1. Jeffrey Mack California State University, Sacramento Chapter 17 Principles of Chemical Reactivity: The Chemistry of Acids and Bases

2. In Chapter 3, you were introduced to two definitions of acids and bases: the Arrhenius and the Brønsted–Lowry definition. Arrhenius acid: Any substance that when dissolved in water increases the concentration of hydrogen ions, H +. Arrhenius base: Any substance that increases the concentration of hydroxide ions, OH , when dissolved in water. A Brønsted–Lowry acid is a proton (H + ) donor. A Brønsted–Lowry base is a proton acceptor. Acids & Bases: A Review

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

4. HNO 3, HCl, H 2 SO 4 and HClO 4 are classified as strong acids. Strong & Weak Acids/Bases

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

6. Weak base:Weak base: less than 100% ionized in water An example of a weak base is ammonia NH 3 (aq) + H 2 O(liq)  NH 4 + (aq) + OH - (aq) Strong & Weak Acids/Bases

7. Weak acids are much less than 100% ionized in water. Example: acetic acid = CH 3 CO 2 H Strong & Weak Acids/Bases

8. Proton donors may be molecular compounds, cations or anions. The Brønsted–Lowry Concept of Acids & Bases Extended

9. Proton acceptors may be molecular compounds, cations or anions. The Brønsted–Lowry Concept of Acids & Bases Extended

10. BASE ACID Using the Brønsted definition, NH 3 is a BASE in water and water is itself an ACID Proton acceptor Proton donor The Brønsted–Lowry Concept of Acids & Bases Extended

11. Acids such as HF, HCl, HNO 3, and CH 3 CO 2 H (acetic acid) are all capable of donating one proton and so are called monoprotic acids. Other acids, called polyprotic acids are capable of donating two or more protons. The Brønsted–Lowry Concept of Acids & Bases Extended

12. A conjugate acid–base pair consists of two species that differ from each other by the presence of one hydrogen ion. Every reaction between a Brønsted acid and a Brønsted base involves two conjugate acid–base pairs Conjugate Acid–Base Pairs

14. Water Autoionization and the Water Ionization Constant, K w : The water autoionization equilibrium lies far to the left side. In fact, in pure water at 25 °C, only about two out of a billion (10 9 ) water molecules are ionized at any instant. Even in pure water, there is a small concentration of ions present at all times. [H 3 O + ] = [OH  ] = 1.00  10  7 Water & the pH Scale

15. H 2 O can function as both an ACID and a BASE. AUTOIONIZATION In pure water there can be AUTOIONIZATION. Equilibrium constant for autoionization = K w K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 °C Water & the pH Scale

16. In a neutral solution, [H 3 O + ] = [OH  ] Both are equal to 1.00  10  7 M In an acidic solution, [H 3 O + ] > [OH  ] [H 3 O + ] > 1.00  10  7 M and [OH  ] < 1.00  10  7 M In a basic solution, [H 3 O + ] < [OH  ] [H 3 O + ] 1.00  10  7 M Water & the pH Scale

17. The pH Scale

18. The pH of a solution is defined as the negative of the base (10) logarithm (log) of the hydronium ion concentration. pH =  log[H 3 O + ] In a similar way, we can define the pOH of a solution as the negative of the base - 10 logarithm of the hydroxide ion concentration. pOH =  log[OH  ] pH + pOH = pK w = 14 The pH Scale

19. The concentration of acid, [H 3 O + ] is found by taking the antilog of the solutions pH. In a similar way, [OH  ] can be found from: The pH Scale

20. Once [H 3 O + ] is known, [OH  ] can be found from: And vice versa. The pH Scale

21. In Chapter 3, it was stated that acids and bases can be divided roughly into those that are strong electrolytes (such as HCl, HNO 3, and NaOH) and those that are weak electrolytes (such as CH 3 CO 2 H and NH 3 ) In this chapter we will discuss the quantitative aspects of dissociation of weak acids and bases. The relative strengths of weak acids and bases can be ranked based on the magnitude of individual equilibrium constants. Equilibrium Constants for Acids & Bases

22. Strong acids and bases almost completely ionize in water (~100%): K strong >> 1 (product favored) Weak acids and bases almost completely ionize in water (<<100%): K weak << 1 (Reactant favored) Equilibrium Constants for Acids & Bases

