Chapter 16 Lecture- Acid/Base Equilibrium

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Chapter 16 lecture for AP chemistry

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Chapter 16 Lecture- Acid/Base Equilibrium

  1. 1. Chapter 16 Acid/Base Equilibrium Sections 16.1 - 16.3 Acid/Base Theories Read pages 669- 678 HOMEWORK Pg 712 #1, 2, 15, 16, 17, 19, 21, 25, 27, 29, 31 Acids and Bases
  2. 2. Some Definitions • Arrhenius Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions. Acids and Bases
  3. 3. Some Definitions • Arrhenius Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions. Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions. Acids and Bases
  4. 4. Acids and Bases
  5. 5. H+ and OH– Acids and Bases
  6. 6. Some Definitions • Brønsted–Lowry Acid: Proton donor Base: Proton acceptor Acids and Bases
  7. 7. A Brønsted–Lowry acid… …must have a removable (acidic) proton. A Brønsted–Lowry base… …must have a pair of nonbonding electrons. Acids and Bases
  8. 8. If it can be either… ...it is amphiprotic. HCO3 − HSO4 − H 2O Acids and Bases
  9. 9. What Happens When an Acid Dissolves in Water? • Water acts as a Brønsted–Lowry base and abstracts a proton (H+) from the acid. Acids and Bases
  10. 10. What Happens When an Acid Dissolves in Water? • Water acts as a Brønsted–Lowry base and abstracts a proton (H+) from the acid. • As a result, the conjugate base of the acid and a hydronium ion are formed. Acids and Bases
  11. 11. Acids and Bases
  12. 12. NH3(aq) Acids and Bases
  13. 13. Is NH3 an acid, base, or could it be both? 1. Acid 2. Base 3. Both Acids and 13 Bases
  14. 14. Is NH3 an acid, base, or could it be both? 1. Acid 2. Base 3. Both In this chapter we learned that a more general definition for a base is a substance that can accept another proton, which is true for NH3 because of Acids and the lone electron pair on the N atom. 14 Bases
  15. 15. Is H2O an acid, base, or could it be both? 1. Acid 2. Base 3. Both Acids and 15 Bases
  16. 16. Is H2O an acid, base, or could it be both? As indicated in the 1. Acid equilibrium below, water is an amphoteric substance 2. Base that can either accept 3. Both another proton or donate a proton. H H + H H + _ H O H H O H O Acids + and 16 Bases
  17. 17. Is the ion PO4 3 an acid, base, or could it be both? - - O O 1. Acid - + - - - O P O 2. Base O P O - 3. Both O O Acids and 17 Bases
  18. 18. Is the ion PO4 3 an acid, base, or could it be both? - - O O 1. Acid - + - - - O P O 2. Base O P O - 3. Both O O Phosphate is a proton acceptor, regardless of which resonance structure Acids and is being considered. 18 Bases
  19. 19. Conjugate Acids and Bases: • From the Latin word conjugare, meaning “to join together.” Acids and Bases
  20. 20. Conjugate Acids and Bases: • From the Latin word conjugare, meaning “to join together.” • Reactions between acids and bases always yield their conjugate bases and acids. Acids and Bases
  21. 21. SAMPLE EXERCISE 16.1 Identifying Conjugate Acids and Bases (a) What is the conjugate base of each of the following acids: HClO4, H2S, PH4+, HCO3– ? (b) What is the conjugate acid of each of the following bases: CN–, SO42–, H2O, HCO3– ?
  22. 22. SAMPLE EXERCISE 16.1 Identifying Conjugate Acids and Bases (a) What is the conjugate base of each of the following acids: HClO4, H2S, PH4+, HCO3– ? (b) What is the conjugate acid of each of the following bases: CN–, SO42–, H2O, HCO3– ? conjugate base = parent substance minus one proton, conjugate acid = parent substance plus one proton.
  23. 23. SAMPLE EXERCISE 16.1 Identifying Conjugate Acids and Bases (a) What is the conjugate base of each of the following acids: HClO4, H2S, PH4+, HCO3– ? (b) What is the conjugate acid of each of the following bases: CN–, SO42–, H2O, HCO3– ? conjugate base = parent substance minus one proton, conjugate acid = parent substance plus one proton. (a)HClO4 less one proton (H+) is ClO4–. The other conjugate bases are HS–, PH3, and CO32–.
  24. 24. SAMPLE EXERCISE 16.1 Identifying Conjugate Acids and Bases (a) What is the conjugate base of each of the following acids: HClO4, H2S, PH4+, HCO3– ? (b) What is the conjugate acid of each of the following bases: CN–, SO42–, H2O, HCO3– ? conjugate base = parent substance minus one proton, conjugate acid = parent substance plus one proton. (a)HClO4 less one proton (H+) is ClO4–. The other conjugate bases are HS–, PH3, and CO32–. (b) CN– plus one proton (H+) is HCN. The other conjugate acids are HSO4–, H3O+, and H2CO3.
  25. 25. SAMPLE EXERCISE 16.1 Identifying Conjugate Acids and Bases (a) What is the conjugate base of each of the following acids: HClO4, H2S, PH4+, HCO3– ? (b) What is the conjugate acid of each of the following bases: CN–, SO42–, H2O, HCO3– ? conjugate base = parent substance minus one proton, conjugate acid = parent substance plus one proton. (a)HClO4 less one proton (H+) is ClO4–. The other conjugate bases are HS–, PH3, and CO32–. (b) CN– plus one proton (H+) is HCN. The other conjugate acids are HSO4–, H3O+, and H2CO3. Hydrogen carbonate ion (HCO3–) is amphiprotic: It can act as either an acid or a base.
  26. 26. SAMPLE EXERCISE 16.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO3–) is amphiprotic. (a) Write an equation for the reaction of HSO3– with water, in which the ion acts as an acid. (b) Write an equation for the reaction of HSO3– with water, in which the ion acts as a base. In both cases identify the conjugate acid-base pairs. Acids and Bases
  27. 27. SAMPLE EXERCISE 16.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO3–) is amphiprotic. (a) Write an equation for the reaction of HSO3– with water, in which the ion acts as an acid. (b) Write an equation for the reaction of HSO3– with water, in which the ion acts as a base. In both cases identify the conjugate acid-base pairs. a) Acids and Bases
  28. 28. SAMPLE EXERCISE 16.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO3–) is amphiprotic. (a) Write an equation for the reaction of HSO3– with water, in which the ion acts as an acid. (b) Write an equation for the reaction of HSO3– with water, in which the ion acts as a base. In both cases identify the conjugate acid-base pairs. a) The conjugate pairs in this equation are HSO3– (acid) and SO32– (conjugate base); and H2O (base) and H3O+ (conjugate acid). (b) Acids and Bases
  29. 29. SAMPLE EXERCISE 16.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO3–) is amphiprotic. (a) Write an equation for the reaction of HSO3– with water, in which the ion acts as an acid. (b) Write an equation for the reaction of HSO3– with water, in which the ion acts as a base. In both cases identify the conjugate acid-base pairs. a) The conjugate pairs in this equation are HSO3– (acid) and SO32– (conjugate base); and H2O (base) and H3O+ (conjugate acid). (b) Acids and Bases
  30. 30. SAMPLE EXERCISE 16.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO3–) is amphiprotic. (a) Write an equation for the reaction of HSO3– with water, in which the ion acts as an acid. (b) Write an equation for the reaction of HSO3– with water, in which the ion acts as a base. In both cases identify the conjugate acid-base pairs. a) The conjugate pairs in this equation are HSO3– (acid) and SO32– (conjugate base); and H2O (base) and H3O+ (conjugate acid). (b) The conjugate pairs in this equation are H2O (acid) and OH– (conjugate base), and HSO3– (base) and H2SO3 (conjugateAcids and acid). Bases
  31. 31. Acid and Base Strength • Substances with negligible acidity do not dissociate in water. Their conjugate bases are exceedingly strong. Acids and Bases
  32. 32. Strong acid Weak acid Negligible acidity Acids and Bases
  33. 33. Strong acid Weak acid Negligible acidity HNO3 is a strong acid, meaning NO3– has negligible base strength. Acids and Bases
  34. 34. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq) Acids and Bases
  35. 35. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq) H2O is a much stronger base than Cl−, so the equilibrium lies so far to the right Acids and K is not measured (K>>1). Bases
  36. 36. Acid and Base Strength C2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Acids and Bases
  37. 37. Acid and Base Strength C2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1). Acids and Bases
  38. 38. SAMPLE EXERCISE 16.3 Predicting the Position of a Proton-Transfer Equilibrium For the following proton-transfer reaction, predict whether the equilibrium lies predominantly to the left (that is, Kc < 1) or to the right (Kc > 1): Acids and Bases
  39. 39. For the following proton-transfer reaction, predict whether the equilibrium lies predominantly to the left (that is, Kc < 1) or to the right (Kc > 1): This is a proton-transfer reaction, and the position of equilibrium will favor the proton going to the stronger of two bases. Solve: CO32– is a stronger base than SO42–. CO32– will get the proton preferentially to become HCO3–, while SO42– will remain mostly unprotonated. The resulting equilibrium will lie to the right, favoring Acids and products (that is, Kc > 1). Bases
  40. 40. Autoionization of Water • As we have seen, water is amphoteric. Acids and Bases
  41. 41. Autoionization of Water • As we have seen, water is amphoteric. • In pure water, a few molecules act as bases and a few act as acids. Acids and Bases
  42. 42. Autoionization of Water • As we have seen, water is amphoteric. • In pure water, a few molecules act as bases and a few act as acids. H2O(l) + H2O(l) H3O+(aq) + OH−(aq) Acids and Bases
  43. 43. Autoionization of Water • As we have seen, water is amphoteric. • In pure water, a few molecules act as bases and a few act as acids. H2O(l) + H2O(l) H3O+(aq) + OH−(aq) • This is referred to as autoionization. Acids and Bases
  44. 44. Ion-Product Constant • The equilibrium expression for this process is Kc = [H3O+] [OH−] Acids and Bases
  45. 45. Ion-Product Constant • The equilibrium expression for this process is Kc = [H3O+] [OH−] • This special equilibrium constant is referred to as the ion-product constant for water, Kw. Acids and Bases
  46. 46. Ion-Product Constant • The equilibrium expression for this process is Kc = [H3O+] [OH−] • This special equilibrium constant is referred to as the ion-product constant for water, Kw. • At 25°C, Kw = 1.0 × 10−14 Acids and Bases
  47. 47. SAMPLE EXERCISE 16.4 Calculating [H+] for Pure Water Calculate the values of [H+] and [OH–] in a neutral solution at 25°C. Acids and Bases
  48. 48. SAMPLE EXERCISE 16.4 Calculating [H+] for Pure Water Calculate the values of [H+] and [OH–] in a neutral solution at 25°C. By definition, [H+] = [OH–] in a neutral solution. We will represent the concentration of [H+] and [OH–] in neutral solution with x. This gives In an acid solution [H+] is greater than 1.0 × 10–7 M ; in a basic solution [H+] is Acids and less than 1.0 × 10–7 M. Bases
  49. 49. PRACTICE EXERCISE Indicate whether solutions with each of the following ion concentrations are neutral, acidic, or basic: (a) [H+] = 4 × 10–9 M; (b) [OH–] = 1 × 10–7 M; (c) [OH–] = 7 × 10–13M. Acids and 38 Bases
  50. 50. PRACTICE EXERCISE Indicate whether solutions with each of the following ion concentrations are neutral, acidic, or basic: (a) [H+] = 4 × 10–9 M; (b) [OH–] = 1 × 10–7 M; (c) [OH–] = 7 × 10–13M. Answers: (a) basic, (b) neutral, (c) acidic Acids and 38 Bases
  51. 51. SAMPLE EXERCISE 16.5 Calculating [H+] from [OH–] Calculate the concentration of H+ (aq) in (a) a solution in which [OH–] is 0.010 M, (b) a solution in which [OH–] is 1.8 × 10–9 M. Acids and Bases
  52. 52. SAMPLE EXERCISE 16.5 Calculating [H+] from [OH–] Calculate the concentration of H+ (aq) in (a) a solution in which [OH–] is 0.010 M, (b) a solution in which [OH–] is 1.8 × 10–9 M. Solve: (a) Acids and Bases
  53. 53. SAMPLE EXERCISE 16.5 Calculating [H+] from [OH–] Calculate the concentration of H+ (aq) in (a) a solution in which [OH–] is 0.010 M, (b) a solution in which [OH–] is 1.8 × 10–9 M. Solve: (a) This solution is basic because Acids and Bases
  54. 54. SAMPLE EXERCISE 16.5 Calculating [H+] from [OH–] Calculate the concentration of H+ (aq) in (a) a solution in which [OH–] is 0.010 M, (b) a solution in which [OH–] is 1.8 × 10–9 M. Solve: (a) This solution is basic because (b) In this instance Acids and Bases
  55. 55. SAMPLE EXERCISE 16.5 Calculating [H+] from [OH–] Calculate the concentration of H+ (aq) in (a) a solution in which [OH–] is 0.010 M, (b) a solution in which [OH–] is 1.8 × 10–9 M. Solve: (a) This solution is basic because (b) In this instance This solution is acidic because Acids and Bases
  56. 56. PRACTICE EXERCISE Calculate the concentration of OH– (aq) in a solution in which (a) [H+] = 2 × 10–6 M; (b) [H+] = [OH–]; (c) [H+] = 100 × [OH–]. Acids and Bases
  57. 57. PRACTICE EXERCISE Calculate the concentration of OH– (aq) in a solution in which (a) [H+] = 2 × 10–6 M; (b) [H+] = [OH–]; (c) [H+] = 100 × [OH–]. Answers: (a) 5 × 10–9 M, (b) 1.0 × 10–7 M, (c) 1.0 × 10–8 M Acids and Bases
  58. 58. Chapter 16 Acid/Base Equilibrium Sections 16.4 - 16.5 pH calcs Read pages 678 - 684 HOMEWORK Pg 714 #33, 35, 37, 39, 41, 43, 45, 47 Acids and Bases
  59. 59. Bellwork Predict the products of the acid-base reactions, and also predict whether the equilibrium lies to the left or the right of the equation. a)HCO3-(aq) + F-(aq) ⇌ b)C2H3O2-(aq) + H3O+(aq) ⇌ c) NH4+(aq) + OH-(aq) ⇌ Acids and Bases
  60. 60. pH pH is defined as the negative base-10 logarithm of the hydronium ion concentration. pH = −log [H3 O+] Acids and Bases
  61. 61. pH • In pure water, Kw = [H3O +] [OH−] = 1.0 × 10−14 Acids and Bases
  62. 62. pH • In pure water, Kw = [H3O +] [OH−] = 1.0 × 10−14 Because [H3O+] = [OH−] in pure water, [H3O+] = (1.0 × 10−14)1/2 = 1.0 × 10−7 Acids and Bases
  63. 63. pH • Therefore, in pure water, pH = −log (1.0 × 10−7) = 7.00 Acids and Bases
  64. 64. pH • Therefore, in pure water, pH = −log (1.0 × 10−7) = 7.00 • An acid has a higher [H3O+] than pure water, so its pH is <7 Acids and Bases
  65. 65. pH • Therefore, in pure water, pH = −log (1.0 × 10−7) = 7.00 • An acid has a higher [H3O+] than pure water, so its pH is <7 • A base has a lower [H3O+] than pure water, so its pH is >7. Acids and Bases
  66. 66. pH These are the pH values for several common substances. Acids and Bases
  67. 67. Acids and Bases
  68. 68. At pH = 7, [H+] = [OH–] Acids and Bases
  69. 69. At pH = 7, [H+] = [OH–] The pH increases as [OH–] increases. Acids and Bases
  70. 70. What is [H+] in an aqueous solution whose pH = 3.72? 1. 1.9 x 10-3 M 2. 5.1 x 10-3 M 3. 1.9 x 10-4 M 4. 5.1 x 10-4 M 5. 5.1 x 10-10 M Acids and 51 Bases
  71. 71. What is [H+] in an aqueous solution whose pH = 3.72? 1. 1.9 x 10-3 M 2. 5.1 x 10-3 M 3. 1.9 x 10-4 M 4. 5.1 x 10-4 M 5. 5.1 x 10-10 M 3.72 = -log[H+] -3.72 = -log[H +] Acids [H +] = 1.9 x 10- 4 M and 52 Bases
  72. 72. What is the pH of an aqueous solution of [OH-] = 6.0 x 10-3 M? 2.22 1. 3.22 2. 7.00 3. 10.78 4. 11.78 Acids and 53 Bases
  73. 73. What is the pH of an aqueous solution of [OH-] = 6.0 x 10-3 M? 2.22 There is more than one method to 1. 3.22 solve this problem. One is presented 2. 7.00 below: 3. 10.78 pOH = -log(6.0 x 10-3) 4. 11.78 = 2.22 pH = 14.00 - pOH = 14.00 - 2.22 Acids = 11.78 and 54 Bases
  74. 74. What is the approximate pH of an aqueous solution of 1x10-12 M HCl? 1. 2 2. 7 3. 12 4. 14 5. None of the above Acids and 55 Bases
  75. 75. What is the approximate pH of an aqueous solution of 1x10-12 M HCl? 1. 2 2. 7 3. 12 4. 14 5. None of the above This is an aqueous solution, therefore water will autoionize. The concentration of protons due to autoionization of water Acids and will dominate that of such a dilute acid. 56 Bases
  76. 76. Other “p” Scales • The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions). Acids and Bases
  77. 77. Other “p” Scales • The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions). • Some similar examples are pOH = −log [OH−] pKw = −log Kw Acids and Bases
  78. 78. Watch This! Because [H3O+] [OH−] = Kw = 1.0 × 10−14, we know that −log [H3O+] + −log [OH−] = −log Kw = 14.0 Acids and Bases
  79. 79. Watch This! Because [H3O+] [OH−] = Kw = 1.0 × 10−14, we know that −log [H3O+] + −log [OH−] = −log Kw = 14.0 Acids and Bases
  80. 80. Watch This! Because [H3O+] [OH−] = Kw = 1.0 × 10−14, we know that −log [H3O+] + −log [OH−] = −log Kw = 14.0 or, in other words, pH + pOH = pKw = 14.00 Acids and Bases
  81. 81. Acids and Bases
  82. 82. pH = 11.00, basic Acids and Bases
  83. 83. How Do We Measure pH? • For less accurate measurements, one can use Litmus paper • “Red” paper turns blue above ~pH = 8 • “Blue” paper turns red below ~pH = 5 An indicator
  84. 84. How Do We Measure pH? • For less accurate measurements, one can use Litmus paper • “Red” paper turns blue above ~pH = 8 • “Blue” paper turns red below ~pH = 5 An indicator
  85. 85. How Do We Measure pH? • For less accurate measurements, one can use Litmus paper • “Red” paper turns blue above ~pH = 8 • “Blue” paper turns red below ~pH = 5 An indicator
  86. 86. How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution. Acids and Bases
  87. 87. Strong Acids • You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. Acids and Bases
  88. 88. Strong Acids • You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. • These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. Acids and Bases
  89. 89. Strong Acids • You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. • These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. • For the monoprotic strong acids, [H3O+] = [acid]. Acids and Bases
  90. 90. SAMPLE EXERCISE 16.8 Calculating the pH of a Strong Acid What is the pH of a 0.040 M solution of HClO4? Acids and Bases
  91. 91. SAMPLE EXERCISE 16.8 Calculating the pH of a Strong Acid What is the pH of a 0.040 M solution of HClO4? HClO4 is a strong acid, so it is completely ionized. [H+] = [ClO4–] = 0.040 M. Because [H+] lies between 1 × 10–2 and 1 × 10–1 the pH will be between 2.0 and 1.0. Acids and Bases
  92. 92. SAMPLE EXERCISE 16.8 Calculating the pH of a Strong Acid What is the pH of a 0.040 M solution of HClO4? HClO4 is a strong acid, so it is completely ionized. [H+] = [ClO4–] = 0.040 M. Because [H+] lies between 1 × 10–2 and 1 × 10–1 the pH will be between 2.0 and 1.0. Acids pH = –log(0.040) = 1.40. and Bases
  93. 93. PRACTICE EXERCISE An aqueous solution of HNO3 has a pH of 2.34. What is the concentration of the acid? Acids and 71 Bases
  94. 94. PRACTICE EXERCISE An aqueous solution of HNO3 has a pH of 2.34. What is the concentration of the acid? Answer: 0.0046 M Acids and 71 Bases
  95. 95. Strong Bases • Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+). Acids and Bases
  96. 96. Strong Bases • Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+). • Again, these substances dissociate completely in aqueous solution. Acids and Bases
  97. 97. SAMPLE EXERCISE 16.9 Calculating the pH of a Strong Base What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2? Acids and Bases
  98. 98. SAMPLE EXERCISE 16.9 Calculating the pH of a Strong Base What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2? (a) NaOH dissociates in water to give Solve: one OH– ion per formula unit. Therefore, the OH– concentration is 0.028 M. Acids and Bases
  99. 99. SAMPLE EXERCISE 16.9 Calculating the pH of a Strong Base What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2? Acids and Bases
  100. 100. SAMPLE EXERCISE 16.9 Calculating the pH of a Strong Base What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2? (b) Ca(OH)2 is a strong base that dissociates in water to give two OH– ions per formula unit. Thus, the concentration of OH–(aq) is 2 × (0.0011M) = 0.0022 M. Acids and Bases
  101. 101. PRACTICE EXERCISE What is the concentration of a solution of (a) KOH for which the pH is 11.89; (b) Ca(OH)2 for which the pH is 11.68? Acids and Bases
  102. 102. PRACTICE EXERCISE What is the concentration of a solution of (a) KOH for which the pH is 11.89; (b) Ca(OH)2 for which the pH is 11.68? Answers: (a) 7.8 × 10–3 M, (b) 2.4 × 10–13 M Acids and Bases
  103. 103. Acids and Bases
  104. 104. – CH3 removes a proton from water – to form CH4 and OH Acids and Bases
  105. 105. Chapter 16 Acid/Base Equilibrium Sections 16.6 - 16.7 Weak acids & bases Read pages 684-696 HOMEWORK Pg 712 #3, 4, 6, 53, 55, 57, 63, 65, 71, 75, 77 Acids and Bases
  106. 106. Bellwork Calculate pH, pOH, [H+], and [OH-] for 1. 0.5M HNO3 2. 0.1 M Sr(OH)2 Acids and Bases
  107. 107. Dissociation Constants • For a generalized acid dissociation, HA(aq) + H2O(l) A−(aq) + H3O+(aq) the equilibrium expression would be [H3O+] [A−] Kc = [HA] Acids and Bases
  108. 108. Dissociation Constants • For a generalized acid dissociation, HA(aq) + H2O(l) A−(aq) + H3O+(aq) the equilibrium expression would be [H3O+] [A−] Kc = [HA] • This equilibrium constant is called the acid-dissociation constant, Ka. Acids and Bases
  109. 109. Dissociation Constants The greater the value of Ka, the stronger the acid. Acids and Bases
  110. 110. The [H+] in an 0.020 M solution of HNO2 is 3.0 x 10 -3 M. What is the N Ka of HNO2? O OH nitrous acid 1. 4.5 x 10-4 2. 6.0 x 10-5 3. 9.0 x 10-6 4. 1.5 x 10-1 5. None of Acids and Bases the above
  111. 111. The [H+] in an 0.020 M solution of HNO2 is 3.0 103 M. What is the Ka N of HNO2? O OH [H + ][A − ] Ka = nitrous acid [HA] [3.0 × 10−3 ]2 9.0 × 10−6 1. 4.5 x 10-4 Ka = = 0.020 0.020 2. 6.0 x 10-5 = 4.5 × 10−4 3. 9.0 x 10-6 Ka 4. 1.5 x 10-1 5. None of Acids and Bases the above
  112. 112. Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. Acids and Bases
  113. 113. Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. • We know that [H3O+] [COO−] Ka = [HCOOH] Acids and Bases
  114. 114. Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. Acids and Bases
  115. 115. Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. • To calculate Ka, we need the equilibrium concentrations of all three things. Acids and Bases
  116. 116. Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. • To calculate Ka, we need the equilibrium concentrations of all three things. • We can find [H3O+], which is the same as [HCOO−], from the pH. Acids and Bases
  117. 117. Calculating Ka from the pH pH = −log [H3O+] 2.38 = −log [H3O+] −2.38 = log [H3O+] 10−2.38 = 10log [H3O+] = [H3O+] 4.2 × 10−3 = [H3O+] = [HCOO−] Acids and Bases
  118. 118. Calculating Ka from pH Now we can set up a table… [HCOOH], M [H3O+], M [HCOO−], M Initially 0.10 0 0 Change −4.2 × 10-3 +4.2 × 10-3 +4.2 × 10−3 At 0.10 − 4.2 × 10−3 4.2 × 10−3 4.2 × 10−3 Equilibrium = 0.0958 = 0.10 Acids and Bases
  119. 119. Calculating Ka from pH [4.2 × 10−3] [4.2 × 10−3] Ka = [0.10] Acids and Bases
  120. 120. Calculating Ka from pH [4.2 × 10−3] [4.2 × 10−3] Ka = [0.10] = 1.8 × 10−4 Acids and Bases
  121. 121. PRACTICE EXERCISE Niacin, one of the B vitamins, has the following molecular structure: A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution? (b) What is the acid-dissociation constant, Ka, for niacin? Acids and Bases
  122. 122. PRACTICE EXERCISE Niacin, one of the B vitamins, has the following molecular structure: A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution? (b) What is the acid-dissociation constant, Ka, for niacin? Acids and Bases Answers: (a) 2.7%, (b) 1.5 × 10–5
  123. 123. Calculating Percent Ionization [H3O+]eq • Percent Ionization = × 100 [HA]initial Acids and Bases
  124. 124. Calculating Percent Ionization [H3O+]eq • Percent Ionization = × 100 [HA]initial • In this example [H3O+]eq = 4.2 × 10−3 M [HCOOH]initial = 0.10 M Acids and Bases
  125. 125. Calculating Percent Ionization 4.2 × 10−3 Percent Ionization = × 100 0.10 Acids and Bases
  126. 126. Calculating Percent Ionization 4.2 × 10−3 Percent Ionization = × 100 0.10 = 4.2% Acids and Bases
  127. 127. The pH of a 0.050 M weak acid is 3.00. What is the percentage ionization? 1. 0.10% 2. 0.20% 3. 1.0% 4. 2.0% 5. 3.0% Acids and 97 Bases
  128. 128. The pH of a 0.050 M weak acid is 3.00. What is the percentage ionization? 1. 0.10% Since pH = 3.00, 2. 0.20% [H+] = 1.0 × 10−3 M, so 3. 1.0% + [H ] eq % ionization = × 100% 4. 2.0% [HA]o 5. 3.0% % ionization = [1.0 ×10 -3 ] ×100% [0.050] % ionization = 2.0% Acids and 98 Bases
  129. 129. Calculating pH from Ka Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25°C. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Ka for acetic acid at 25°C is 1.8 × 10−5. Acids and Bases
  130. 130. Calculating pH from Ka The equilibrium constant expression is Acids and Bases
  131. 131. Calculating pH from Ka The equilibrium constant expression is [H3O+] [C2H3O2−] Ka = [HC2H3O2] Acids and Bases
  132. 132. Calculating pH from Ka We next set up a table… [C2H3O2], M [H3O+], M [C2H3O2−], M Initially 0.30 0 0 Change −x +x +x At 0.30 − x ≈ 0.30 x x Equilibrium Acids and Bases
  133. 133. Calculating pH from Ka We next set up a table… [C2H3O2], M [H3O+], M [C2H3O2−], M Initially 0.30 0 0 Change −x +x +x At 0.30 − x ≈ 0.30 x x Equilibrium We are assuming that x will be very small Acids compared to 0.30 and can, therefore, beBases and ignored.
  134. 134. Calculating pH from Ka Now, (x)2 1.8 × 10−5 = (0.30) Acids and Bases
  135. 135. Calculating pH from Ka Now, (x)2 1.8 × 10−5 = (0.30) (1.8 × 10−5) (0.30) = x2 5.4 × 10−6 = x2 2.3 × 10−3 = x Acids (Check to ensure that ionization is < 5% or the “simplifying and Bases assumption” is not valid
  136. 136. Calculating pH from Ka pH = −log [H3O+] pH = −log (2.3 × 10−3) pH = 2.64 Acids and Bases
  137. 137. Acids and Bases
  138. 138. because weak acids typically undergo very little ionization, often less than 1% in solution. Acids and Bases
  139. 139. PRACTICE EXERCISE The Ka for niacin (Practice Exercise 16.10) is 1.5 × 10–5. What is the pH of a 0.010 M solution of niacin? Acids and Bases
  140. 140. PRACTICE EXERCISE The Ka for niacin (Practice Exercise 16.10) is 1.5 × 10–5. What is the pH of a 0.010 M solution of niacin? Answer: 3.42 Acids and Bases
  141. 141. SAMPLE EXERCISE 16.12 Using Ka to Calculate Percent Ionization Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution. Acids and Bases
  142. 142. SAMPLE EXERCISE 16.12 Using Ka to Calculate Percent Ionization Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution. Acids and Bases
  143. 143. SAMPLE EXERCISE 16.12 Using Ka to Calculate Percent Ionization Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution. The equilibrium-constant expression is Acids and Bases
  144. 144. SAMPLE EXERCISE 16.12 Using Ka to Calculate Percent Ionization Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution. The equilibrium-constant expression is When we try solving this equation using the approximation 0.10 – x = 0.10 (that is, by neglecting the Acids concentration of acid that ionizes in comparison with the and initial concentration), we obtain Bases
  145. 145. SAMPLE EXERCISE 16.12 continued Because this value is greater than 5% of 0.10 M, we should work the problem without the approximation, using an equation-solving calculator or the quadratic formula. Rearranging our equation and writing it in standard quadratic form, we have Acids and Bases
  146. 146. SAMPLE EXERCISE 16.12 continued Because this value is greater than 5% of 0.10 M, we should work the problem without the approximation, using an equation-solving calculator or the quadratic formula. Rearranging our equation and writing it in standard quadratic form, we have This equation can be solved using the standard quadratic formula. Acids and Bases
  147. 147. SAMPLE EXERCISE 16.12 continued Because this value is greater than 5% of 0.10 M, we should work the problem without the approximation, using an equation-solving calculator or the quadratic formula. Rearranging our equation and writing it in standard quadratic form, we have This equation can be solved using the standard quadratic formula. Substituting the appropriate numbers gives Acids and Bases
  148. 148. SAMPLE EXERCISE 16.12 continued Because this value is greater than 5% of 0.10 M, we should work the problem without the approximation, using an equation-solving calculator or the quadratic formula. Rearranging our equation and writing it in standard quadratic form, we have This equation can be solved using the standard quadratic formula. Substituting the appropriate numbers gives Of the two solutions, only the one that gives a positive value for x is Acids and chemically reasonable. Thus, Bases
  149. 149. SAMPLE EXERCISE 16.12 continued From our result, we can calculate the percent of molecules ionized: Acids and Bases
  150. 150. SAMPLE EXERCISE 16.12 continued From our result, we can calculate the percent of molecules ionized: (b) Proceeding similarly for the 0.010 M solution, we have Acids and Bases
  151. 151. SAMPLE EXERCISE 16.12 continued From our result, we can calculate the percent of molecules ionized: (b) Proceeding similarly for the 0.010 M solution, we have Solving the resultant quadratic expression, we obtain Acids and Bases
  152. 152. SAMPLE EXERCISE 16.12 continued From our result, we can calculate the percent of molecules ionized: (b) Proceeding similarly for the 0.010 M solution, we have Solving the resultant quadratic expression, we obtain The percentage of molecules ionized is Acids and Bases
  153. 153. SAMPLE EXERCISE 16.12 continued From our result, we can calculate the percent of molecules ionized: (b) Proceeding similarly for the 0.010 M solution, we have Solving the resultant quadratic expression, we obtain The percentage of molecules ionized is Comment: Notice that if we do not use the quadratic formula to solve the problem properly, we calculate 8.2% ionization for (a) and 26% ionization for (b). Notice also that in diluting the solution by a factor of 10, the percentage of molecules ionized increases by a factor of 3. This result is in accord with what we see in Figure 16.9. It is also what we would expect from Le Châtelier’s principle. • Acids (Section 15.6) There are more “particles” or reaction components on the right side of the equation than and on the left. Dilution causes the reaction to shift in the direction of the larger number of particles Bases because this counters the effect of the decreasing concentration of particles.
  154. 154. PRACTICE EXERCISE In Practice Exercise 16.10, we found that the percent ionization of niacin (Ka = 1.5 × 10–5) in a 0.020 M solution is 2.7%. Calculate the percentage of niacin molecules ionized in a solution that is (a) 0.010 M, (b) 1.0 × 10–3 M. Acids and Bases
  155. 155. PRACTICE EXERCISE In Practice Exercise 16.10, we found that the percent ionization of niacin (Ka = 1.5 × 10–5) in a 0.020 M solution is 2.7%. Calculate the percentage of niacin molecules ionized in a solution that is (a) 0.010 M, (b) 1.0 × 10–3 M. Answers: (a) 3.8%, (b) 12% Acids and Bases
  156. 156. Polyprotic Acids • Have more than one acidic proton. Acids and Bases
  157. 157. Polyprotic Acids • Have more than one acidic proton. • If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation. Acids and Bases
  158. 158. Acids and Bases
  159. 159. This is the acid dissociation constant for the 3rd and final proton from H3PO4. Acids and Bases
  160. 160. SAMPLE EXERCISE 16.13 Calculating the pH of a Polyprotic Acid Solution The solubility of CO2 in pure water at 25°C and 0.1 atm pressure is 0.0037 M. The common practice is to assume that all of the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced by reaction between the CO2 and H2O: What is the pH of a 0.0037 M solution of H2CO3? Acids and Bases
  161. 161. SAMPLE EXERCISE 16.13 Calculating the pH of a Polyprotic Acid Solution The solubility of CO2 in pure water at 25°C and 0.1 atm pressure is 0.0037 M. The common practice is to assume that all of the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced by reaction between the CO2 and H2O: What is the pH of a 0.0037 M solution of H2CO3? H2CO3 is a diprotic acid; the two acid-dissociation constants, Ka1 and Ka2 (Table 16.3), differ by more than a factor of 103. Consequently, the pH can be determined by considering only Ka1, thereby Acids and treating the acid as if it were a monoprotic acid. Bases
  162. 162. The solubility of CO2 in pure water at 25°C and 0.1 atm pressure is 0.0037 M. The common practice is to assume that all of the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced by reaction between the CO2 and H2O: What is the pH of a 0.0037 M solution of H2CO3? Acids and Bases
  163. 163. The solubility of CO2 in pure water at 25°C and 0.1 atm pressure is 0.0037 M. The common practice is to assume that all of the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced by reaction between the CO2 and H2O: What is the pH of a 0.0037 M solution of H2CO3? Acids and Bases
  164. 164. The solubility of CO2 in pure water at 25°C and 0.1 atm pressure is 0.0037 M. The common practice is to assume that all of the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced by reaction between the CO2 and H2O: What is the pH of a 0.0037 M solution of H2CO3? Acids and Bases
  165. 165. Solving this equation using an equation-solving calculator, we get Acids and Bases
  166. 166. Solving this equation using an equation-solving calculator, we get Because Ka1 is small, we can make the simplifying approximation that x is small, so that Acids and Bases
  167. 167. Solving this equation using an equation-solving calculator, we get Because Ka1 is small, we can make the simplifying approximation that x is small, so that Acids and Bases
  168. 168. Solving this equation using an equation-solving calculator, we get Because Ka1 is small, we can make the simplifying approximation that x is small, so that Acids and Bases
  169. 169. Solving this equation using an equation-solving calculator, we get Because Ka1 is small, we can make the simplifying approximation that x is small, so that The small value of x indicates that our simplifying assumption was justified. Acids and Bases
  170. 170. Comment: If we were asked to solve for [CO32–], we would need to use Ka2. Using the values of [HCO3–] and [H+] calculated above, and setting [CO32–] = y
  171. 171. Comment: If we were asked to solve for [CO32–], we would need to use Ka2. Using the values of [HCO3–] and [H+] calculated above, and setting [CO32–] = y Assuming that y is small compared to 4.0 × 10–5, we have
  172. 172. Comment: If we were asked to solve for [CO32–], we would need to use Ka2. Using the values of [HCO3–] and [H+] calculated above, and setting [CO32–] = y Assuming that y is small compared to 4.0 × 10–5, we have The value calculated for y is indeed very small compared to 4.0 × 10–5, showing that our assumption was justified. It also shows that the ionization of HCO3– is negligible compared to that of H2CO3, as far as production of H+ is concerned. However, it is the only source of CO32–, which has a very low concentration in the solution. Our calculations thus tell us that in a solution of carbon dioxide in water, most of the CO2 is in the form of CO2 or H2CO3, a small fraction ionizes to form H+ and HCO3–, and an even smaller fraction ionizes to give CO32–. Notice also that [CO32–] is numerically equal to Ka2.
