH + pOH = 14.00 pH = 14.00 – pOH pH = 14.00 – 3.54 = 10.46

46. Acids and Bases 2009-201046 Calculate the pH of a 0.020 M Ba(OH) 2 solution. Ba(OH) 2 (aq)Ba +2 (aq)OH - (aq) I (M) 0.0200.00 C (M) -0.020+0.0202(+0.020) E (M) 0.000.0200.040 Thus, [OH - ] = 0.040 M pOH = -log 0.040 = 1.40 pH = 14.00 – pOH pH = 14.00 – 1.40 = 12.60 Ba(OH) 2 is a strong base; each Ba(OH) 2 unit produces two OH-ions: Ba(OH) 2 (aq)  Ba +2 (aq) + 2OH - (aq)

47. 47 To determine the hydroxide ion when given the pOH, you need to use the formula 10 -pOH = [OH - ] What is the molarity of a NaOH solution that has a pH of 11.30? Because NaOH is a strong base, the [OH - ] at the end is equal to the initial concentration of NaOH. The molarity of the NaOH = 2.0 x 10 -3 M 14.00 – pH = pOH 14.00 – 11.30 = 2.70 10 -2.70 = 2.0 x 10 -3 = [OH - ]

48. Acids and Bases 2009-201048 Most acids are weak acids, which ionize only to a limited extent in water. At equilibrium, aqueous solutions of weak acids contain a mixture of nonionized acid molecules, H 3 O + ions, and the conjugate base. The limited ionization of weak acids is related to the equilibrium constant for ionization, which is represented as k a. Weak Acids and Acid Ionization Constants

49. Acids and Bases 2009-201049 Consider a weak monoprotic acid, HA. Its ionization in water is represented by HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) Weak acid conjugate base or simply HA(aq) H + (aq) + A - (aq)

50. Acids and Bases 2009-201050 Write the equilibrium expression for the ionization of HA. Ka=Ka= [H + ][A - ] [HA] k a, the acid ionization constant, is the equilibrium constant for the ionization of an acid.

51. Acids and Bases 2009-201051 At a given temperature, the strength of the acid HA is measured quantitatively by the magnitude of k a. The larger k a, the stronger the acid – that is, the greater the concentration of H + ions at equilibrium due to its ionization. Keep in mind, however, that only weak acids have k a values associated with them.

52. Acids and Bases 2009-201052 Percent Ionization Another measure of the strength of an acid is its percent ionization, which is defined as Percent ionization = H + concentration at equilibrium Initial concentration of acid X 100

53. Acids and Bases 2009-201053 The stronger the acid, the greater the percent ionization, for example, a strong acid like HCl ionizes 100%. The extent to which a weak acid ionizes depends on the initial concentration of the acid. The more dilute the solution, the greater the percent ionization.

54. Acids and Bases 2009-201054 You have a reference sheet that lists a number of weak acids and their k a values at 25 o C. Although all of the acids on that sheet are weak, within the group there is great variation in their strengths. For example, k a for HF (6.8 x 10 -4 ) is about 1.5 million times greater than that for HCN (6.2 x 10 -10 ).

55. Acids and Bases 2009-201055 Generally, we can calculate the hydrogen ion concentration or pH of an acid solution at equilibrium, given the initial concentration of the acid and its k a value. Alternatively, if we know the pH of a weak acid solution and its initial concentration, we can determine its k a.

56. Acids and Bases 2009-201056 Suppose we are asked to calculate the pH of a 0.50 M HF solution at 25 o C. The ionization of HF is given by HF(aq) H+(aq) + F-(aq) From your reference sheet we can write k a = [H + ][F - ] [HF] = 6.8 x 10 -4

57. Acids and Bases 2009-201057 The first step is to identify all the species present in solution that may affect its pH. Because weak acids ionize to a small extent, at equilibrium the major species present are nonionized HF and some H + and F - ions. Another major species is H 2 O, but its very small k w (1.0 x 10 -14 ) means that water is not a significant contributor to the H + ion concentration. Therefore, unless otherwise stated, we will always ignore the H + or OH - ions produced by the autoionization of water.

58. Acids and Bases 2009-201058 HF(aq)H + (aq)F - (aq) I0.500.00 C-x+x E0.50 – xxx HF(aq) H+(aq) + F-(aq) We can summarize the changes in the concentrations of HF, H +, and F - in the table below:

59. Acids and Bases 2009-201059 The equilibrium concentrations of HF, H + and F -, expressed in terms of the unknown x, are substituted into the ionization constant expression to give k a = (x)(x) 0.50 - x = 6.8 x 10 -4 Rearranging this expression, we write x 2 + 6.8 x 10 -4 x – 3.4 x 10 -4 = 0

60. Acids and Bases 2009-201060 This is a quadratic equation which can be solved using the quadratic formula. Or, we can try using a shortcut to solve for x. Because HF is a weak acid and weak acids ionize only to a slight extent, we reason that x must be small compared to 0.50 Therefore we can make the approximation 0.50 – x ≈ 0.50

61. Acids and Bases 2009-201061 Now the ionization constant expression becomes x 2 0.50 - x ≈ x 2 0.50 = 6.8 x 10 -4 Rearranging, we get x 2 = (0.50)(6.8 x 10 -4 ) = 3.4 x 10 -4 x = √3.4 x 10 -4 = 0.018 M

62. Acids and Bases 2009-201062 Thus we have solved for x without having to use the quadratic equation. At equilibrium, we have: [HF] = [H + ] = [F - ] = And the pH of the solution is (0.50 – 0.018) M = 0.48 M 0.018 M pH = -log(0.018) = 1.74 This is determined by going back to the ICE chart

63. 63 How good is this approximation? The approximation is valid if the following expression is equal to or less than 5% 0.018 M 0.50 M X 100 = 3.6% Molarity of H + at equilibrium Initial concentration of weak acid If this percent ionization is greater than 5%, you must use the quadratic formula

64. Acids and Bases 2009-201064 The Quadratic Equation -b± √b 2 – 4ac 2a x = If the percent ionization had been greater than 5%, we would have had to use the quadratic equation. x 2 + 6.8 x 10 -4 x – 3.4 x 10 -4 = 0 a = 1; b = 6.8 x 10 -4 ; c = -3.4 x 10 -4

65. 65 -b± √b 2 – 4ac 2a x = -6.8 x 10 -4 ± √ (6.8 x 10 -4 ) 2 – 4(1)(-3.4 x 10 -4 ) 2(1) x = -6.8 x 10 -4 ± .0014 2(1) x = x =.018 M or -.018 M The second solution is physically impossible because the concentration of ions produced as a result of ionization cannot be negative. pH = -log(0.018) = 1.74 Note – this is the same value as we estimated earlier!

66. Acids and Bases 2009-201066 If the percent ionization for.50 M HF is 3.6%, what is the percent ionization of HF if you decrease the molarity to 0.050M? HF(aq) H+(aq) + F-(aq) HF(aq)H + (aq)F - (aq) I0.0500.00 C-x+x E0.050 – xxx

67. 67 k a = 6.8 x 10 -4 = (x)(x) 0.050 - x 3.4 x 10 -5 – 6.8 x 10 -4 x = x 2 0 = x 2 + 6.8 x 10 -4 x - 3.4 x 10 -5 using the quadratic equation, x=.0055, 0.0055 M 0.050 M X 100 = 11% Notice, the percent ionization of HF increased as the molarity decreased

68. Acids and Bases 2009-201068 Weak bases, like weak acids, are weak electrolytes. B + H 2 O HB + + OH - weak base conjugate acid Ammonia is a weak base that ionizes only to a limited extent in water: NH 3 (aq) + H 2 O(l) NH 4 + + OH - (aq) Weak Bases and Base Ionization Constants

69. Acids and Bases 2009-201069 The equilibrium constant is given by k b = [NH 4 + ][OH - ] [NH 3 ] Where k b is called the base ionization constant.

70. Acids and Bases 2009-201070 Follow the same procedures you used with weak acids when solving problems involving weak bases. The main difference is that you will calculate [OH - ] first, rather than [H + ].