23. ionization constantThe relative strength of an acid or base can also be expressed quantitatively with an equilibrium constant, often called an ionization constant. For the general acid HA, we can write: ConjugateacidConjugatebase Equilibrium Constants for Acids & Bases

24. ionization constantThe relative strength of an acid or base can also be expressed quantitatively with an equilibrium constant, often called an ionization constant. For the general base B, we can write: ConjugatebaseConjugateAcid Equilibrium Constants for Acids & Bases

25. Acids ConjugateBases Increase strength Ionization Constants for Acids/Bases

26. The strongest acids are at the upper left. They have the largest K a values. K a values become smaller on descending the chart as the acid strength declines. The strongest bases are at the lower right. They have the largest K b values. K b values become larger on descending the chart as base strength increases. Equilibrium Constants for Acids & Bases

27. The weaker the acid, the stronger its conjugate base: The smaller the value of K a, the larger the value of K b. Aqueous acids that are stronger than H 3 O + are completely ionized. Their conjugate bases (such as NO 3  ) do not produce meaningful concentrations of OH  ions, their K b values are “very small.” Similar arguments follow for strong bases and their conjugate acids. Equilibrium Constants for Acids & Bases

28. AcidHCO 3  HClOHF KaKa 4.8  10  11 3.5  10  8 7.2  10  4 BaseCO 3 2  ClO  FF KbKb 2.1  10  4 2.9  10  7 1.4  10  11 Equilibrium Constants for Acids & Bases

30. K a Values for Polyprotic Acids In general, each successive dissociation produces a weaker acid. Equilibrium Constants for Acids & Bases

31. Logarithmic Scale of Relative Acid Strength, pK a Many chemists use a logarithmic scale to report and compare relative acid strengths. pK a =  log(K a ) The lower the pK a, the stronger the acid. AcidHCO 3  HClOHF pK a 10.327.463.14 Equilibrium Constants for Acids & Bases

32. Relating the Ionization Constants for an Acid and Its Conjugate Base Equilibrium Constants for Acids & Bases

33. Relating the Ionization Constants for an Acid and Its Conjugate Base Equilibrium Constants for Acids & Bases

34. Relating the Ionization Constants for an Acid and Its Conjugate Base Equilibrium Constants for Acids & Bases

35. Relating the Ionization Constants for an Acid and Its Conjugate Base Equilibrium Constants for Acids & Bases

36. Relating the Ionization Constants for an Acid and Its Conjugate Base When adding equilibria, multiply the K values. Equilibrium Constants for Acids & Bases

37. Acid–Base Properties of Salts

38. Anions that are conjugate bases of strong acids (for examples, Cl  or NO 3 . These species are such weak bases that they have no effect on solution pH. Acid–Base Properties of Salts

39. Anions such as CO 3  that are the conjugate bases of weak acids will raise the pH of a solution. Hydroxide ions are produced via “Hydrolysis”. Acid–Base Properties of Salts

40. Anions such as CO 3  that are the conjugate bases of weak acids will raise the pH of a solution. Hydroxide ions are produced via “Hydrolysis”. A partially deprotonated anion (such as HCO 3  ) is amphiprotic. Its behavior will depend on the other species in the reaction. Acid–Base Properties of Salts

41. Alkali metal and alkaline earth cations have no measurable effect on solution pH. Since these cations are conjugate acids of strong bases, hydrolysis does not occur. Acid–Base Properties of Salts

42. Basic cations are conjugate bases of acidic cations such as [Al(H 2 O) 6 ] 3+. Acidic cations fall into two categories: (a) metal cations with 2 + and 3 + charges and (b) ammonium ions (and their organic derivatives). All metal cations are hydrated in water, forming ions such as [M(H 2 O) 6 ] n+. Acid–Base Properties of Salts

43. Acid–Base Properties of Salts: Practice

45. According to the Brønsted–Lowry theory, all acid– base reactions can be written as equilibria involving the acid and base and their conjugates. All proton transfer reactions proceed from the stronger acid and base to the weaker acid and base. Predicting the Direction of Acid– Base Reactions

46. When a weak acid is in solution, the products are a stronger conjugate acid and base. Therefore equilibrium lies to the left. All proton transfer reactions proceed from the stronger acid and base to the weaker acid and base. Predicting the Direction of Acid– Base Reactions

47. Will the following acid/base reaction occur spontaneously? Predicting the Direction of Acid– Base Reactions

48. Will the following acid/base reaction occur spontaneously? K a = 7.5  10  5 K a = 1.8  10  5 K b = 5.6  10  10 K b = 1.3  10  12 Predicting the Direction of Acid– Base Reactions

49. Will the following acid/base reaction occur spontaneously? Equilibrium lies to the right since all proton transfer reactions proceed from the stronger acid and base to the weaker acid and base. K a = 7.5  10  5 K a = 1.8  10  5 K b = 5.6  10  10 K b = 1.3  10  12 Stronger Acid + Stronger BaseWeaker Base + Weaker Acid Predicting the Direction of Acid– Base Reactions

50. Strong acid (HCl) + Strong base (NaOH) Net ionic equation Mixing equal molar quantities of a strong acid and strong base produces a neutral solution. Types Acids–Base Reactions

51. Weak acid (HCN) + Strong base (NaOH) Mixing equal amounts (moles) of a strong base and a weak acid produces a salt whose anion is the conjugate base of the weak acid. The solution is basic, with the pH depending on K b for the anion. Types Acids–Base Reactions

52. Strong acid (HCl) + Weak base (NH 3 ) Mixing equal amounts (moles) of a weak base and a strong acid produces a conjugate acid of the weak base. The solution is basic, with the pH depending on K a for the acid. Types Acids–Base Reactions

53. Weak acid (CH 3 CO 2 H) + Weak base (NH 3 ) Mixing equal amounts (moles) of a weak acid and a weak base produces a salt whose cation is the conjugate acid of the weak base and whose anion is the conjugate base of the weak acid. The solution pH depends on the relative K a and K b values. Types Acids–Base Reactions

54. Weak acid + Weak base Product cation = conjugate acid of weak base. Product anion = conjugate base of weak acid. pH of solution depends on relative strengths of cation and anion. Types Acids–Base Reactions

55. Types Acids–Base Reactions Summary

56. Determining K from Initial Concentrations and pH [H 2 S][H 3 O + ][HS  ] Initial0.10 Change Equilibrium Calculations with Equilibrium Constants

57. Determining K from Initial Concentrations and pH [H 2 S][H 3 O + ][HS  ] Initial0.1000 Change0.10 - x+ x Equilibrium0.10 - xxx Calculations with Equilibrium Constants

58. Determining K from Initial Concentrations and pH [H 2 S][H 3 O + ][HS  ] Initial0.1000 Change0.10 - x+ x Equilibrium0.10 - xxx Calculations with Equilibrium Constants

59. Determining K from Initial Concentrations and pH [H 3 O + ] = [NO 2  ] = 6.76  10  3 Calculations with Equilibrium Constants

60. Determining K from Initial Concentrations and pH [H 2 S][H 3 O + ][HS  ] Initial1.00 Change Equilibrium Calculations with Equilibrium Constants

61. [H 2 S][H 3 O + ][HS  ] Initial1.0000 Change- x+ x Equilibrium1.00 - xxx Calculations with Equilibrium Constants Determining K from Initial Concentrations and pH

62. x = 3.2  10  4 pH = 3.50 Calculations with Equilibrium Constants

63. Determining K from Initial Concentrations and pH In general, the approximation that [HA] equilibrium = [HA] initial  x  [HA] initial is valid whenever [HA] initial is greater than or equal to 100  K a. If this is not the case, the quadratic equation must by used. Calculations with Equilibrium Constants

64. Determining pH after an acid/base reaction: Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. Calculations with Equilibrium Constants

65. Determining pH after an acid/base reaction: Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. Solution: Solution: From the volume and concentration of each solution, the moles of acid and base can be calculated. Knowing the moles after the reaction and the equilibrium constants, the concentration of H 3 O + and pH can be calculated.

66. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. All of the acetic acid is converted to acetate ion.

67. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. All of the acetic acid is converted to acetate ion.

68. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. [CH 3 CO 2 - ][CH 3 CO 2 H][OH - ] Initial0.07500 Change Equilibrium

69. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. [CH 3 CO 2 - ][CH 3 CO 2 H][OH - ] Initial0.07500 Change- x+ x Equilibrium0.075 - xxx

70. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. [CH 3 CO 2 - ][CH 3 CO 2 H][OH - ] Initial0.07500 Change- x+ x Equilibrium0.075 - xxx

71. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH. Since K b  100 > [CH 3 CO 2 - ] initial, the quadratic equation is not needed. [CH 3 CO 2 - ][CH 3 CO 2 H][OH - ] Initial0.07500 Change- x+ x Equilibrium0.075 - xxx

72. Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 mL of 0.15 M NaOH.

73. Because polyprotic acids are capable of donating more than one proton they present us with additional challenges when predicting the pH of their solutions. For many inorganic polyprotic acids, the ionization constant for each successive loss of a proton is about 10 4 to 10 6 smaller than the previous step. This implies that the pH of many inorganic polyprotic acids depends primarily on the hydronium ion generated in the first ionization step. The hydronium ion produced in the second step can be neglected. Polyprotic Acuids & Bases

74. Sulfurous acid, H 2 SO 3, is a weak acid capable of providing two H + ions. (a) What is the pH of a 0.45 M solution of H 2 SO 3 ? (b) What is the equilibrium concentration of the sulfite ion, SO 3 2- in the 0.45 M solution of H 2 SO 3 ? Polyprotic Acids & Bases

75. Sulfurous acid, H 2 SO 3, is a weak acid capable of providing two H + ions. (a) What is the pH of a 0.45 M solution of H 2 SO 3 ? (b) What is the equilibrium concentration of the sulfite ion, SO 3 2- in the 0.45 M solution of H 2 SO 3 ? Polyprotic Acids & Bases

76. Sulfurous acid, H 2 SO 3, is a weak acid capable of providing two H + ions. (a) What is the pH of a 0.45 M solution of H 2 SO 3 ? (b) What is the equilibrium concentration of the sulfite ion, SO 3 2- in the 0.45 M solution of H 2 SO 3 ? Since 100  K a is not << 0.45M, the quadratic equation must be used Polyprotic Acids & Bases

77. Sulfurous acid, H 2 SO 3, is a weak acid capable of providing two H + ions. (a) What is the pH of a 0.45 M solution of H 2 SO 3 ? (b) What is the equilibrium concentration of the sulfite ion, SO 3 2- in the 0.45 M solution of H 2 SO 3 ? (a) x = [H 3 O + ] = 0.0677 M pH = 1.17

78. Sulfurous acid, H 2 SO 3, is a weak acid capable of providing two H + ions. (a) What is the pH of a 0.45 M solution of H 2 SO 3 ? (b) What is the equilibrium concentration of the sulfite ion, SO 3 2- in the 0.45 M solution of H 2 SO 3 ? (a) in part a we found that x = [H 3 O + ] = 0.0677 M (b) Polyprotic Acids & Bases

79. Halide Acid Strengths HF << HCl < HBr < HIExperiments show that the acid strength increases in the order: HF << HCl < HBr < HI. Stronger acids result when the H  X bond is readily broken (as signaled by a smaller, positive value of  H for bond dissociation) and a more negative value for the electron attachment enthalpy of X. Molecular Structure, Bonding, & Acid–Base Behavior

80. Comparing Oxoacids: HNO 2 and HNO 3 In all the series of related oxoacid compounds, the acid strength increases as the number of oxygen atoms bonded to the central element increases. Thus, nitric acid (HNO 3 ) is a stronger acid than nitrous acid (HNO 2 ). Molecular Structure, Bonding, & Acid–Base Behavior

81. Why Are Carboxylic Acids Brønsted Acids? There is a large class of organic acids, all like acetic acid (CH 3 CO 2 H) have the carboxylic acid group,  CO 2 H carboxylic acidsThey are collectively called carboxylic acids. Molecular Structure, Bonding, & Acid–Base Behavior

82. Why Are Carboxylic Acids Brønsted Acids? The carboxylate anion is stabilized by resonance. Molecular Structure, Bonding, & Acid–Base Behavior

83. Why Are Carboxylic Acids Brønsted Acids? The acidity of carboxylic acids is enhanced if electronegative substituents replace the hydrogen atoms in the alkyl (–CH 3 or –C 2 H 5 ) groups. Compare, for example, the pKa values of a series of acetic acids in which hydrogen is replaced sequentially by the more electronegative element chlorine. Molecular Structure, Bonding, & Acid–Base Behavior

84. Trichloroacetic acid is a much stronger acid owing to the high electronegativity of Cl. Cl withdraws electrons from the rest of the molecule. This makes the O—H bond highly polar. The H of O—H is very positive. Trichloroacetic acid is a much stronger acid owing to the high electronegativity of Cl. Cl withdraws electrons from the rest of the molecule. This makes the O—H bond highly polar. The H of O—H is very positive. Acetic acidTrichloroacetic acid K a = 1.8 x 10 -5 K a = 0.3