  173. 173. PRACTICE EXERCISE (a) Calculate the pH of a 0.020 M solution of oxalic acid (H2C2O4). (See Table 16.3 for Ka1 and Ka2.) (b) Calculate the concentration of oxalate ion, [C2O42–], in this solution. Acids and Bases
  174. 174. PRACTICE EXERCISE (a) Calculate the pH of a 0.020 M solution of oxalic acid (H2C2O4). (See Table 16.3 for Ka1 and Ka2.) (b) Calculate the concentration of oxalate ion, [C2O42–], in this solution. Answers: (a) pH = 1.80, (b) [C2O42–] = 6.4 × 10–5 M Acids and Bases
  175. 175. Weak Bases Bases react with water to produce hydroxide ion. Acids and Bases
  176. 176. Weak Bases The equilibrium constant expression for this reaction is [HB] [OH−] Kb = [B−] where Kb is the base-dissociation constant. Acids and Bases
  177. 177. Weak Bases Kb can be used to find [OH−] and, through it, pH. Acids and Bases
  178. 178. pH of Basic Solutions What is the pH of a 0.15 M solution of NH3? NH3(aq) + H2O(l) NH4+(aq) + OH−(aq) Acids and Bases
  179. 179. pH of Basic Solutions What is the pH of a 0.15 M solution of NH3? NH3(aq) + H2O(l) NH4+(aq) + OH−(aq) [NH4+] [OH−] Kb = = 1.8 × 10−5 [NH3] Acids and Bases
  180. 180. pH of Basic Solutions Tabulate the data. [NH3], M [NH4+], M [OH−], M Initially 0.15 0 0 At Equilibrium 0.15 - x ≈ 0.15 x x Acids and Bases
  181. 181. pH of Basic Solutions (x)2 1.8 × 10−5 = (0.15) (1.8 × 10−5) (0.15) = x2 2.7 × 10−6 = x2 1.6 × 10−3 = x2 Acids and Bases
  182. 182. pH of Basic Solutions Therefore, [OH−] = 1.6 × 10−3 M pOH = −log (1.6 × 10−3) pOH = 2.80 pH = 14.00 − 2.80 pH = 11.20 The value obtained for x is only about 1% of the NH3 concentration, 0.15 M. Therefore, neglecting Acids and x relative to 0.15 was justified. Bases
  183. 183. SAMPLE EXERCISE 16.15 Using pH to Determine the Concentration of a Salt A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a pH of 10.50. Calculate the number of moles of NaClO that were added to the water. ClO– ion is a weak base with Kb = 3.33 × 10–7 Acids and Bases
  184. 184. SAMPLE EXERCISE 16.15 Using pH to Determine the Concentration of a Salt A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a pH of 10.50. Calculate the number of moles of NaClO that were added to the water. ClO– ion is a weak base with Kb = 3.33 × 10–7 Acids and Bases
  185. 185. SAMPLE EXERCISE 16.15 Using pH to Determine the Concentration of a Salt Acids and Bases
  186. 186. SAMPLE EXERCISE 16.15 Using pH to Determine the Concentration of a Salt This concentration is high enough that we can neglect any OH– produced by the autoionization of H2O. Acids and Bases
  187. 187. SAMPLE EXERCISE 16.15 continued Acids and Bases
  188. 188. SAMPLE EXERCISE 16.15 continued Acids and Bases
  189. 189. SAMPLE EXERCISE 16.15 continued We say that the solution is 0.30 M in NaClO, even though some of the ClO– ions have reacted with water. Because the solution is 0.30 M in NaClO and the total volume of solution is 2.00 L, 0.60 mol Acids of NaClO is the amount of the salt that wasBases and added to the water.
  190. 190. PRACTICE EXERCISE A solution of NH3 in water has a pH of 11.17. What is the molarity of the solution? Acids and Bases
  191. 191. PRACTICE EXERCISE A solution of NH3 in water has a pH of 11.17. What is the molarity of the solution? Answer: 0.12 M Acids and Bases
  192. 192. Chapter 16 Acid/Base Equilibrium Sections 16.8 - 16.9 Ka from Kb pH of Salt solutions Read pages 696-702 HOMEWORK Pg 712 #5, 7, 79, 81, 83, 85, 87, 89 Acids and Bases
  193. 193. Ka and Kb Acids and Bases
  194. 194. Ka and Kb Ka and Kb are related in this way: Ka × Kb = Kw Therefore, if you know one of them, you can calculate the other. Acids and Bases
  195. 195. SAMPLE EXERCISE 16.16 Calculating Ka or Kb for a Conjugate Acid-Base Pair Calculate (a) the base-dissociation constant, Kb, for the fluoride ion (F–); (b) the acid-dissociation constant, Ka, for the ammonium ion (NH4+). Acids and Bases
  196. 196. SAMPLE EXERCISE 16.16 Calculating Ka or Kb for a Conjugate Acid-Base Pair Calculate (a) the base-dissociation constant, Kb, for the fluoride ion (F–); (b) the acid-dissociation constant, Ka, for the ammonium ion (NH4+). (a) Ka for the weak acid, HF, is given in Table 16.2 and Solve: Appendix D as Ka = 6.8 × 10–4 . We can use Equation 16.40 to calculate Kb for the conjugate base, F–: Acids and Bases
  197. 197. SAMPLE EXERCISE 16.16 Calculating Ka or Kb for a Conjugate Acid-Base Pair Calculate (a) the base-dissociation constant, Kb, for the fluoride ion (F–); (b) the acid-dissociation constant, Ka, for the ammonium ion (NH4+). (a) Ka for the weak acid, HF, is given in Table 16.2 and Solve: Appendix D as Ka = 6.8 × 10–4 . We can use Equation 16.40 to calculate Kb for the conjugate base, F–: Kb for NH3 is listed in Table 16.4 and in Appendix D as (b) Kb = 1.8 × 10–5. Using Equation 16.40, we can calculate Ka for the conjugate acid, NH4+: Acids and Bases
  198. 198. SAMPLE EXERCISE 16.16 continued PRACTICE EXERCISE (a)Which of the following anions has the largest base- dissociation constant: NO2–, PO43– , or N3– ? (b) The base quinoline has the following structure: Acids and Bases
  199. 199. SAMPLE EXERCISE 16.16 continued PRACTICE EXERCISE (a)Which of the following anions has the largest base- dissociation constant: NO2–, PO43– , or N3– ? (b) The base quinoline has the following structure: Its conjugate acid is listed in handbooks as having a pKa of 4.90. What is the base-dissociation constant for quinoline? Acids and Bases
  200. 200. SAMPLE EXERCISE 16.16 continued PRACTICE EXERCISE (a)Which of the following anions has the largest base- dissociation constant: NO2–, PO43– , or N3– ? (b) The base quinoline has the following structure: Its conjugate acid is listed in handbooks as having a pKa of 4.90. What is the base-dissociation constant for quinoline? Answers: (a) PO43–(Kb = 2.4 × 10–2), (b) 7.9 × 10–10 Acids and Bases
  201. 201. Reactions of Anions with Water Acids and Bases
  202. 202. Reactions of Anions with Water • Anions are bases. Acids and Bases
  203. 203. Reactions of Anions with Water • Anions are bases. • As such, they can react with water in a hydrolysis reaction to form OH− and the conjugate acid: X−(aq) + H2O(l) HX(aq) + OH−(aq) Acids and Bases
  204. 204. Reactions of Cations with Water Acids and Bases
  205. 205. Reactions of Cations with Water • Cations with acidic protons (like NH4+) will lower the pH of a solution. Acids and Bases