71. Acids and Bases 2009-201071 The Relationship Between the Ionization Constants of Acids and Their Conjugate Bases For any conjugate acid-base pair it is always true that k a k b = k w

72. Acids and Bases 2009-201072 Calculate the k b of the conjugate base of acetic acid CH 3 COOH H + + CH 3 COO - Conjugate base k a k b = k w (1.8 x 10 -5 )(k b ) = 1.0 x 10 -14 k b = 5.6x 10 -10

73. Acids and Bases 2009-201073 Diprotic and Polyprotic Acids The treatment of diprotic and polyprotic acids is more involved than that of monoprotic acids because these substances yield more than one hydrogen atom per molecule. These acids ionize in a stepwise manner, that is, they lose one proton at a time. An ionization constant expression should be written for each ionization step.

74. Acids and Bases 2009-201074 Oxalic acid (H 2 C 2 O 4 ) is a poisonous substance used chiefly as a bleaching and cleansing agent (for example, to remove bathtub rings). Calculate the concentrations of all the species present at equilibrium in a 0.10 M solution.

75. Acids and Bases 2009-201075 H 2 C 2 O 4 H + + HC 2 O 4 - H 2 C 2 O 4 (aq)H + (aq)HC 2 O 4 -(aq) I (M)0.100.00 C (M)-x+x E (M)0.10 – xxx

76. Acids and Bases 2009-201076 K a = [H + ][HC 2 O 4 - ] [H 2 C 2 O 4 ] = 5.6 x 10 -2 K a = (x)(x) 0.10 - x = 5.6 x 10 -2 Let me save you some work, you need to use the quadratic formula for this one x 2 + 5.6 x 10 -2 x - 5.6 x 10 -3 = 0 x = 0.052

77. Acids and Bases 2009-201077 When the equilibrium for the first stage of ionization is reached, the concentrations are: [H + ] = [HC 2 O 4 - ] = [H 2 C 2 O 4 ] = 0.052 M (0.10 - 0.052) M = 0.048 M

78. Acids and Bases 2009-201078 Next we consider the second stage of ionization. At this stage, the major species will be HC 2 O 4 -, (this serves as the acid in the second stage), H +, and C 2 O 4 -2 (the conjugate base).

79. Acids and Bases 2009-201079 HC 2 O 4 - H + + C 2 O 4 -2 HC 2 O 4 - (aq)H + (aq)C 2 O 4 - 2 (aq) I (M)0.052 0.00 C (M)-y+y E (M)0.052 – y0.052 + yy

80. Acids and Bases 2009-201080 K a = [H + ][C 2 O 4 -2 ] [HC 2 O 4 - ] = 5.4 x 10 -5 K a = (0.052 + y)(y) 0.052 - y = 5.4 x 10 -5 Let me save you some work, you DON’T need to use the quadratic formula for this one, 0.052 + y and 0.052 – y ≈ 0.052 K a = (0.052)(y) 0.052 = 5.4 x 10 -5 y = 5.4 x 10 -5

81. Acids and Bases 2009-201081 Testing the approximation - checking the 5% rule 5.4 x 10 -5 M 0.052 M X 100 =.10% The approximation is valid.

82. Acids and Bases 2009-201082 At equilibrium [H 2 C 2 O 4 ] = [HC 2 O 4 - ] = [H + ] = [C 2 O 4 -2 ] = 0.048 M (0.052 – 5.4 x 10 -5 ) M = 0.052 M (0.052 + 5.4 x 10 -5 ) M = 0.052 M 5.4 x 10 -5 M

83. Acids and Bases 2009-201083 This example shows that for diprotic acids, if K a1 » K a2, then we can assume that the concentration of H + ions is the product of only the first stage of ionization.

84. Acids and Bases 2009-201084 Salts (which are one of the products of an acid-base neutralization reaction) are strong electrolytes that completely dissociate into ions in water. The term salt hydrolysis describes the reaction of an anion or a cation of a salt, or both, with water. Salt hydrolysis usually affects the pH of a solution. Acid – Base Properties of Salts

85. Acids and Bases 2009-201085 Salts that Produce Neutral Solutions NaNO 3 Na + + NO 3 - The cation of this salt came from a strong base (NaOH) The anion of this salt came from a strong acid (HNO 3 ) H2OH2O

86. Acids and Bases 2009-201086 Salts that Produce Neutral Solutions NaNO 3 Na + + NO 3 - H2OH2O The Na + ion and the OH - from the water would not stay together (NaOH would be a strong base and would dissociate) The NO 3 - ion and the H + from the water would not stay together, (remember, HNO 3 is a strong acid)

87. Acids and Bases 2009-201087 Salts that Produce Neutral Solutions NaNO 3 Na + + NO 3 - H2OH2O Consequently, NaNO 3 and other salts formed from a strong acid and a strong base do not affect the pH of a solution.

88. Acids and Bases 2009-201088 Salts that Produce Basic Solutions NaCH 3 COO Na + + CH 3 COO - H2OH2O The cation of this salt came from a strong base (NaOH) The anion of this salt came from a weak acid (CH 3 COOH)

89. Acids and Bases 2009-201089 Salts that Produce Basic Solutions NaCH 3 COO Na + + CH 3 COO - H2OH2O The Na + ion and the OH - from the water would not stay together (NaOH would be a strong base and would dissociate) The CH 3 COO - ion and the H + from the water WOULD stay together, (remember, CH 3 COOH is a weak acid, meaning it only ionizes slightly)

90. Acids and Bases 2009-201090 Salts that Produce Basic Solutions CH 3 COO - OH - + CH 3 COOH H2OH2O Because the CH 3 COO - ions would bond with the H + ions, the OH - ions (which are left behind when the H+ ions come off of the water molecules) affect the pH of the solution. In other words, solutions produced by salts made from strong bases and weak acids will be basic in nature.

91. Acids and Bases 2009-201091 Salts that Produce Acidic Solutions NH 4 Cl NH 4 + + Cl - H2OH2O The cation of this salt came from a weak base, NH 3 The anion of this salt came from a strong acid

92. Acids and Bases 2009-201092 Salts that Produce Acidic Solutions NH 4 Cl NH 4 + + Cl - H2OH2O NH 4 + is a conjugate acid, and it will dissociate producing excess H + ions. In other words, solutions produced by salts made from strong acids and weak bases will be acidic in nature. NH 3 + H +

93. Acids and Bases 2009-201093 NO 2 - has an affinity for the H + produced by the autoionization of water, leaving an excess of OH - behind. The NH 4 + dissociates producing H + ions When the cation of a salt comes from a weak base and the anion comes from a weak acid, you need to compare the K a and K b values to determine if the solution will be acidic or basic. For example, NH 4 NO 2 NH 4 + + NO 2 - K a = 5.7 x 10 -10 NH 4 + NH 3 + H + K b = 1.4 x 10 -11 NO 2 - HNO 2 + OH - therefore an aqueous solution of NH 4 NO 2 will be acidic. If K b > K a, the solution will be basic, if K a > K b, the solution will be acidic..

94. Acids and Bases 2009-201094 Oxides can be classified as acidic, basic, or amphoteric. Most alkali metal oxides and all alkaline earth metal oxides form bases when they react with water. These oxides are sometimes called basic anhydrides. Beryllium oxide and several metallic oxides in the boron family (Group 3A) and carbon family (Group 4A) are amphoteric. Na 2 O + H 2 O  2NaOH 2Ba(OH) 2 BaO + H 2 O 

95. Acids and Bases 2009-201095 Nonmetalic oxides that react with water to form acids are sometimes referred to as acidic anhydrides. CO 2 + H 2 O  SO 3 + H 2 O  N 2 O 5 + H 2 O  P 4 O 10 + H 2 O  Cl 2 O 7 + H 2 O  H 2 CO 3 H 2 SO 4 2HNO 3 4H 3 PO 4 2HClO 4

96. Acids and Bases 2009-201096 Buffer Solutions A buffer solution is a solution of (1) a weak acid or a weak base and (2) its salt; both components must be present. The solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Buffers are very important to chemical and biological systems. The pH in the human body varies greatly from one fluid to another; for example, the pH of blood is about 7.4, whereas the gastric juice in our stomach has a pH of about 1.5. These pH values, which are crucial for proper enzyme function and the balance of osmotic pressure, are maintained by buffers in most cases.

97. Acids and Bases 2009-201097 NaCH 3 COO CH 3 COO - + Na + H2OH2O CH 3 COOH CH 3 COO - + H + If you add a base to this solution, the OH- will be neutralized by the acetic acid in the buffer, therefore you will not notice a significant difference in the pH of the solution. Weak acid Conjugate base

98. Acids and Bases 2009-201098 NaCH 3 COO CH 3 COO - + Na + H2OH2O If you add an acid to this solution, the acetate ion will bond to most of the added H +, so no appreciable change in pH will be observed. CH 3 COOH CH 3 COO - + H +

99. Acids and Bases 2009-201099 The buffering capacity, that is, the effectiveness of the buffer solution, depends on the amount of acid and conjugate base from which the buffer is made. The larger the amount, the greater the buffering capacity.

100. Acids and Bases 2009-2010100 In general, a buffer system can be represented as weak acid/salt or weak acid/conjugate base. Thus the sodium acetate/acetic acid buffer system we discussed can be written as CH 3 COOH/CH 3 COONa or simply CH 3 COOH/CH 3 COO -

101. Acids and Bases 2009-2010101 Which of the following are buffer systems? KF/HF NaClO 4 /HClO 4

102. Acids and Bases 2009-2010102 (a)Calculate the pH of a buffer system containing 1.0 M CH 3 COOH and 1.0 M CH 3 COONa. (b)What is the pH of the buffer system after the addition of 0.10 mole of gaseous HCl to 1 L of the solution. Assume that the volume of the solution does not change when the HCl is added.

103. Acids and Bases 2009-2010103 CH 3 COOHCH 3 COO - H+H+ I 1.0 0.0 C -x+ xx E 1.0 - x1.0 + x0.0 + x Assume ionization is negligible

104. Acids and Bases 2009-2010104 CH 3 COOHCH 3 COO - H+H+ I 1.0 0.0 C -x+ xx E 1.0 x

105. Acids and Bases 2009-2010105 k a = CH 3 COOH CH 3 COO - + H + [H + ] [CH 3 COO-] [CH 3 COOH] [H + ] = 1.8 x 10 -5 = [H + ] (1.0) 1.0 x 10 -5

106. Acids and Bases 2009-2010106 [H + ] = -log[H+] =pH = When the concentration of the acid and the conjugate base are the same, the pH of the buffer is equal to the pK a, which is determined by taking the –log of the k a, of the acid pH = pk a = -log k a 1.0 x 10 -5 4.74

107. Acids and Bases 2009-2010107 (b)After the addition of HCl, complete ionization of the 1.0 M HCl acid occurs; HCl  H + + Cl - 0.10 mol 0.10 mol This 0.10 mol of H + ions will bond with 0.10 mol of the CH 3 COO - ions, DECREASING the amount of CH 3 COO - by.10 mol and INCREASING the amount of CH 3 COOH by.10 mol

108. Acids and Bases 2009-2010108 So now instead of 1.0 mol of CH 3 COO - ions, there will only be 0.90 mol (1.0 -.10) And since there were originally 1.0 mol of CH 3 COOH, you now have 1.10 mol of CH 3 COOH with the added.10 mol [CH 3 COO - ] =.90 M [CH 3 COOH] =1.10 M

109. Acids and Bases 2009-2010109 k a = [H + ][CH 3 COO - ] [CH 3 COOH] [H + ] = 1.8 x 10 -5 = [H + ](.90) (1.1) 2.2 x 10 -5

110. Acids and Bases 2009-2010110 [H + ] = pH = -log(2.2 x 10 -5 ) pH = This is a very slight change in pH, before the HCl was added, the [H + ] = 1.8 x 10 -5, after the addition of HCl, the [H + ] = 2.2 x 10 -5, this is an increase by a factor of 1.2 with a pH change from 4.74 to 4.66 2.2 x 10 -5 4.66

111. Acids and Bases 2009-2010111 To appreciate the effectiveness of the CH 3 COONa/CH 3 COOH buffer, let us find out what would happen if 0.10 mol HCl were added to 1 L of water, and compare the increase in H + ion concentration.

112. Acids and Bases 2009-2010112 Before the addition of HCl: After the addition of HCl: 0.10 M 1.0 x 10 -7 M = 1.0 x 10 6 This would be a millionfold increase as the pH changed from 7.00 to 1.00! [H + ] = 1.0 x 10 -7 [H + ] = 0.10 M

113. Acids and Bases 2009-2010113 A "very convenient" equation for dealing with buffer solutions is the Henderson- Hasselback equation. pH = pK a + log [A - ] [HA]

114. Acids and Bases 2009-2010114 log 1 = 0 Previous question - Calculate the pH of a buffer system containing 1.0 M CH 3 COOH and 1.0 M CH 3 COONa. pH = pk a + log [A - ] [HA] pH = pk a + log [1.0] pH = pk a pH = -log(k a )pH = -log(1.8 x 10 -5 ) pH = 4.74

115. Acids and Bases 2009-2010115 Just as we did previously, lets calculate the pH of this buffer system after the addition of 0.10 mole of gaseous HCl to 1 L of the solution. Assume that the volume of the solution does not change when the HCl is added. pH = pk a + log [A - ] [HA] pH = 4.65 Remember, the.10 M H + from the HCl would shift the equilibrium to the left decreasing the amount of A - to.90, and increasing the amount of HA to 1.10 pH = pk a + log [.90] [1.10] pH = -log(1.8 x 10 -5 ) + -.087

116. Acids and Bases 2009-2010116 Calculate the pH of a buffer system containing.50 M C 6 H 5 COOH and 1.0 M C 6 H 5 COOK. pH = pk a + log [A - ] [HA] pH = 4.49 pH = pk a + log [1.0] [.50] pH = -log(6.5 x 10 -5 ) +.30

117. Acids and Bases 2009-2010117 What is the pH of the previous buffer system after the addition of 0.10 mole of gaseous NaOH to 1 L of the solution. Assume that the volume of the solution does not change when the base is added. Adding a.10 mol base will neutralize.10 mol of the acid, C 6 H 5 COOH, decreasing the molarity of the acid to.40M, and, for every.10 mol of C 6 H 5 COOH you neutralize, you will form.10 mol more of C 6 H 5 COO - [C 6 H 5 COO - ] =1.10 M [CH 3 COOH] =.40 M

118. Acids and Bases 2009-2010118 [C 6 H 5 COO - ] =1.10 M [CH 3 COOH] =.40 M pH = pk a + log [A - ] [HA] pH = 4.63 pH = pk a + log [1.10] [.40] pH = -log(6.5 x 10 -5 ) +.44

119. Acids and Bases 2009-2010119 Acid – Base Titrations Quantitative studies of acid-base neutralization reactions are most conveniently carried out using a technique known as titration. In titration, a solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete.

120. Acids and Bases 2009-2010120 If we know the volumes of the standard solution, we can calculate the concentration of the unknown solution. Sodium hydroxide is one of the bases commonly used in the laboratory. However, it is difficult to obtain solid sodium hydroxide in a pure form because it is hygroscopic, (it has a tendency to absorb water from air), and its solution reacts with carbon dioxide. For these reasons, a solution of sodium hydroxide must be standardized before it can be used in accurate analytical work.

121. Acids and Bases 2009-2010121 We can standardize the sodium hydroxide solution by titrating it against an acid solution of accurately known concentration. The acid often chosen for this task is a monoprotic acid called potassium hydrogen phthalate (abbreviated as KHP), for which the molecular formula is KHC 8 H 4 O 4.

