2. 1 HOMEWORK Page 704 Day 1 #29, 31, 35, 37, 39, 43, 44, 45 Day 2 #47, 49, 51, 53, 57, 61, 63 Day 3 #65, 67, 69, 71, 75, 77, 79,83, 97,89 Day 4 #93,95,97.99.101,111,113,117

3. 2 The Chemistry of Acids and Bases

4. 3 Acid and Bases

5. 4

6. 5

7. 6 Acids Have a sour taste. Vinegar is a solution of acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Have a bitter taste. Feel slippery. Many soaps contain bases. Bases

8. 7 Some Properties of Acids þ Produce H + (as H 3 O + ) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) þ Taste sour þ Corrode metals þ Electrolytes þ React with bases to form a salt and water þ pH is less than 7 þ Turns blue litmus paper to red “Blue to Red A-CID”

9. 8 Some Properties of Bases  Produce OH - ions in water  Taste bitter, chalky  Are electrolytes  Feel soapy, slippery  React with acids to form salts and water  pH greater than 7  Turns red litmus paper to blue “Basic Blue”

10. 9 Some Common Bases NaOHsodium hydroxidelye KOHpotassium hydroxideliquid soap Ba(OH) 2 barium hydroxidestabilizer for plastics Mg(OH) 2 magnesium hydroxide“MOM” Milk of magnesia Al(OH) 3 aluminum hydroxideMaalox (antacid) Al(OH) 3 aluminum hydroxideMaalox (antacid)

11. 10 Acid/Base definitions Definition #1: Arrhenius (traditional) Acids – produce H + ions (or hydronium ions H 3 O + ) Bases – produce OH - ions (problem: some bases don’t have hydroxide ions!)

12. 11 Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water

13. 12 Acid/Base Definitions Definition #2: Brønsted – Lowry Acids – proton donor Bases – proton acceptor A “proton” is really just a hydrogen atom that has lost it’s electron!

14. 13 A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor acid conjugate base base conjugate acid

15. 14 ACID-BASE THEORIES The Brønsted definition means NH 3 is a BASE in water — and water is itself an ACID

16. 15 Conjugate Pairs

17. 16 Learning Check! Label the acid, base, conjugate acid, and conjugate base in each reaction: HCl + OH -  Cl - + H 2 O H 2 O + H 2 SO 4  HSO 4 - + H 3 O +

18. 17 Acids & Base Definitions Lewis acid - a substance that accepts an electron pair Lewis base - a substance that donates an electron pair Definition #3 – Lewis

19. 18 Formation of hydronium ion is also an excellent example. Lewis Acids & Bases Electron pair of the new O-H bond originates on the Lewis base.Electron pair of the new O-H bond originates on the Lewis base.

20. 19 Lewis Acid/Base Reaction

21. 20 Lewis Acid-Base Interactions in Biology The heme group in hemoglobin can interact with O 2 and CO.The heme group in hemoglobin can interact with O 2 and CO. The Fe ion in hemoglobin is a Lewis acidThe Fe ion in hemoglobin is a Lewis acid O 2 and CO can act as Lewis basesO 2 and CO can act as Lewis bases Heme group

22. 21 The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H + (or OH - ) ion. Under 7 = acid 7 = neutral Over 7 = base

23. 22 pH Scale

24. 23 pH of Common Substances

25. 24 Strong Acids: Perchloric HClO4 Chloric, HClO 3 Hydrobromic, HBr Hydrochloric, HCl Hydroiodic, HI Nitric, HNO 3 Sulfuric, H 2 SO 4 Strong Acids: 100% ionized (completely dissociated) in water. HCl + H 2 O  H 3 O + + Cl -

26. 25 What is a strong Base? A base that is completely dissociated in water (highly soluble). NaOH(s)  Na + + OH - Strong Bases: Group 1A metal hydroxides (LiOH, NaOH, KOH, RbOH, CsOH) Heavy Group 2A metal hydroxides [Ca(OH) 2, Sr(OH) 2, and Ba(OH) 2 ]

27. 26 Weak Acids & Bases: “The Rest”

28. 27 Strong Acids: 100% ionized (completely dissociated) in water. HCl + H 2 O  H 3 O + + Cl - Note the “one way arrow”. Weak Acids: Only a small % (dissociated) in water. HC 2 H 3 O 2 + H 2 O  H 3 O + + C 2 H 3 O 2 - Note the “2-way” arrow. Why are they different?

