1. 1 The Chemistry of Acids and Bases Chemistry I – Chapter 19 SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print "Handouts" instead of "Slides" in the print setup. Also, turn off the backgrounds (Tools>Options>Print>UNcheck "Background Printing")!

2. 2 Acid and Bases

3. 3

4. 4

5. 5 Acids Have a sour taste. Vinegar is a solution of acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Have a bitter taste. Feel slippery. Many soaps contain bases. Bases

6. 6 Some Properties of Acids þ Produce H + (as H 3 O + ) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) þ Taste sour þ Corrode metals þ Electrolytes þ React with bases to form a salt and water þ pH is less than 7 þ Turns blue litmus paper to red “Blue to Red A-CID”

7. 7 Acid Nomenclature Review No Oxygen  w/Oxygen An easy way to remember which goes with which… “In the cafeteria, you ATE something ICky”

8. 8 Acid Nomenclature Flowchart

9. 9 HBr (aq)HBr (aq) H 2 CO 3H 2 CO 3 H 2 SO 3H 2 SO 3  hydrobromic acid  carbonic acid  sulfurous acid Acid Nomenclature Review

10. 10 Name ‘Em! HI (aq)HI (aq) HCl (aq)HCl (aq) H 2 SO 4H 2 SO 4 HNO 3HNO 3 HIO 4HIO 4 Hydroiodic acid Hydrochloric acid Sulfuric Acid Nitric Acid Periodic Acid

11. 11 Some Properties of Bases  Produce OH - ions in water  Taste bitter, chalky  Are electrolytes  Feel soapy, slippery  React with acids to form salts and water  pH greater than 7  Turns red litmus paper to blue “Basic Blue”

12. 12 Some Common Bases NaOHsodium hydroxidelye KOHpotassium hydroxideliquid soap Ba(OH) 2 barium hydroxidestabilizer for plastics Mg(OH) 2 magnesium hydroxide“MOM” Milk of magnesia Al(OH) 3 aluminum hydroxideMaalox (antacid) Al(OH) 3 aluminum hydroxideMaalox (antacid)

13. 13 Acid/Base definitions you must know both Definition #1: Arrhenius (traditional) Acids – produce H + ions (or hydronium ions H 3 O + ) Bases – produce OH - ions (problem: some bases don’t have hydroxide ions!)

14. 14 Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water

15. 15 Acid/Base Definitions Definition #2: Brønsted – Lowry Acids – proton donor (H + donated) Bases – proton acceptor(H + accepted) A “proton” is really just a hydrogen atom that has lost it’s electron!

16. 16 A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor acid conjugate base base conjugate acid

17. 17 ACID-BASE THEORIES The Brønsted definition means NH 3 is a BASE in water — and water is itself an ACID

18. 18 Conjugate Pairs

19. 19 Learning Check! Label the acid, base, conjugate acid, and conjugate base in each reaction: HCl + OH -  Cl - + H 2 O H 2 O + H 2 SO 4  HSO 4 - + H 3 O +

20. 20 The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H + (or OH - ) ion. Under 7 = acid 7 = neutral Over 7 = base The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H + (or OH - ) ion. pouvoir hydrogène (Fr.) “hydrogen power” Under 7 = acid 7 = neutral Over 7 = base

21. 21 pH of Common Substances

22. 22 pH indicators

23. 23 Calculating the pH pH = - log [H+] (Remember that the [ ] mean Molarity) Example: If [H + ] = 1 X 10 -10 pH = - log 1 X 10 -10 pH = - (- 10) pH = 10 Example: If [H + ] = 1.8 X 10 -5 pH = - log 1.8 X 10 -5 pH = - (- 4.74) pH = 4.74

24. 24 Try These! Find the pH of these: 1) A 0.15 M solution of Hydrochloric acid 2) A 3.00 X 10 -7 M solution of Nitric acid

25. 25 pH calculations – Solving for H+ If the pH of Coke is 3.12, [H + ] = ??? Because pH = - log [H + ] then - pH = log [H + ] - pH = log [H + ] Take antilog (10 x ) of both sides and get 10 -pH = [H + ] [H + ] = 10 -3.12 = 7.6 x 10 -4 M *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button

26. 26 pH calculations – Solving for H+ A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution?A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution? pH = - log [H + ] 8.5 = - log [H + ] -8.5 = log [H + ] Antilog -8.5 = antilog (log [H + ]) 10 -8.5 = [H + ] 3.16 X 10 -9 = [H + ] pH = - log [H + ] 8.5 = - log [H + ] -8.5 = log [H + ] Antilog -8.5 = antilog (log [H + ]) 10 -8.5 = [H + ] 3.16 X 10 -9 = [H + ]

27. 27 More About Water H 2 O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION Equilibrium constant for water = K w K w = [H + ] [OH - ] = 1.00 x 10 -14 at 25 o C

28. 28 More About Water K w = [H + ] [OH - ] = 1.00 x 10 -14 at 25 o C In a neutral solution [H + ] = [OH - ] so K w = [H + ] 2 = [OH - ] 2 and so [H + ] = [OH - ] = 1.00 x 10 -7 M Autoionization

29. 29 Self ionization of water

30. 30 pOH Since acids and bases are opposites, pH and pOH are opposites!Since acids and bases are opposites, pH and pOH are opposites! pOH does not really exist, but it is useful for changing bases to pH.pOH does not really exist, but it is useful for changing bases to pH. pOH looks at the perspective of a basepOH looks at the perspective of a base pOH = - log [OH - ] Since pH and pOH are on opposite ends, pH + pOH = 14

31. 31

32. 32 [H 3 O + ], [OH - ] and pH What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10 -3 M) pOH = - log 0.0010 pOH = - log 0.0010 pOH = 3 pOH = 3 pH = 14 – 3 = 11 OR K w = [H + ] [OH - ] [H + ] = 1.0 x 10 -11 M pH = - log (1.0 x 10 -11 ) = 11.00

33. 33 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H + ion concentration of the rainwater? The OH - ion concentration of a blood sample is 2.5 x 10 -7 M. What is the pH of the blood?

34. 34 [OH - ] [H + ] pOH pH 10 -pOH 10 -pH -Log[H + ] 14 - pOH 14 - pH 1.0 x 10 -14 [OH - ] [OH - ] 1.0 x 10 -14 [H + ] [H + ]

35. 35 Calculating [H 3 O + ], pH, [OH - ], and pOH Problem 1: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H + ], pH, [OH - ], and pOH of the two solutions at 25°C. Problem 2: What is the [H + ], [OH - ], and pOH of a solution with pH = 3.67? Is this an acid, base, or neutral? Problem 3: Problem #2 with pH = 8.05?

36. 36 HNO 3, HCl, H 2 SO 4 and HClO 4 are among the only known strong acids. Also HBr and HI Strong and Weak Acids/Bases The strength of an acid (or base) is determined by the amount of IONIZATION or dissociation.

37. 37 Knowledge check The [H + ] is______ to the concentration of the strong acid. The [OH - ] is _______ to the concentration of the strong base.

38. 38 Strong and Weak Acids/Bases Generally divide acids and bases into STRONG or WEAK ones.Generally divide acids and bases into STRONG or WEAK ones. STRONG ACID: HNO 3 (aq) + H 2 O (l)  H 3 O + (aq) + NO 3 - (aq) HNO 3 is about 100% dissociated in water. movie

39. 39 Weak acids are much less than 100% ionized in water.Weak acids are much less than 100% ionized in water. One of the best known is acetic acid = CH 3 CO 2 H Strong and Weak Acids/Bases

40. 40 Strong Base: 100% dissociated in water.Strong Base: 100% dissociated in water. NaOH (aq) ---> Na + (aq) + OH - (aq) NaOH (aq) ---> Na + (aq) + OH - (aq) Strong and Weak Acids/Bases Other common strong bases include KOH and Ca(OH) 2. CaO (lime) + H 2 O --> Ca(OH) 2 (slaked lime) Ca(OH) 2 (slaked lime) CaO

41. 41 Weak base: less than 100% ionized in waterWeak base: less than 100% ionized in water One of the best known weak bases is ammonia NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq) Strong and Weak Acids/Bases

42. 42 Weak Bases

43. 43 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Does the fact that a weak acid-strong base titration is basic mean that there is some base “left over”

44. 44 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Does the fact that a weak acid-strong base titration is basic mean that there is some base “left over” No. All of the acid and base has been converted to a salt solution at the equivalence point. The solution is basic because the salt hydrolyzes. The salt has negative ions that attract hydrogen ions from water.

45. 45 This is not in your book Refer to ICE charts in resources

46. 46 Equilibria Involving Weak Acids and Bases Consider acetic acid, HC 2 H 3 O 2 (HOAc) HC 2 H 3 O 2 + H 2 O  H + + C 2 H 3 O 2 - Acid Conj. base (K is designated K a for ACID) K gives the ratio of ions (split up) to molecules (don’t split up) [Products] coefficient /[reactants] coeffi.

47. 47 Ionization Constants for Acids/Bases Acids ConjugateBases Increase strength

48. 48 Equilibrium Constants for Weak Acids Weak acid has K a < 1 Leads to small [H 3 O + ] and a pH of 2 - 7

49. 49 Equilibrium Constants for Weak Bases Weak base has K b < 1 Leads to small [OH - ] and a pH of 12 - 7

50. 50 Relation of K a, K b, [H 3 O + ] and pH

51. 51 Record steps and examples. You will use the same steps for all equilibrium problems.

52. 52 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc][H 3 O + ][OAc - ] [HOAc][H 3 O + ][OAc - ]initialchangeequilib 1.0000 -x+x+x 1.00-xxx

53. 53 Equilibria Involving A Weak Acid Step 2. Write K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. This is a quadratic. Solve using quadratic formula. or you can make an approximation if x is very small! (Rule of thumb: 10 -5 or smaller is ok)

54. 54 Equilibria Involving A Weak Acid Step 3. Solve K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. First assume x is very small because K a is so small. Now we can more easily solve this approximate expression.

55. 55 Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. x = [ H 3 O + ] = [ OAc - ] = 4.2 x 10 -3 M pH = - log [ H 3 O + ] = -log (4.2 x 10 -3 ) = 2.37

56. 56 Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO 2 H. HCO 2 H + H 2 O  HCO 2 - + H 3 O + HCO 2 H + H 2 O  HCO 2 - + H 3 O + K a = 1.8 x 10 -4 Approximate solution [H 3 O + ] = 4.2 x 10 -4 M, pH = 3.37 [H 3 O + ] = 4.2 x 10 -4 M, pH = 3.37 Exact Solution [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M pH = 3.47 pH = 3.47

57. 57 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib 0.01000 -x+x +x 0.010 - xx x

58. 58 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib 0.01000 -x+x+x 0.010 - xx x

59. 59 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 2. Solve the equilibrium expression Assume x is small, so x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M and [NH 3 ] = 0.010 - 4.2 x 10 -4 ≈ 0.010 M The approximation is valid!

60. 60 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 3. Calculate pH [OH - ] = 4.2 x 10 -4 M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = 10.63

61. 61 Find Ka The pH of a 0.063 M solution of hydrobromous acid (HBrO) is 4.95. Calculate the Ka Weak acid HBrO  H+ + BrO- The pH = 4.95 [H+]= 10 -4.95 [BrO-] = [H+] [HBrO] =0.063M Solve For Ka

62. 62 Types of Acid/Base Reactions: Summary

63. 63 pH testing There are several ways to test pHThere are several ways to test pH –Blue litmus paper (red = acid) –Red litmus paper (blue = basic) –pH paper (multi-colored) –pH meter (7 is neutral, 7 base) –Universal indicator (multi-colored) –Indicators like phenolphthalein –Natural indicators like red cabbage, radishes

64. 64 Paper testing Paper tests like litmus paper and pH paperPaper tests like litmus paper and pH paper –Put a stirring rod into the solution and stir. –Take the stirring rod out, and place a drop of the solution from the end of the stirring rod onto a piece of the paper –Read and record the color change. Note what the color indicates. –You should only use a small portion of the paper. You can use one piece of paper for several tests.

65. 65 pH paper

66. 66 pH meter Tests the voltage of the electrolyteTests the voltage of the electrolyte Converts the voltage to pHConverts the voltage to pH Very cheap, accurateVery cheap, accurate Must be calibrated with a buffer solutionMust be calibrated with a buffer solution

67. 67 pH indicators Indicators are dyes that can be added that will change color in the presence of an acid or base. Some indicators only work in a specific range of pH Once the drops are added, the sample is ruined Some dyes are natural, like radish skin or red cabbage

68. 68 ACID-BASE REACTIONS Titrations H 2 C 2 O 4 (aq) + 2 NaOH(aq) ---> acid base acid base Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) Carry out this reaction using a TITRATION. Oxalic acid, H 2 C 2 O 4

69. 69 Setup for titrating an acid with a base

70. 70 TitrationTitration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3.Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) This is called NEUTRALIZATION. This is called NEUTRALIZATION.

71. 71 35.62 mL of NaOH is neutralized with 25.2 mL of 0.0998 M HCl by titration to an equivalence point. What is the concentration of the NaOH? Moles of H+ = Moles of OH- LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.

72. 72 For Neutralizations : M a V a = M b V b moles H+ = moles OH- Check stoichiometry of balanced equation

73. 73 6 grams of HF takes 6.00 ml (.006 L) of NaOH to neutralize. What is the molarity of NaOH?

74. 74 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!

75. 75 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add?

76. 76 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do ? How much water is added? The important point is that ---> moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution moles of NaOH in FINAL solution

77. 77 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Amount of NaOH in original solution = M V = M V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH) / (0.50 M) = 0.30 L or 300 mL

78. 78 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

79. 79 A shortcut A shortcut M 1 V 1 = M 2 V 2 Preparing Solutions by Dilution

80. 80 You try this dilution problem You have a stock bottle of hydrochloric acid, which is 12.1 M. You need 400 mL of 0.10 M HCl. How much of the acid and how much water will you need? How much 2.5 M KOH is necessary to neutralize the diluted solution?

81. 81 Normality or Equivalents This is not in your book

82. 82 Normality/ Equivalents Where Molarity (M) represents the concentration of an ion. Normality (N) goes one step further and represents the molar concentration only of the acid component (usually the H+ ion in an acid solution) or only the base component (usually the OH- ion in a base solution). 1Mole of HCl has one equilvalent = 1 Normality 1Mole of H 2 SO 4 has 2 equivalents = 2 Normality HCl + NaOH  H 2 O + NaCl 1 mole of HCl will neutralize 1 mole of NaOH.

83. 83 Normality (N) = Equiv / L 2HCl + Ca(OH) 2  2H 2 O + CaCl 2 2 moles of acid are required to neutralize 1 mole of base. Mole:mole ratios 2 moles of HCl contain 2 equivalents of acid (H+) 1 mole of Ca(OH) 2 contains 2 equivalents (OH-) of base.

84. 84 Example What is the Normality of the following solutions: 2 M HCl 0.1 M HC 2 H 3 O 2 0.3 M H 3 PO 4 0.25 M H 2 SO 4.09M NaOH 9M Ba(OH) 2

85. 85 For Titrations N a V a = N b V b How many ml of a 0.67 M HCl are needed to neutralize 125 mL of a 0.11 M Ba(OH) 2 ? Multiply the Molarity by number of equivalents.

86. 86 Another problem What is the Molar concentration of 29 mL of H 3 PO 4 if it takes 25mL of 0.85 M NaOH to neutralize the acid? N a x 29 mL = 25ml x 0.85 N a = 21.25/29 normality 0.73 eq/L x 1mol/3eq M =.24 M

87. 87 How many equivalents are in 2.6 L of 0.5N H 2 SO 4 ? Number of equivalents = V(L) x N = 2.6 L x 0.50 equiv = 1.3 equiv L

88. 88 How many equivalents are in the following A. 0.55 L of 1.8 N NaOH B. 1.6 L of 0.5 N H 3 PO 4 C. 250 ml of 0.28 N H 2 SO 4 0.99 Eq 0.80 Eq 0.070 eq

89. 89 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 19 Acids, Bases, and Salts 19.1 Acid-Base Theories 19.2 Hydrogen Ions and Acidity 19.3 Strengths of Acids and Bases 19.4 Neutralization Reactions 19.5 Salts in Solution

90. 90 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Salt Hydrolysis When is the solution of a salt acidic or basic? Salt Hydrolysis

91. 91 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. A salt is one of the products of a neutralization reaction. A salt consists of an anion from an acid and a cation from a base. The solutions of many salts are neutral. Salt Hydrolysis

92. 92 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Some salts form acidic or basic solutions. Salt Hydrolysis Universal indicator was added to these 0.10M aqueous salt solutions. Based on the indicator color, the solutions can be classified as follows: Ammonium chloride, NH 4 Cl(aq), is acidic (pH of about 5.3). Sodium ethanoate, CH 3 COONa(aq), is basic (pH of about 8.7). Sodium chloride, NaCl(aq), is neutral (pH of 7).

93. 93 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Salts that produce acidic solutions have positive ions that release hydrogen ions to water. Salts that produce basic solutions have negative ions that attract hydrogen ions from water. