1. 1 Chapter 10 Acids and Bases

2. 2 Arrhenius acids  Produce H + ions in water. H 2 O HCl H + (aq) + Cl - (aq)  Are electrolytes.  Have a sour taste.  Corrode metals.  React with bases to form salts and water. Acids

3. 3 Arrhenius bases  Produce OH - ions in water.  Taste bitter or chalky.  Are electrolytes.  Feel soapy and slippery.  React with acids to form salts and water. Bases

4. 4 Comparing Acids and Bases

5. 5 Identify each as a characteristic of an A) acid or B) base ____1. Has a sour taste. ____2. Produces OH - in aqueous solutions. ____3. Has a chalky taste. ____4. Is an electrolyte. ____5. Produces H + in aqueous solutions. Learning Check

6. 6 Identify each as a characteristic of an A) acid or B) base A 1. Has a sour taste. B 2. Produces OH - in aqueous solutions. B 3. Has a chalky taste. A, B 4. Is an electrolyte. A 5. Produces H + in aqueous solutions. Solution

7. 7 Names of Acids Acids with H and one nonmetal are named with the prefix hydro- and end with -ic acid. HClhydrochloric acid Acids with H and a polyatomic ion are named by changing the end of an –ate ion to -ic acid and an –ite ion to -ous acid. HClO 3 chloric acid HClO 2 chlorous acid

8. 8 Some Acids and Their Anions

9. 9 Name each of the following as acids: A. HBr 1. bromic acid 2. bromous acid 3. hydrobromic acid B. H 2 CO 3 1. carbonic acid 2. hydrocarbonic acid 3. carbonous acid Learning Check

10. 10 A. HBr3. hydrobromic acid The name of an acid with H and one nonmetal begins with the prefix hydro- and ends with -ic acid. B. H 2 CO 3 1. carbonic acid An acid with H and a polyatomic ion is named by changing the end of an –ate ion to -ic acid. Solution

11. 11  Bases with OH - ions are named as the hydroxide of the metal in the formula. NaOHsodium hydroxide KOH potassium hydroxide Ba(OH) 2 barium hydroxide Al(OH) 3 aluminum hydroxide Fe(OH) 3 iron (III) hydroxide Some Common Bases

12. 12 Match the formulas with the names: A. ___ HNO 2 1) hydrochloric acid B. ___Ca(OH) 2 2) sulfuric acid C. ___H 2 SO 4 3) sodium hydroxide D. ___HCl4) nitrous acid E. ___NaOH5) calcium hydroxide Learning Check

13. 13 Match the formulas with the names: A. 4 HNO 2 4) nitrous acid B. 5 Ca(OH) 2 5) calcium hydroxide C. 2 H 2 SO 4 2) sulfuric acid D. 1 HCl 1) hydrochloric acid E. 3 NaOH 3) sodium hydroxide Solution

14. 14 According to the Br Ø nsted-Lowry theory, Acids are hydrogen ion (H + ) donors. Bases are hydrogen ion (H + ) acceptors. donor acceptor hydronium ion HCl + H 2 O H 3 O + + Cl - + - + + Br Ø nsted-Lowry Acids and Bases

15. 15 When NH 3 dissolves in water, a few NH 3 molecules react with water to form ammonium ion NH 4 + and a hydroxide ion. NH 3 + H 2 O NH 4 + (aq) + OH - (aq) acceptor donor + - + + NH 3, A Bronsted-Lowry Base

16. 16 Conjugate Acids and Bases An acid that donates H + forms a conjugate base. A base that accepts a H + forms a conjugate acid. In an acid-base reaction, there are two conjugate acid-base pairs. acid 1 conjugate base 1 + + base 2 conjugate acid 2 HF H2OH2O H3O+H3O+ F -

17. 17 Conjugate Acid-Base Pairs A conjugate acid-base pair is two substances related by a loss or gain of H +. + + acid 1– conjugate base 1 base 2– conjugate acid 2 HF H2OH2O H3O+H3O+ F - HF, F - H 2 O, H 3 O +

18. 18 Learning Check A. Write the conjugate base of the following: 1. HBr 2. H 2 S 3. H 2 CO 3 B. Write the conjugate acid of the following: 1. NO 2 - 2. NH 3 3. OH -

19. 19 Solution A. Write the conjugate base of the following: 1. HBrBr - 2. H 2 SHS - 3. H 2 CO 3 HCO 3 - B. Write the conjugate acid of the following: 1. NO 2 - HNO 2 2. NH 3 NH 4 + 3. OH - H 2 O

20. 20 Learning Check Identify the following that are acid-base conjugate pairs. 1. HNO 2, NO 2 - 2. H 2 CO 3, CO 3 2- 3. HCl, ClO 4 - 4. HS -, H 2 S 5. NH 3, NH 4 +

21. 21 Solution Identify the following that are acid-base conjugate pairs. 1. HNO 2, NO 2 - 4. HS -, H 2 S 5. NH 3, NH 4 +

22. 22 Learning Check A. The conjugate base of HCO 3 - is 1. CO 3 2- 2. HCO 3 - 3. H 2 CO 3 B. The conjugate acid of HCO 3 - is 1. CO 3 2- 2. HCO 3 - 3. H 2 CO 3 C. The conjugate base of H 2 O is 1. OH - 2. H 2 O 3. H 3 O + D. The conjugate acid of H 2 O is 1. OH - 2. H 2 O 3. H 3 O +

23. 23 Solution A. The conjugate base of HCO 3 - is 1. CO 3 2- B. The conjugate acid of HCO 3 - is 3. H 2 CO 3 C. The conjugate base of H 2 O is 1. OH - D. The conjugate acid of H 2 O is 3. H 3 O +

24. 24 Strong acids completely ionize (100%) in aqueous solutions. HCl + H 2 O H 3 O + (aq) + Cl - (aq) (100 % ions)  Strong bases completely (100%) dissociate into ions in aqueous solutions. H 2 O NaOH Na + (aq) + OH - (aq) (100 % ions) Strengths of Acids and Bases

25. 25 Strong and Weak Acids In an HCl solution, the strong acid HCl dissociates 100%. A solution of the weak acid CH 3 COOH contains mostly molecules and a few ions.

26. 26 Only a few acids are strong acids. The conjugate bases of strong acids are weak bases. Strong Acids

27. 27 Most acids are weak acids. The conjugate bases of weak acids are strong bases. Weak Acids

28. 28 Most bases in Groups 1A and 2A are strong bases. They include LiOH, NaOH, KOH, and Ca(OH) 2 Most other bases are weak bases. Strong Bases

29. 29 Learning Check Identify each of the following as a strong or weak acid or base. A. HBr B. HNO 2 C. NaOH D. H 2 SO 4 E. Cu(OH) 2

30. 30 Solution Identify each of the following as a strong or weak acid or base. A. HBrstrong acid B. HNO 2 weak acid C. NaOHstrong base D. H 2 SO 4 strong acid E. Cu(OH) 2 weak base

31. 31 Learning Check A. Identify the stronger acid in each pair. 1. HNO 2 or H 2 S 2. HCO 3 - or HBr 3. H 3 PO 4 or H 3 O + B. Identify the strong base in each pair. 1. NO 3 - or F - 2. CO 3 2- or NO 2 - 3. OH - or H 2 O

32. 32 Solution A. Identify the stronger acid in each pair. 1. HNO 2 2. HBr 3. H 3 O + B. Identify the stronger base in each pair. 1. F - 2. CO 3 2- 3. OH -

33. 33 Strong Acids In water, the dissolved molecules of a strong acid are essentially all separated into ions. The concentrations of H 3 O + and the anion (A - ) are large.

34. 34 Weak Acids In weak acids, The equilibrium favors the undissociated (molecular) form of the acid. The concentrations of the H 3 O + and the anion (A - ) are small.

35. 35 Dissociation Constants In a weak acid, the rate of the dissociation of the acid is equal to the rate of the association. HA + H 2 O H 3 O + + A - The equilibrium expression for a weak acid is K eq = [H 3 O + ][A - ] [HA][H 2 O]

36. 36 Acid Dissociation Constants The [H 2 O] is large and remains unchanged. It is combined with the K eq to write the acid dissociation constant, K a. K a = K eq [H 2 O] = [H 3 O + ][A - ] [HA]

37. 37 Writing K a for a Weak Acid Write the K a for H 2 S. 1. Write the equation for the dissociation of H 2 S. H 2 S + H 2 O H 3 O + + HS - 2. Set up the K a expression K a = [H 3 O + ][HS - ] [H 2 S]

38. 38 Learning Check Write the K a for HCN.