29. 28 Strong Acids: HCl ADD WATER to MOLECULAR ACID (H 2 O)

30. 29 Strong Acids: (H 2 O) H3O+H3O+ H3O+H3O+ H3O+H3O+ H3O+H3O+ H3O+H3O+ Cl - Note: No HCl molecules remain in solution, all have been ionized in water.

31. 30 HC 2 H 3 O 2  H 3 0 + C 2 H 3 O 2 - (H 2 O) Weak Acid Ionization: Note: At any given time only a small portion of the acid molecules are ionized and since reactions are running in BOTH directions the mixture composition stays the same. This gives rise to an Equilbrium expression, K a H 3 0 + C 2 H 3 O 2 - HC 2 H 3 O 2

32. 31 HNO 3, HCl, H 2 SO 4 and HClO 4 are among the only known strong acids. Strong and Weak Acids/Bases The strength of an acid (or base) is determined by the amount of IONIZATION.

33. 32 Strong and Weak Acids/Bases Generally divide acids and bases into STRONG or WEAK ones.Generally divide acids and bases into STRONG or WEAK ones. STRONG ACID: HNO 3 (aq) + H 2 O (l) ---> H 3 O + (aq) + NO 3 - (aq) HNO 3 is about 100% dissociated in water.

34. 33 Weak acids are much less than 100% ionized in water.Weak acids are much less than 100% ionized in water. One of the best known is acetic acid = CH 3 CO 2 H Strong and Weak Acids/Bases

35. 34 Strong Base: 100% dissociated in water.Strong Base: 100% dissociated in water. NaOH (aq) ---> Na + (aq) + OH - (aq) NaOH (aq) ---> Na + (aq) + OH - (aq) Strong and Weak Acids/Bases Other common strong bases include KOH and Ca(OH) 2. CaO (lime) + H 2 O --> Ca(OH) 2 (slaked lime) Ca(OH) 2 (slaked lime) CaO

36. 35 More About Water H 2 O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION Equilibrium constant for water = K w K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C

37. 36 More About Water K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C In a neutral solution [H 3 O + ] = [OH - ] so K w = [H 3 O + ] 2 = [OH - ] 2 and so [H 3 O + ] = [OH - ] = 1.00 x 10 -7 M Autoionization

38. 37 Write equations for the following 1. sulfuric acid in water 2. ammonia in water 3. calcium hydroxide in water 4. hydroflouric acid in water 5. hydrogen sulfide in water

39. 38 Calculating the pH pH = - log [H+] (Remember that the [ ] mean Molarity) Example: If [H + ] = 1 X 10 -10 pH = - log 1 X 10 -10 pH = - (- 10) pH = 10 Example: If [H + ] = 1.8 X 10 -5 pH = - log 1.8 X 10 -5 pH = - (- 4.74) pH = 4.74

40. 39 Try These! Find the pH of these: 1) A 0.15 M solution of Hydrochloric acid 2) A 3.00 X 10 -7 M solution of Nitric acid

41. 40 pH calculations – Solving for H+ If the pH of Coke is 3.12, [H + ] = ??? Because pH = - log [H + ] then - pH = log [H + ] - pH = log [H + ] Take antilog (10 x ) of both sides and get 10 -pH = [H + ] [H + ] = 10 -3.12 = 7.6 x 10 -4 M *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button

42. 41 pH calculations – Solving for H+ A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution?A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution?

43. 42 pOH Since acids and bases are opposites, pH and pOH are opposites!Since acids and bases are opposites, pH and pOH are opposites! pOH does not really exist, but it is useful for changing bases to pH.pOH does not really exist, but it is useful for changing bases to pH. pOH looks at the perspective of a basepOH looks at the perspective of a base pOH = - log [OH - ] Since pH and pOH are on opposite ends, pH + pOH = 14

44. 43 pH [H + ] [OH - ] pOH

45. 44 [H 3 O + ], [OH - ] and pH What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10 -3 M) pOH = - log 0.0010 pOH = - log 0.0010 pOH = 3 pOH = 3 pH = 14 – 3 = 11 OR K w = [H 3 O + ] [OH - ] [HO + ] = 1.0 x 10 -11 M [H 3 O + ] = 1.0 x 10 -11 M pH = - log (1.0 x 10 -11 ) = 11.00

46. 45 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H + ion concentration of the rainwater? The OH - ion concentration of a blood sample is 2.5 x 10 -7 M. What is the pH of the blood?