85. The concept of acid–base behavior advanced by Brønsted and Lowry in the 1920’s works well for reactions involving proton transfer. However, a more general acid– base concept, was developed by Gilbert N. Lewis in the 1930’s. A Lewis acid is a substance that can accept a pair of electrons from another atom to form a new bond. A Lewis base is a substance that can donate a pair of electrons to another atom to form a new bond. The Lewis Concept of Acids & Bases

86. coordinate covalent bondThe product is often called an acid–base adduct. In Section 8.3, this type of chemical bond was called a coordinate covalent bond. Lewis acid-base reactions are very common. In general, they involve Lewis acids that are cations or neutral molecules with an available, empty valence orbital and bases that are anions or neutral molecules with a lone electron pair. The Lewis Concept of Acids & Bases

87. Lewis acid a substance that accepts an electron pair Lewis base a substance that donates an electron pair The Lewis Concept of Acids & Bases

88. New bond formed using electron pair from the Lewis base. Coordinate covalent bondCoordinate covalent bond Notice geometry change on reaction. Reaction of a Lewis Acid & Lewis Base

89. The formation of a hydronium ion is an example of a Lewis acid / base reaction H H H BASE O—H O—H H + ACID The H + is an electron pair acceptor. Water with it’s lone pairs is a Lewis acid donor. The Lewis Concept of Acids & Bases

90. Lewis Acid  Base Reactions

91. Metal cations often act as Lewis acids because of open d-orbitals. Lewis Acids & Bases

92. The combination of metal ions (Lewis acids) with Lewis bases such as H 2 O and NH 3 leads to Coordinate Complex ions. Lewis Acids & Bases

93. Aqueous solutions of Fe 3+, Al 3+, Cu 2+, Pb 2+, etc. are acidic through hydrolysis. This interaction weakens this bond Another H 2 O pulls this H away as H + [Al(H 2 O) 6 ] 3+ (aq) + H 2 O(l)  [Al(H 2 O) 5 (OH)] 2+ (aq) + H 3 O + (aq) Lewis Acids & Bases

94. Because oxygen is more electronegative than C, the C  O bonding electrons in CO 2 are polarized away from carbon and toward oxygen. This causes the carbon atom to be slightly positive, and it is this atom that the negatively charged Lewis base OH  can attack to give, ultimately, the bicarbonate ion. Molecular Lewis Acids

95. Ammonia is the parent compound of an enormous number of compounds that behave as Lewis and Brønsted bases. These molecules all have an electronegative N atom with a partial negative charge surrounded by three bonds and a lone pair of electrons. This partially negative N atom can extract a proton from water. Molecular Lewis Acids

96. Many complex ions containing water undergo HYDROLYSIS to give acidic solutions. 2+ + + Lewis Acids & Bases

97. Reaction of NH 3 with Cu 2+ (aq)

98. The heme group in hemoglobin can interact with O 2 and CO. The Fe ion in hemoglobin is a Lewis acid O 2 and CO can act as Lewis bases Heme group Lewis Acid–Base Interactions in Biology

99. Slides from Chang Book 99 pH – A Measure of Acidity pH = -log [H + ] [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic [H + ] = 1 x 10 -7 [H + ] > 1 x 10 -7 [H + ] < 1 x 10 -7 pH = 7 pH < 7 pH > 7 At 25 0 C pH[H + ]

100. 100 percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% For a monoprotic acid HA Percent ionization = [H + ] [HA] 0 x 100% [HA] 0 = initial concentration % Ionization =

101. 101 What is the pH of a 2 x 10 -3 M HNO 3 solution? HNO 3 is a strong acid – 100% dissociation. HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) pH = -log [H + ] = -log [H 3 O + ] = -log(0.002) = 2.7 Start End 0.002 M 0.0 M What is the pH of a 1.8 x 10 -2 M Ba(OH) 2 solution? Ba(OH) 2 is a strong base – 100% dissociation. Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) Start End 0.018 M 0.036 M0.0 M pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6

102. 102 What is the pH of a 0.5 M HF solution (at 25 0 C)? HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] = 7.1 x 10 -4 HF (aq) H + (aq) + F - (aq) Initial (M) Change (M) Equilibrium (M) 0.500.00 -x-x+x+x 0.50 - x 0.00 +x+x xx K a = x2x2 0.50 - x = 7.1 x 10 -4 Ka  Ka  x2x2 0.50 = 7.1 x 10 -4 0.50 – x  0.50 K a << 1 x 2 = 3.55 x 10 -4 x = 0.019 M [H + ] = [F - ] = 0.019 M pH = -log [H + ] = 1.72 [HF] = 0.50 – x = 0.48 M

103. 103 When can I use the approximation? 0.50 – x  0.50 K a << 1 When x is less than 5% of the value from which it is subtracted. x = 0.019 0.019 M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. What is the pH of a 0.05 M HF solution (at 25 0 C)? Ka  Ka  x2x2 0.05 = 7.1 x 10 -4 x = 0.006 M 0.006 M 0.05 M x 100% = 12% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or method of successive approximations. K a *100 << [HA] 0

104. 104 What is the pH of a 0.122 M monoprotic acid whose K a is 5.7 x 10 -4 ? HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx K a = x2x2 0.122 - x = 5.7 x 10 -4 Ka  Ka  x2x2 0.122 = 5.7 x 10 -4 0.122 – x  0.122 K a << 1 x 2 = 6.95 x 10 -5 x = 0.0083 M 0.0083 M 0.122 M x 100% = 6.8% More than 5% Approximation not ok.

105. 105 K a = x2x2 0.122 - x = 5.7 x 10 -4 x 2 + 0.00057x – 6.95 x 10 -5 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = x = 0.0081x = - 0.0081 HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx [H + ] = x = 0.0081 M pH = -log[H + ] = 2.09

106. 106 Acid-Base Properties of Salts Neutral Solutions: Salts containing an alkali metal or alkaline earth metal ion (except Be 2+ ) and the conjugate base of a strong acid (e.g. Cl -, Br -, and NO 3 - ). NaCl (s) Na + (aq) + Cl - (aq) H2OH2O Basic Solutions: Salts derived from a strong base and a weak acid. NaCH 3 COOH (s) Na + (aq) + CH 3 COO - (aq) H2OH2O CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq)

107. 107 Acid-Base Properties of Salts Acid Solutions: Salts derived from a strong acid and a weak base. NH 4 Cl (s) NH 4 + (aq) + Cl - (aq) H2OH2O NH 4 + (aq) NH 3 (aq) + H + (aq) Salts with small, highly charged metal cations (e.g. Al 3+, Cr 3+, and Be 2+ ) and the conjugate base of a strong acid. Al(H 2 O) 6 (aq) Al(OH)(H 2 O) 5 (aq) + H + (aq) 3+2+

108. 108 Acid-Base Properties of Salts Solutions in which both the cation and the anion hydrolyze: K b for the anion > K a for the cation, solution will be basic K b for the anion < K a for the cation, solution will be acidic K b for the anion  K a for the cation, solution will be neutral

109. 109 What is the molarity of an NH 4 NO 3 (aq) solution that has a pH = 4.80? Strategy Ammonium nitrate is the salt of a strong acid (HNO 3 ) and a weak base (NH 3 ). In NH 4 NO 3 (aq), NH 4 + hydrolyzes and NO 3 – does not. The ICE format must be based on the hydrolysis equilibrium for NH 4 + (aq). In that format [H 3 O + ], derived from the pH, will be a known quantity, and the initial concentration of NH 4 + will be the unknown. Solution We begin by writing the equation for the hydrolysis equilibrium and the equation for K a in terms of K w and K b. As usual, we can calculate [H 3 O + ] from the pH of the solution. log[H 3 O + ] = –pH = –4.80 [H 3 O + ] = 10 –4.80 = 1.6 x 10 –5 M

110. 110 Example 15.15 continued Solution continued If we assume that all the hydronium ion comes from the hydrolysis reaction, we can set up an ICE format in which x represents the unknown initial concentration of NH 4 +. We now substitute equilibrium concentrations into the ionization constant expression for the hydrolysis reaction. We can assume that the ammonium ion is mostly nonhydrolyzed and that the change in [NH 4 + ] is much smaller than the initial [NH 4 + ], so that 1.6 x 10 – 5 << x and we can replace (x – 1.6 x 10 –5 ) by x. Then we can solve for x. The solution is 0.46 M NH 4 NO 3.