39. 39 Solution Write the K a for HCN. Write the equation for the dissociation of HCN. HCN+ H 2 O H 3 O + + CN - Set up the K a expression K a = [H 3 O + ][CN - ] [HCN] Note: K a = K eq [H 2 O]

40. 40 Chapter 10 Acids and Bases 10.5 Ionization of Water

41. 41  In water, H + is transferred from one H 2 O molecule to another.  One water molecule acts as an acid, while another acts as a base.  H 2 O + H 2 O H 3 O + + OH -........ : O : H + : O : H H : O : H + + : O : H -........ H H H water molecules hydronium hydroxide ion (+) ion (-) Ionization of Water

42. 42  In pure water, the ionization of molecules produces small, but equal quantities of H 3 O + and OH - ions. H 2 O + H 2 O H 3 O + + OH -  Molar concentrations are indicated as [H 3 O + ] and [OH - ]. [H 3 O + ] = 1.0 x 10 -7 M [OH - ] = 1.0 x 10 -7 M Pure Water is Neutral

43. 43  Adding an acid to pure water increases the [H 3 O + ].  In acids, [H 3 O + ] exceeds 1.0 x 10 -7 M.  As [H 3 O + ] increases, [OH - ] decreases. Acidic Solutions

44. 44  Adding a base to pure water increases the [OH - ].  In bases, [OH - ] exceeds 1.0 x 10 -7 M.  As [OH - ] increases, [H 3 O + ] decreases. Basic Solutions

45. 45 Comparison of [H 3 O + ] and [OH - ]

46. 46 The ion product constant, K w, for water is the product of the concentrations of the hydronium and hydroxide ions. K w = [ H 3 O + ][ OH - ] We can obtain the value of K w using the concentrations in pure water. K w = [1.0 x 10 -7 ][ 1.0 x 10 -7 ] = 1.0 x 10 -14 Ion Product of Water, K w

47. 47 K w in Acids and Bases In neutral, acidic, and basic solutions, the K w is equal to 1.0 x 10 -14.

48. 48 What is the [H 3 O + ] of a solution if [OH - ] is 1.0 x 10 -8 M? Rearrange the K w expression for [H 3 O + ]. K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 [H 3 O + ] = 1.0 x 10 -14 [OH - ] [H 3 O + ] = 1.0 x 10 -14 = 1.0 x 10 -6 M 1.0 x 10 - 8 Calculating [H 3 O + ]

49. 49 The [H 3 O + ] of lemon juice is 1.0 x 10 -3 M. What is the [OH - ] of the solution? 1) 1.0 x 10 3 M 2) 1.0 x 10 -11 M 3) 1.0 x 10 11 M Learning Check

50. 50 The [H 3 O + ] of lemon juice is 1.0 x 10 - 3 M. What is the [OH - ] of the solution? 2) 1.0 x 10 -11 M Rearrange the K w to solve for [OH - ] K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 [OH - ] = 1.0 x 10 -14 = 1.0 x 10 -11 M 1.0 x 10 - 3 Solution

51. 51 The [OH - ] of a solution is 5 x 10 -2 M. What is the [H 3 O + ] of the solution? 1) 2 x 10 - 2 M 2) 1 x 10 2 M 3) 2 x 10 -13 M Learning Check

52. 52 The [OH - ] of a water solution is 5 x 10 -2 M. What is the [H 3 O + ] of the solution? 3) 2 x 10 -13 M [ H 3 O + ] = 1.0 x 10 -14 5 x 10 - 2 Solution

53. 53 A. What is the [OH - ] if the [H 3 O + ] is 1 x 10 - 4 M? Identify the solution as acidic, basic, or neutral. 1) 1 x 10 -6 M 2) 1 x 10 -8 M 3) 1 x 10 -10 M B. What is [H 3 O + ] if the [OH - ] is 5 x 10 -9 M? Identify the solution as acidic, basic, or neutral. 1) 1 x 10 - 6 M 2) 2 x 10 - 6 M 3) 2 x 10 -7 M Learning Check

54. 54 K w = [H 3 O + ][OH - ] = 1.0 x 10 14 A. 3) [OH - ] = 1.0 x 10 -14 = 1.0 x 10 -10 1.0 x 10 - 4 acidic B. 2) [H 3 O + ] = 1.0 x 10 -14 = 2 x 10 - 6 5 x 10 - 9 acidic Solution

55. 55 Chapter 10 Acids and Bases 10.6 The pH Scale

56. 56 pH Scale The pH scale:  Is used to indicate the acidity of a solution.  Has values that usually range from 0 to 14.  Indicates an acidic solution when the values are less than 7.  Indicates a neutral solution with a pH of 7.  Indicates a basic solution when the values are greater than 7.

57. 57 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Neutral [H 3 O + ]>[OH - ] [H 3 O + ] = [OH - ] [H 3 O + ]<[OH - ] Acidic Basic pH Range

58. 58

59. 59 Identify each solution as 1. acidic 2. basic3. neutral A. ___ HCl with a pH = 1.5 B. ___ Pancreatic fluid [H 3 O + ] = 1 x 10 -8 M C. ___ Sprite soft drink pH = 3.0 D. ___ pH = 7.0 E. ___ [OH - ] = 3 x 10 -10 M F. ___ [H 3 O + ] = 5 x 10 -12 Learning Check

60. 60 Identify each solution as 1. acidic 2. basic3. neutral A. 1 HCl with a pH = 1.5 B. 2 Pancreatic fluid [H 3 O + ] = 1 x 10 -8 M C. 1 Sprite soft drink pH = 3.0 D. 3 pH = 7.0 E. 1 [OH - ] = 3 x 10 -10 M F. 2 [H 3 O + ] = 5 x 10 -12 Solution

61. 61 Testing the pH of Solutions The pH of solutions can be determined using a a) pH meter, b) pH paper, and c) indicators that have specific colors at different pHs.

62. 62 pH is:  Defined as the negative log of the hydrogen ion concentration. pH = - log [H 3 O + ]  The value of the negative exponent of [H 3 O + ] for concentrations with a coefficient of 1. [H 3 O + ] =1 x 10 -4 pH = 4.0 [H 3 O + ] =1 x 10 -11 pH = 11.0 pH Scale

63. 63 A calculator can be used to find the pH of a solution with a [H 3 O + ] of 1 x 10 -3 as follows. 1. Enter 1 x 10 -3 by pressing 1 (EE) 3 (+/-). The EE key gives an exponent of 10 and the +/- key changes the sign. 2. Press the log key to obtain log 1 x 10 -3. log (1 x 10 -3 ) = -3 3. Press the +/- key to multiply by –1. (-3) +/- = 3 Calculations of pH

64. 64 Significant Figures in pH When expressing log values, the number of decimal places in the pH is equal to the number of significant figures in the coefficient of [H 3 O + ]. [H 3 O + ] = 1 x 10 -4 pH = 4.0 [H 3 O + ] = 1.0 x 10 -6 pH = 6.00 [H 3 O + ] = 2.4 x 10 -8 pH = 7.62

65. 65 A. The [H 3 O + ] of tomato juice is 2 x 10 -4 M. What is the pH of the solution? 1) 4.02) 3.73) 10.3 B. The [OH - ] of a solution is 1.0 x 10 -3 M. What is the pH of the solution? 1) 3.002) 11.003) -11.00 Learning Check

66. 66 A. 2) 3.7 pH = - log [ 2 x 10 -4 ] = 3.7 2 (EE) 4 (+/-) log (+/-) B. 2) 11.00 Use the K w to obtain [H 3 O + ] = 1.0 x 10 -11 pH = - log [1.0 x 10 - 11 ] = - 11.00 (+/-) = 11.00 Solution

67. 67 Calculating [H 3 O + ] from pH The [H 3 O + ] can be expressed with the pH as the negative power of 10. [H 3 O + ] = 1 x 10 -pH If the pH is 3.0, the [H 3 O + ] = 1 x 10 -3 On a calculator 1. Enter the pH value 3.0 2. Use (+/-) to change sign 3.0 (+/-) = -3.0 3. Use the inverse log key (or 10 x ) to obtain the [H 3 0 + ]. =1 x 10 -3 M

68. 68 A. What is the [H 3 O + ] of a solution with a pH of 10.0? 1) 1 x 10 -4 M2) 1 x 10 10 M 3) 1 x 10 -10 M B. What is the [H 3 O + ] of a solution with a pH of 5.7? 1) 2 x 10 - 6 M2) 5 x 10 5 M 3) 1 x 10 -5 M Learning Check

69. 69 A. What is the [H 3 O + ] of a solution with a pH of 10.0? 3) 1 x 10 -10 M1 x 10 -pH B. What is the [H 3 O + ] of a solution with a pH of 5.7? 1) 2 x 10 - 6 M 5.7 (+/-) inv log (or 10 x ) = 2 x 10 -6 Solution

70. 70 pOH The pOH of a solution:  Is related to the [OH - ]. Is similar to pH. pOH = - log [OH - ] Added to the pH value is equal to 14.0. pH + pOH = 14.0

71. 71 pH and pOH The sum of the pH and pOH values is equal to 14.0 for any aqueous solution.

72. 72 Learning Check What is the pH and the pOH of coffee if the [H 3 O + ] is 1 x 10 -5 M? 1) pH = 5.0 pOH =7.0 2) pH = 7.0 pOH = 9.0 3) pH = 5.0pOH = 9.0

73. 73 Solution What is the pH and the pOH of coffee if the [H 3 O + ] is 1 x 10 -5 M? 3) pH = 5.0pOH = 9.0 pH = -log [1 x 10 -5 ] = -(-5.0) = 5.0 pH + pOH = 14.0 pOH = 14.0 – pH = 14.0 – 5.0 = 9.0 or [OH - ] = 1 x 10 -9 pOH = - log [1 x 10 -9 ] = -(-9.0) = 9.0

74. 74 Chapter 10 Acids and Bases 10.7 Reactions of Acids and Bases 10.8 Acidity of Salt Solutions

75. 75 Acids and Metals Acids react with metals such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn to produce hydrogen gas and the salt of the metal. 2K + 2HCl 2KCl + H 2 (g) Zn + 2HCl ZnCl 2 + H 2 (g)

76. 76 Acids and Carbonates Acids react with carbonates and hydrogen carbonates to produce carbon dioxide gas, a salt, and water. 2HCl + CaCO 3 CO 2 + CaCl 2 + H 2 O HCl + NaHCO 3 CO 2 + NaCl + H 2 O

77. 77 Learning Check Write the products of the following reactions of acids: A. Zn + 2 HCl B. MgCO 3 + 2HCl

78. 78 Solution Write the products of the following reactions of acids: A. Zn(s) + 2 HCl(aq) ZnCl 2 (aq) + H 2 (g) B. MgCO 3 + 2HCl MgCl 2 (aq) + CO 2 (g) + H 2 O(l)

79. 79 An acid such as HCl produces H 3 O + and a base such as NaOH provides OH -. HCl + H 2 OH 3 O + + Cl - NaOHNa + + OH - In a neutralization reaction, the H 3 O + and the OH - combine to form water. H 3 O + + OH - 2 H 2 O The positive ion (metal) and the negative ion (nonmetal) are the ions of a salt. Neutralization Reactions

80. 80 In the equation for neutralization, an acid and a base produce a salt and water. NaOH + HCl NaCl + H 2 O base acid salt water Ca(OH) 2 + 2 HCl CaCl 2 + 2H 2 O base acid salt water Neutralization Equations

81. 81 Write the balanced equation for the neutralization of magnesium hydroxide and nitric acid. 1. Write the formulas of the acid and base. Mg(OH) 2 + HNO 3 2. Balance to give equal OH - and H +. Mg(OH) 2 + 2 HNO 3 3. Write the products (a salt and water) and balance: Mg(OH) 2 + 2HNO 3 Mg(NO 3 ) 2 + 2H 2 O Balancing Neutralization Reactions

82. 82 Write strong acids, bases, and salts as ions. H + + Cl - + Na + + OH - Na + + Cl - + H 2 O Cross out matched ions. H + + Cl - + Na + + OH - Na + + Cl - + H 2 O Write a net ionic reaction. H + + OH - H 2 O Ionic Equations for Neutralization

83. 83 Select the correct group of coefficients for the following neutralization equations. A. __ HCl + __ Al(OH) 3 __AlCl 3 + __ H 2 O 1) 1, 3, 3, 1 2) 3, 1, 1, 1 3) 3, 1, 1 3 B. __ Ba(OH) 2 + __H 3 PO 4 __Ba 3 (PO 4 ) 2 + __ H 2 O 1) 3, 2, 2, 2 2) 3, 2, 1, 6 3) 2, 3, 1, 6 Learning Check

84. 84 A. 3) 3, 1, 1 3 3HCl + 1Al(OH) 3 1AlCl 3 + 3H 2 O B. 2) 3, 2, 1, 6 3Ba(OH) 2 + 2H 3 PO 4 1Ba 3 (PO 4 ) 2 + 6H 2 O Solution

85. 85  Antacids neutralize stomach acid (HCl). Alka-Seltzer NaHCO 3, citric acid, aspirin Chooz, Tums CaCO 3 Di-gel CaCO 3 and Mg(OH) 2 Gelusil, Maalox, Al(OH) 3 and Mg(OH) 2 Mylanta Milk of Magnesia Mg(OH) 2 Rolaids AlNa(OH) 2 CO 3 Antacids

86. 86 Learning Check Write the neutralization reactions for stomach acid HCl and Mylanta.

87. 87 Solution Write the neutralization reactions for stomach acid HCl and Mylanta. Mylanta: Al(OH) 3 and Mg(OH) 2 3HCl + Al(OH) 3 AlCl 3 + 3H 2 O 2HCl + Mg(OH) 2 MgCl 2 + 2H 2 O

88. 88 Some Salt Solutions are Neutral A salt containing the ions of a strong acid and a strong base forms a neutral solution. The metal ion does not produce H + and the negative ion does not attract H + from water. For example, KNO 3 contains ions from a strong base (KOH) and a strong acid (HNO 3 ). A solution of KNO 3 would be neutral.

89. 89 Some Salt Solutions are Basic A salt containing the ions of a weak acid and a strong base forms a basic solution. The ion of the weak acid attracts H + from water to form a basic solution. For example, KHCO 3 contains ions from a strong base (KOH) and a weak acid (H 2 CO 3 ). A solution of KHCO 3 is basic. HCO 3 - + H 2 O H 2 CO 3 + OH -

90. 90 Some Salt Solutions are Acidic A salt containing the ions of a strong acid and a weak base forms an acidic solution. The ion of the weak base produces H + in water to form an acidic solution. For example, NH 4 Cl contains ions from a weak base (NH 4 OH) and a strong acid (HCl). A solution of NH 4 Cl is acidic. NH 4 + + H 2 O H 3 O + + NH 3

91. 91 Properties of Salts in Water

92. 92 Learning Check Predict whether a solution of each salt will be 1) acidic2) basicor 3) neutral A. Li 2 S B. Mg(NO 3 ) 2 C. NH 4 Cl

93. 93 Solution Predict whether a solution of each salt will be 1) acidic2) basicor 3) neutral A. Li 2 S 2) basic; S 2- forms HS - and OH - B. Mg(NO 3 ) 2 3) neutral C. NH 4 Cl 1) acidic; NH 4 + produces H 3 O +

94. 94 Chapter 10 Acids and Bases 10.9 Buffers

95. 95 When an acid or base is added to water, the pH changes drastically. A buffer solution resists a change in pH when an acid or base is added. Buffers

96. 96 Buffers Absorb H 3 O + or OH - from foods and cellular processes to maintain pH. Are important in the proper functioning of cells and blood. In blood maintain a pH close to 7.4. A change in the pH of the blood affects the uptake of oxygen and cellular processes. Buffers

97. 97 A buffer solution Contains a combination of acid-base conjugate pairs. Contains a weak acid and a salt of the conjugate base of that acid. Typically has equal concentrations of a weak acid and its salt. May also contain a weak base and a salt with the conjugate acid. Components of a Buffer

98. 98 The acetic acid/acetate buffer contains acetic acid (CH 3 COOH) and sodium acetate (CH 3 COONa). The salt produces sodium and acetate ions. CH 3 COONa CH 3 COO - + Na + The salt provides a higher concentration of the conjugate base CH 3 COO - than the weak acid. CH 3 COOH + H 2 O CH 3 COO - + H 3 O + Large amount Large amount Buffer Action

99. 99 The function of the weak acid is to neutralize a base. The acetate ion in the product adds to the available acetate. CH 3 COOH + OH - CH 3 COO - + H 2 O Function of the Weak Acid

100. 100 The function of the acetate ion CH 3 COO - (conjugate base) is to neutralize H 3 O + from acids. The weak acid product adds to the weak acid available. CH 3 COO - + H 3 O + CH 3 COOH + H 2 O Function of the Conjugate Base

101. 101 The weak acid in a buffer neutralizes base. The conjugate base in the buffer neutralizes acid. The pH of the solution is maintained. Summary of Buffer Action

102. 102 Learning Check Does each combination make a buffer solution? 1) yes2) noExplain. A. HCl and KCl B. H 2 CO 3 and NaHCO 3 C. H 3 PO 4 and NaCl D. CH 3 COOH and CH 3 COOK

103. 103 Solution Does each combination make a buffer solution? 1) yes2) no Explain. A. HCl + KClno; HCl is a strong acid. B. H 2 CO 3 + NaHCO 3 yes; weak acid and its salt. C. H 3 PO 4 + NaCl no; NaCl does not contain a conjugate base of H 3 PO 4. D. CH 3 COOH + CH 3 COOK yes; weak acid and its salt.

104. 104 pH of a Buffer The [H 3 O + ] in the K a expression is used to determine the pH of a buffer. Weak acid + H 2 O H 3 O + + Conjugate base K a = [H 3 O + ][conjugate base] [weak acid] [H 3 O + ] = K a x [weak acid] [conjugate base] pH = -log [H 3 O + ]

105. 105 Calculation of Buffer pH The weak acid H 2 PO 4 - in a blood buffer H 2 PO 4 - /HPO 4 2- has K a = 6.2 x 10 -8. What is the pH of the buffer if it is 0.20 M in both H 2 PO 4 - and HPO 4 2- ? [H 3 O + ] = K a x [H 2 PO 4 - ] [HPO 4 2- ] [H 3 O + ] = 6.2 x 10 -8 x [0.20 M] = 6.2 x 10 -8 [0.20 M] pH = -log [6.2 x 10 -8 ] = 7.17

106. 106 Learning Check What is the pH of a H 2 CO 3 buffer that is 0.20 M H 2 CO 3 and 0.10 M HCO 3 - ? K a (H 2 CO 3 ) = 4.3 x 10 -7

107. 107 Solution What is the pH of a H 2 CO 3 buffer that is 0.20 M H 2 CO 3 and 0.10 M HCO 3 - ? K a (H 2 CO 3 ) = 4.3 x 10 -7 [H 3 O + ] = K a x [H 2 CO 3 ] [HCO 3 - ] [H 3 O + ] = 4.3 x 10 -7 x [0.20 M] = 8.6 x 10 -7 [0.10 M] pH = -log [8.6 x 10 -7 ] = 6.07

108. 108 Chapter 10 Acids and Bases 10.10Dilutions 10.11Acid-Base Titration

109. 109 Dilution Diluting a solution Is the addition of water. Decreases concentration. ConcentratedDilutedSolution

110. 110 Diluting a Solution In a dilution, the amount (moles or grams) of solute does not change. Amount solute (initial) =Amount solute (dilute) A dilution increases the volume of the solution. Amount solute Volume (initial) Volume (increased) As a result, the concentration of the solution decreases.