122. Acids and Bases 2009-2010122 The procedure for the titration of KHP and NaOH is as follows: 1.Add a known amount of KHP to an Erlenmeyer flask. Add some distilled water to make up a solution. 2.Next, carefully add NaOH solution from a buret until the equivalence point is reached. The equivalence point is the point at which the acid has completely reacted with or been neutralized by the base.

123. 123 The equivalence point is usually signaled by a sharp change in the color of an indicator. In an acid-base titration, indicators are substances that have distinctly different colors in acidic and basic solutions. One common indicator is phenolphthalein. Phenolphthalein indicates the presence of a/n _________. Phenolphthalein is ________ in acidic solutions, _______ in neutral solutions and ________ in basic solutions. base colorless pink colorless

124. Acids and Bases 2009-2010124 At the equivalence point, all the KHP present has been neutralized by the added NaOH and the solution is still colorless. However, if we add just one more drop of NaOH solution from the buret, the solution will immediately turn pink because the solution is now basic.

125. 125 Apparatus for acid-base titration. A NaOH solution is added from the buret to a KHP solution in an Erlenmeyer flask. A faint pink color appears when the equivalence point is reached. If your solution turns fuchsia, you have gone past the equivalence point

126. Acids and Bases 2009-2010126 The neutralization reaction between NaOH and KHP is one of the simplest types of acid-base neutralization known. A neutralization reaction is a reaction between an acid and a base. Aqueous strong acid-strong base reactions produce water and a salt, (an ionic compound made up of a cation other than H + and an anion other than OH - or O -2 )

127. Acids and Bases 2009-2010127 The reaction between KHP and sodium hydroxide is: KHC 8 H 4 O 4 (aq) + NaOH(aq)  KNaC 8 H 4 O 4 (aq) + H 2 O(l) acidbasesaltwater

128. Acids and Bases 2009-2010128 You can use the following format to solve acid-base neutralization problems MMmol LL AcidBase

129. Acids and Bases 2009-2010129 MMmol LL AcidBase What is the molarity of the acid if 16.1 mL of 0.610 M NaOH was required to neutralize 20.0 mL of H 2 SO 4 ?.0161.610.0200

130. Acids and Bases 2009-2010130 MMmol LL AcidBase.0161.610.0200.0161 L NaOH x.00982 1 mol H 2 SO 4 2 mol NaOH =.00491 mol H 2 SO 4.610 mol NaOH 1L NaOH x.00491 H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O

131. Acids and Bases 2009-2010131 MMmol LL AcidBase.0161.610.0200.00982.00491 M =.00491 mol =.0200 L.0246 M.0246

132. Acids and Bases 2009-2010132 The reaction between HCl, a strong acid, and NaOH, a strong base, can be represented by HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) The pH profile of a titration of this neutralization reaction is known as a titration curve. Titration Curve

133. 133 Consider the addition of 0.10 M NaOH solution (from a buret) to an Erlenmeyer flask containing 25 mL of 0.10 M HCl. Before the addition of NaOH, the pH of the acid is 1.00 When the NaOH is added, the pH of the solution increases slowly at first Near the equivalence point, the pH begins to rise steeply, and at the equivalence point the curve rises almost vertically Beyond the equivalence point, the pH again increases slowly with the addition of NaOH

134. Acids and Bases 2009-2010134 It is possible to calculate the pH of a solution at every stage of titration. What is the pH of the solution after the addition of 10.0 mL of 0.10 M NaOH to 25.0 mL of 0.10 M HCl?

135. Acids and Bases 2009-2010135 0.10 mol OH - = 1 L = 0.00100 mol OH - 0.00250 mol H + - 0.00100 mol OH - 0.00150 mol H + Excess acid 0.0100 L OH - x 0.10 mol H + = 1 L = 0.00250 mol H + 0.0250 L H + x Because both NaOH and HCl completely dissociate…

136. Acids and Bases 2009-2010136 To determine the pH, you need to calculate [H + ].0350 L.00150 mol H + =.043M = [H + ] pH = -log (.043) pH = 1.37 Total volume of the original HCl and NaOH

137. Acids and Bases 2009-2010137 At the equivalence point of a titration between a weak acid and a strong base, the pH will be greater than 7 because…