47. 46 [OH - ] [H + ] pOH pH 10 -pOH 10 -pH -Log[H + ] Log[OH - ] -Log[OH - ] 14 - pOH 14 - pH 1.0 x 10 -14 [OH - ] [OH - ] 1.0 x 10 -14 [H + ] [H + ]

48. 47 Calculating [H 3 O + ], pH, [OH - ], and pOH A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H 3 O + ], pH, [OH - ], and pOH of the two solutions at 25°C.

49. 48 Calculating [H 3 O + ], pH, [OH - ], and pOH What is the [H 3 O + ], [OH - ], and pOH of a solution with pH = 3.67? Is this an acid, base, or neutral?

50. 49 Weak base: less than 100% ionized in waterWeak base: less than 100% ionized in water One of the best known weak bases is ammonia NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq) Strong and Weak Acids/Bases

51. 50 Weak Bases

52. 51 Equilibria Involving Weak Acids and Bases Consider acetic acid, HC 2 H 3 O 2 (HOAc) HC 2 H 3 O 2 + H 2 O  H 3 O + + C 2 H 3 O 2 - Acid Conj. base (K is designated K a for ACID) K gives the ratio of ions (split up) to molecules (don’t split up)

53. 52 Ionization Constants for Acids/Bases Acids ConjugateBases Increase strength

54. 53 Equilibrium Constants for Weak Acids (Yes, all weak acids do this – DO NOT endeavor to make this complicated!) Weak acid has K a < 1 Leads to small [H 3 O + ] and a pH of 2 - 7 Ka = X 2 / or-x x = H + and conjugate base

55. 54 Calculate the H + and pH for 0.250 M HC 2 H 3 O 3

56. 55 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH.

57. 56 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc][H 3 O + ][OAc - ] [HOAc][H 3 O + ][OAc - ]initialchangeequilib 1.0000 -x+x+x 1.00-xxx

58. 57 Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H 2 O  BH + + OH - (Yes, all weak bases do this – DO NOT endeavor to make this complicated!) Kb = X 2 /or-x X = OH - and conjuate acid

59. 58 Equilibrium Constants for Weak Bases Weak base has K b < 1 Leads to small [OH - ] and a pH of 12 - 7

60. 59 NOTE Ka x Kb = Kw pH + pOH = 14 H 3 O + x OH - = 1 x 10 -14

61. 60 NOTE On the AP test you might get the Ka but need the Kb. If you are given an acid, you need Ka. If given a base, you need Kb. You must know if you are dealing with an acid or base.

62. 61 Relation of K a, K b, [H 3 O + ] and pH

63. 62 Equilibria Involving A Weak Acid Step 2. Write K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. This is a quadratic. Solve using quadratic formula. or you can make an approximation if x is very small! (Rule of thumb: 10 -5 or smaller is ok)

64. 63 Equilibria Involving A Weak Acid Step 3. Solve K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. First assume x is very small because K a is so small. Now we can more easily solve this approximate expression.

65. 64 Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. x = [ H 3 O + ] = [ OAc - ] = 4.2 x 10 -3 M pH = - log [ H 3 O + ] = -log (4.2 x 10 -3 ) = 2.37

66. 65 Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO 2 H.

67. 66 Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO 2 H. HCO 2 H + H 2 O  HCO 2 - + H 3 O + HCO 2 H + H 2 O  HCO 2 - + H 3 O + K a = 1.8 x 10 -4 Approximate solution [H 3 O + ] = 4.2 x 10 -4 M, pH = 3.37 [H 3 O + ] = 4.2 x 10 -4 M, pH = 3.37 Exact Solution [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M pH = 3.47 pH = 3.47

68. 67 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH -

69. 68 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib 0.01000 -x+x+x 0.010 - xx x

70. 69 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib 0.01000 -x+x+x 0.010 - xx x

71. 70 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib 0.01000 -x+x+x 0.010 - xx x

72. 71 Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Kb = [NH 4 + ][OH - ] [NH 3 ]

73. 72 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 2. Solve the equilibrium expression Assume x is small, so x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M and [NH 3 ] = 0.010 - 4.2 x 10 -4 ≈ 0.010 M The approximation is valid!