  206. 206. Reactions of Cations with Water • Cations with acidic protons (like NH4+) will lower the pH of a solution. • Most metal cations that are hydrated in solution also lower the pH of the solution. Acids and Bases
  207. 207. Effect of Cations and Anions Acids and Bases
  208. 208. Effect of Cations and Anions 1. An anion that is the conjugate base of a strong acid will not affect the pH. Acids and Bases
  209. 209. Effect of Cations and Anions 1. An anion that is the conjugate base of a strong acid will not affect the pH. 2. An anion that is the conjugate base of a weak acid will increase the pH. Acids and Bases
  210. 210. Effect of Cations and Anions 1. An anion that is the conjugate base of a strong acid will not affect the pH. 2. An anion that is the conjugate base of a weak acid will increase the pH. 3. A cation that is the conjugate acid of a weak base will decrease the pH. Acids and Bases
  211. 211. Effect of Cations and Anions Acids and Bases
  212. 212. Effect of Cations and Anions 4. Cations of the strong Arrhenius bases will not affect the pH. Acids and Bases
  213. 213. Effect of Cations and Anions 4. Cations of the strong Arrhenius bases will not affect the pH. 5. Other metal ions will cause a decrease in pH. Acids and Bases
  214. 214. Effect of Cations and Anions 4. Cations of the strong Arrhenius bases will not affect the pH. 5. Other metal ions will cause a decrease in pH. 6. When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the Ka and Kb values. Acids and Bases
  215. 215. Acids and Bases
  216. 216. NO3– does not change the pH. Acids and Bases
  217. 217. NO3– does not change the pH. CO32– raises the pH. Acids and Bases
  218. 218. Acids and Bases
  219. 219. K+ Acids and Bases
  220. 220. SAMPLE EXERCISE 16.17 Predicting the Relative Acidity of Salt Solutions List the following solutions in order of increasing pH: (i) 0.1 M Ba(C2H3O2)2, (ii) 0.1 M NH4Cl, (iii) 0.1 M NH3CH3Br, (iv) 0.1 M KNO3. Acids and Bases
  221. 221. SAMPLE EXERCISE 16.17 Predicting the Relative Acidity of Salt Solutions List the following solutions in order of increasing pH: (i) 0.1 M Ba(C2H3O2)2, (ii) 0.1 M NH4Cl, (iii) 0.1 M NH3CH3Br, (iv) 0.1 M KNO3. Solve: Solution (i) contains barium ions and acetate ions. Ba2+ is an ion of one of the heavy alkaline earth metals and will therefore not affect the pH (summary point 4). The anion, C2H3O2–, is the conjugate base of the weak acid HC2H3O2 and will hydrolyze to produce OH– ions, thereby making the solution basic (summary point 2). Solutions (ii) and (iii) both contain cations that are conjugate acids of weak bases and anions that are conjugate bases of strong acids. Both solutions will therefore be acidic. Solution (i) contains NH4+, which is the conjugate acid of NH3 (Kb = 1.8 × 10–5). Solution (iii) contains NH3CH3+, which is the conjugate acid of NH2CH3 (Kb = 4.4 × 10–4). Because NH3 has the smaller Kb and is the weaker of the two bases, NH4+ will be the stronger of the two conjugate acids. Solution (ii) will therefore be the more acidic of the two. Solution (iv) contains the K+ ion, which is the cation of the strong base KOH, and the NO3– ion, which is the conjugate base of the strong acid HNO3. Neither Acids and of the ions in solution (iv) will react with water to any appreciable extent, making Bases the solution neutral. Thus, the order of pH is 0.1 M NH4Cl < 0.1 M NH3CH3Br <
  222. 222. Acids and Bases
  223. 223. PRACTICE EXERCISE In each of the following, indicate which salt will form the more acidic (or less basic) 0.010 M solution: (a) NaNO3, Fe(NO3)3; (b) KBr, KBrO; (c) CH3NH3Cl, BaCl2, (d) NH4NO2, NH4NO3. Acids and Bases
  224. 224. PRACTICE EXERCISE In each of the following, indicate which salt will form the more acidic (or less basic) 0.010 M solution: (a) NaNO3, Fe(NO3)3; (b) KBr, KBrO; (c) CH3NH3Cl, BaCl2, (d) NH4NO2, NH4NO3. Acids and Answers: (a) Fe(NO3)3, (b) KBr, (c) CH3NH3Cl, (d) NH4NO3 Bases
  225. 225. SAMPLE EXERCISE 16.18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the salt Na2HPO4 will form an acidic solution or a basic solution on dissolving in water. Acids and Bases
  226. 226. SAMPLE EXERCISE 16.18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the salt Na2HPO4 will form an acidic solution or a basic solution on dissolving in water. Na+ ion will not affect pH, what will HPO4- do? Acids and Bases
  227. 227. SAMPLE EXERCISE 16.18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the salt Na2HPO4 will form an acidic solution or a basic solution on dissolving in water. Na+ ion will not affect pH, what will HPO4- do? The reaction with the larger equilibrium constant will dominate and determine whether the solution is acidic or basic. Acids and Bases
  228. 228. SAMPLE EXERCISE 16.18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the salt Na2HPO4 will form an acidic solution or a basic solution on dissolving in water. Na+ ion will not affect pH, what will HPO4- do? The reaction with the larger equilibrium constant will dominate and determine whether the solution is acidic or basic. From appendix, the value of Ka for the first equation is 4.2 × 10–13. Calculate the value of Kb for Equation #2 from the value of Ka for its conjugate acid, H2PO4– , using Ka × Kb = Kw Acids and Bases
  229. 229. SAMPLE EXERCISE 16.18 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the salt Na2HPO4 will form an acidic solution or a basic solution on dissolving in water. Na+ ion will not affect pH, what will HPO4- do? The reaction with the larger equilibrium constant will dominate and determine whether the solution is acidic or basic. From appendix, the value of Ka for the first equation is 4.2 × 10–13. Calculate the value of Kb for Equation #2 from the value of Ka for its conjugate acid, H2PO4– , using Ka × Kb = Kw Ka for H2PO4– is 6.2 × 10–8, so Kb = 1.6x10-7. The second equation has a bigger K so it will dominate and Acids and solution is basic. Bases
  230. 230. Chapter 16 Acid/Base Equilibrium Sections 16.10 - 16.11 Acid/Base structure vs. properties Lewis acids and bases Read pages 702-712 HOMEWORK Pg 716 # 8, 9, 10, 91, 93, 95, 99, 101,103 Acids and Bases
  231. 231. Bellwork Predict whether the dipotassium salt of citric acid (K2HC6H5O7) will form an acidic or basic solution in water
  232. 232. Bellwork Predict whether the dipotassium salt of citric acid (K2HC6H5O7) will form an acidic or basic solution in water Answer: acidic
  233. 233. Acid-Base Behavior & Chemical Structure FOR ANY H-X BOND • Polar bonds where H has a ∂+ are acidic. (X= non-metal) • Polar bonds where H has a ∂- are basic. (X= metal) • Non-polar bonds with H have no acid- base properties. (C-H bond) Acids and 144 Bases
  234. 234. Factors Affecting Acid Strength Acids and Bases
  235. 235. Factors Affecting Acid Strength • The weaker the H-X bond, the more acidic the compound. Acids and Bases
  236. 236. Factors Affecting Acid Strength • The weaker the H-X bond, the more acidic the compound. • Acidity increases from left to right across a row (↑ polarity) and from top to bottom down Acids and a group (↑atom size =↓ bond strength). Bases
  237. 237. Factors Affecting Acid Strength Acids and 146 Bases
  238. 238. Factors Affecting Acid Strength • The stability of the conjugate base will also affect the strength of an acid. Acids and 146 Bases
  239. 239. Factors Affecting Acid Strength • The stability of the conjugate base will also affect the strength of an acid. • Strong acids have very stable conjugate bases. Acids and 146 Bases
  240. 240. Factors Affecting Acid Strength • The stability of the conjugate base will also affect the strength of an acid. • Strong acids have very stable conjugate bases. Acids and 146 Bases
  241. 241. Factors Affecting Acid Strength • The stability of the conjugate base will also affect the strength of an acid. • Strong acids have very stable conjugate bases. • Acid strength is determined by all three factors: bond polarity, bond strength, and conjugate stability. Acids and 146 Bases
  242. 242. Acids and Bases
  243. 243. Moving down a column, H-X bond strength decreases; Acids and Bases
  244. 244. Moving down a column, H-X bond strength decreases; Moving left-to-right across a period, electronegativity increases. Acids and Bases
  245. 245. Compound that have -OH bound to an atom of low electronegativity (a metal) are bases that produce OH-. Other compounds have -OH groups where only the H+ comes off. They are oxyacids and they involve non-metals. Acids Sulfuric acid and 148 Bases
  246. 246. Factors Affecting Acid Strength In oxyacids, in which an OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid. Acids and Bases
  247. 247. Factors Affecting Acid Strength For a series of oxyacids, acidity increases with the number of oxygens. More EN means more polar Acids and Bases HX bond and more stable anion.
  248. 248. SAMPLE EXERCISE 16.19 Predicting Relative Acidities from Composition and Structure Arrange the compounds in each of the following series in order of increasing acid strength: (a) AsH3, HI, NaH, H2O; (b) H2SeO3, H2SeO4, H2O.
  249. 249. SAMPLE EXERCISE 16.19 Predicting Relative Acidities from Composition and Structure Arrange the compounds in each of the following series in order of increasing acid strength: (a) AsH3, HI, NaH, H2O; (b) H2SeO3, H2SeO4, H2O. Solve: (a) The elements from the left side of the periodic table form the most basic binary hydrogen compounds because the hydrogen in these compounds carries a negative charge. Thus NaH should be the most basic compound on the list. Because arsenic is less electronegative than oxygen, we might expect that AsH3 would be a weak base toward water. That is also what we would predict by an extension of the trends shown in Figure 16.13. Further, we expect that the binary hydrogen compounds of the halogens, as the most electronegative element in each period, will be acidic relative to water. In fact, HI is one of the strong acids in water. Thus the order of increasing acidity is NaH < AsH3 < H2O < HI.
  250. 250. SAMPLE EXERCISE 16.19 Predicting Relative Acidities from Composition and Structure Arrange the compounds in each of the following series in order of increasing acid strength: (a) AsH3, HI, NaH, H2O; (b) H2SeO3, H2SeO4, H2O. Solve: (a) The elements from the left side of the periodic table form the most basic binary hydrogen compounds because the hydrogen in these compounds carries a negative charge. Thus NaH should be the most basic compound on the list. Because arsenic is less electronegative than oxygen, we might expect that AsH3 would be a weak base toward water. That is also what we would predict by an extension of the trends shown in Figure 16.13. Further, we expect that the binary hydrogen compounds of the halogens, as the most electronegative element in each period, will be acidic relative to water. In fact, HI is one of the strong acids in water. Thus the order of increasing acidity is NaH < AsH3 < H2O < HI. (b) The acidity of oxyacids increases as the number of oxygen atoms bonded to the central atom increases. Thus, H2SeO4 will be a stronger acid than H2SeO3; in fact, the Se atom in H2SeO4 is in its maximum positive oxidation state, and so we expect it to be a comparatively strong acid, much like H2SO4. H2SeO3 is an oxyacid of a nonmetal that is similar to H2SO3. As such, we expect that H2SeO3 is able to donate a proton to H2O, indicating that H2SeO3 is a stronger acid than H2O. Thus, the order of increasing acidity is H2O < H2SeO3 < H2SeO4.
  251. 251. PRACTICE EXERCISE In each of the following pairs choose the compound that leads to the more acidic (or less basic) solution: (a) HBr, HF; (b) PH3, H2S; (c) HNO2, HNO3; (d) H2SO3, H2SeO3.
  252. 252. PRACTICE EXERCISE In each of the following pairs choose the compound that leads to the more acidic (or less basic) solution: (a) HBr, HF; (b) PH3, H2S; (c) HNO2, HNO3; (d) H2SO3, H2SeO3. Answers: (a) HBr, (b) H2S, (c) HNO3, (d) H2SO3
  253. 253. Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic. -COOH = carboxyl group Acids and Bases
  254. 254. Acids and Bases
  255. 255. –COOH Acids and Bases
  256. 256. Lewis Acids Acids and Bases
  257. 257. Lewis Acids • Lewis acids are defined as electron-pair acceptors. Acids and Bases
  258. 258. Lewis Acids • Lewis acids are defined as electron-pair acceptors. • Atoms with an empty valence orbital can be Lewis acids. Acids and Bases
  259. 259. Lewis Bases Acids and Bases
  260. 260. Lewis Bases • Lewis bases are defined as electron-pair donors. Acids and Bases
  261. 261. Lewis Bases • Lewis bases are defined as electron-pair donors. • Anything that could be a Brønsted–Lowry base is a Lewis base. Acids and Bases
  262. 262. Lewis Bases • Lewis bases are defined as electron-pair donors. • Anything that could be a Brønsted–Lowry base is a Lewis base. • Lewis bases can interact with things other than protons, however. Acids and Bases
  263. 263. Reactions of Cations with Water Acids and Bases

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