138. 138 CH 3 COOH + NaOH  CH 3 COONa + H 2 O CH 3 COONaNa + + CH 3 COO - This acetate ion has an affinity for the H + ion in the water, (thus leaving the OH - behind, making the solution basic.) at neutralization

139. Acids and Bases 2009-2010139 The flat portion of the titration curve before the equivalence point is called the buffer region. In this part of the titration, the acid and conjugate base are both present in significant concentrations and the solution resists changes in pH. The pk a of a weak acid can be determined experimentally

140. Acids and Bases 2009-2010140 In the middle of the buffer region lies the half-equivalence point. Here the volume of base added is half that required to reach the equivalence point (half the weak acid has been converted to the conjugate base). This means that the concentrations of weak acid and conjugate base are equal. If we examine the equilibrium expression at the half-equivalence point, we find something interesting… pH = pk a 25.0 mL 12.5 mL

141. Acids and Bases 2009-2010141 25.0 mL 12.5 mL At half-way to the equivalence point, pH = pk a Taking the negative log of both sides yields 10 -pH = k a This gives us an experimental way to determine the k a of a weak acid, and using a k a table, the identity of an unknown weak acid. 4.74 = pk a 10 -4.74 = k a k a = 1.8 x 10 -5 Using a data table, you could match the k a of 1.8 x 10 -5 to CH 3 COOH

142. Acids and Bases 2009-2010142 A slightly different curve results when you titrate a strong acid vs a weak base. At the equivalence point of a titration between a strong acid and a weak base, the pH will be less than 7, because…

143. Acids and Bases 2009-2010143 A weak base, like NH 3 would form an acidic salt when titrated vs HCl HCl + NH 3  NH 4 Cl NH 4 ClNH 3 + H + resulting in a pH of less than 7 is due to the presence of H + ions formed by the hydrolysis of NH 4 + and, at neutralization the NH 4 Cl would dissociate

144. 144 Acid-Base Indicators An indicator is usually a weak organic acid or base that has distinctly different colors in its nonionized (molecular) form and ionized form. The end point of a titration occurs when the indicator changes color. However, not all indicators change color at the same pH, so the choice of indicator for a particular titration depends on the nature of the acid and base used in the titration (that is, whether they are strong or weak). By choosing the proper indicator for a titration, we can use the end point to determine the equivalence point.

145. Acids and Bases 2009-2010145 Let us consider a weak monoprotic acid that we will call HIn. To be an effective indicator, HIn and its conjugate base, ___, must have distinctly different colors. In- HIn H + + In - One color A different color

146. Acids and Bases 2009-2010146 If the indicator is in a sufficiently acidic environment, the equilibrium, according to Le Chatelier’s principle, shifts to the __________ and the predominant color will be that of ______, left HIn HIn H + + In -

147. Acids and Bases 2009-2010147 In a basic environment, the equilibrium shifts to the right because… HIn H + + In - The H + and the OH - will form water, thus removing the H + from the system and therefore shifting the equilibrium to the right. The predominant color will then be that of In -.

148. Acids and Bases 2009-2010148 The end point of an indicator does not occur at a specific pH; rather, there is a range of pH within which the end point will occur. In practice, we choose an indicator whose end point lies on the steep part of the titration curve. Because the equivalence point also lies on the steep part of the curve, this choice ensures that the pH at the equivalence point will fall within the range over which the indicator changes color.

149. Acids and Bases 2009-2010149 Some Common Acid-Base Indicators IndicatorIn acidIn basepH range Thymol blueRedYellow1.2 - 2.8 Bromophenol blueYellowBluish purple3.0 – 4.6 Methyl orangeOrangeYellow3.1 – 4.4 Methyl redRedYellow4.2 – 6.3 Chlorophenol blueYellowRed4.8 – 6.4 Bromothymol blueYellowBlue6.0 – 7.6 Cresol redYellowRed7.2 – 8.8 PhenolphthaleinColorlessReddish pink8.3 – 10.0 The pH range is defined as the range over which the indicator changes from the acid color to the base color.