111. 111 Calculations with Dilutions What is the molarity of an HCl solution prepared by adding 0.50 L of water to 0.10 L of a 12 M HCl. 1. Calculate the moles of HCl. 0.10 L x 12 moles HCl = 1.2 moles HCl 1 L 2. Determine the volume after dilution. 0.10 L + 0.50 L = 0.60 L solution 3. Calculate the molarity of the diluted HCl. 1.2 moles HCl = 2.0 M HCl 0.60 L

112. 112 Dilution Equation The dilution calculation can be expressed as an equation. C 1 V 1 = C 1 V 2 C = concentration For molar concentration, the equation is M 1 V 1 = M 2 V 2 (moles) =(moles after dilution ) For percent concentration, % 1 V 1 = % 2 V 2 (grams) =(grams after dilution )

113. 113 Calculation How many milliliters of 6.0 M NaOH solution are needed to prepare 1.0 L of a 0.15 M NaOH solution? M 1 = 6.0 M M 2 = 1.5 M V 1 = ???V 2 = 1.0 L Rearrange the dilution equation for V 1 V 1 = M 2 V 2 = 0.15 M x 1.0 L M 1 6.0 M = 0.025 L x 1000 mL= 25 mL 1 L

114. 114 Learning Check What volume (mL) of a 3.0 M H 2 SO 4 solution can be prepared from 15 mL of 18 M H 2 SO 4 ? 1)90. mL 2) 27 mL 3) 15 mL

115. 115 Solution 1)90. mL M 1 = 18.0 M M 2 = 3.0 M V 1 = 15 mL (0.015 L)V 2 = ??? V 2 = M 1 V 1 = 18 M x 0.015 L = 0.090 L M 2 3.0 M 0.090 L x 1000 mL = 90. mL 1L

116. 116 Acid-Base Titration Titration is a laboratory procedure used to determine the molarity of an acid. In a titration, a base such as NaOH is added to a specific volume of an acid. Base (NaOH) Acid solution

117. 117 Indicator A few drops of an indicator is added to the acid in the flask. The indicator changes color when the base (NaOH) has neutralized the acid.

118. 118 End Point of Titration At the end point, the indicator has a permanent color. The volume of the base used to reach the end point is measured. The molarity of the acid is calculated using the neutralization equation for the reaction.

119. 119 Calculating Molarity What is the molarity of an HCl solution if 18.0 mL of a 0.25 M NaOH are required to neutralize 10.0 mL of the acid? 1. Write the neutralization equation. HCl + NaOH NaCl + H 2 O 2. Calculate the moles of NaOH. 18.0 mL NaOH x 1 L x 0.25 mole NaOH 1000 mL 1 L = 0.0045 moles NaOH

120. 120 Calculating Molarity (continued) 3. Calculate the moles of HCl. HCl + NaOH NaCl + H 2 O 0.0045 mole NaOH x 1 mole HCl 1 mole NaOH = 0.0045 moles HCl 4. Calculate the molarity of HCl. 10.0 mL HCl = 0.010 L HCl 0.0045 mole HCl = 0.45 M HCl 0.0100 L HCl

121. 121 Calculate the mL of 2.00 M H 2 SO 4 required to neutralize 50.0 mL of 1.00 M KOH. H 2 SO 4 + 2KOH K 2 SO 4 + 2H 2 O 1)12.5 mL 2) 50.0 mL 3) 200. mL Learning Check

122. 122 Calculate the mL of 2.00 M H 2 SO 4 required to neutralize 50.0 mL of 1.00 M KOH. H 2 SO 4 + 2KOH K 2 SO 4 + 2H 2 O 1)12.5 mL 0.0500 L x 1.00 mole KOH x 1 mole H 2 SO 4 x 1 L 2 mole KOH 1 L x 1000 mL = 12.5 mL 2 mole H 2 SO 4 1 L Solution

123. 123 A 25 mL sample of phosphoric acid is neutralized by 40. mL of 1.5 M NaOH. What is the molarity of the phosphoric acid solution? 3NaOH + H 3 PO 4 Na 3 PO 4 + 3H 2 O 1) 0.45 M 2) 0.80 M 3) 7.2 M Learning Check

124. 124 2) 0.80 M 0.040 L x 1.5 mole NaOH x 1 mole H 3 PO 4 1 L 3 mole NaOH = 0.020 mole H 3 PO 4 0.020 mole H 3 PO 4 = 0.80 mole/L = 0.80 M 0.025 L Solution