74. 73 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 3. Calculate pH [OH - ] = 4.2 x 10 -4 M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = 10.63

75. 74 Types of Acid/Base Reactions: Summary

76. 75 pH testing There are several ways to test pHThere are several ways to test pH –Blue litmus paper (red = acid) –Red litmus paper (blue = basic) –pH paper (multi-colored) –pH meter (7 is neutral, 7 base) –Universal indicator (multi-colored) –Indicators like phenolphthalein –Natural indicators like red cabbage, radishes

77. 76 Paper testing Paper tests like litmus paper and pH paperPaper tests like litmus paper and pH paper –Put a stirring rod into the solution and stir. –Take the stirring rod out, and place a drop of the solution from the end of the stirring rod onto a piece of the paper –Read and record the color change. Note what the color indicates. –You should only use a small portion of the paper. You can use one piece of paper for several tests.

78. 77 pH paper

79. 78 Behavior of Salts (Acidic or Basic) Cations are acidic unless they come from a strong base. Anions are basic unless they come from a strong acid.

80. 79 Behavior of Salts Acidic, basic, or neutral? NaF NaOCl KBr NH 4 F Li 2 S Li 2 CO 3 Fe(NO 3 ) 3

81. 80 NOTE Ka > Kb acidic Kb < Ka basic Ka = Kb neutral

82. 81 Behavior of Salts WRITE EQUATIONS to show WHY THE FOLLOWING ARE ACIDIC, BASIC, OR NEUTRAL. NaF NaOCl KBr NH 4 NO 2 NH 4 F Li 2 S ZnCl 2 Li 2 CO 3

83. 82 Amphoteric Salts HPO4 2- HS -

84. 83 % DISSOCIATION OF ACIDS & Bases % Diss of an acid = H + / original conc x 100 % Diss of a base = OH - / original conc x 100

85. 84 Determine the pH, pOH, hydrogen ion concentration, and hydroxide ion concentration for 0.125 M Na 2 CO 3.

86. 85 Determine the pH for 0.015 M sodium fluoride.

87. 86 Determine the percent ionization for 0.250 M KNO 2.

88. 87 What about polyprotic acids?

89. 88 What about polyprotic acids? For example sulfuric acid: 11. Determine the pH for 1.2 M H 2 SO 4

90. 89 What about polyprotic acids? For example sulfuric acid: 11. Determine the pH for 0.012 M H 2 SO 4

91. 90 pH meter Tests the voltage of the electrolyteTests the voltage of the electrolyte Converts the voltage to pHConverts the voltage to pH Very cheap, accurateVery cheap, accurate Must be calibrated with a buffer solutionMust be calibrated with a buffer solution

92. 91 pH indicators Indicators are dyes that can be added that will change color in the presence of an acid or base. Some indicators only work in a specific range of pH Once the drops are added, the sample is ruined Some dyes are natural, like radish skin or red cabbage

93. 92 ACID-BASE REACTIONS Titrations H 2 C 2 O 4 (aq) + 2 NaOH(aq) ---> acid base acid base Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) Carry out this reaction using a TITRATION. Oxalic acid, H 2 C 2 O 4

94. 93 Setup for titrating an acid with a base

95. 94 TitrationTitration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3.Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) This is called NEUTRALIZATION. This is called NEUTRALIZATION.

96. 95 35.62 mL of NaOH is neutralized with 25.2 mL of 0.0998 M HCl by titration to an equivalence point. What is the concentration of the NaOH? LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.

97. 96 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!

98. 97 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add?

99. 98 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do ? How much water is added? The important point is that ---> moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution moles of NaOH in FINAL solution

100. 99 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Amount of NaOH in original solution = M V = M V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH) / (0.50 M) = 0.30 L or 300 mL

101. 100 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

102. 101 A shortcut A shortcut M 1 V 1 = M 2 V 2 Preparing Solutions by Dilution

103. 102 You try this dilution problem You have a stock bottle of hydrochloric acid, which is 12.1 M. You need 400 mL of 0.10 M HCl. How much of the acid and how much water will you need?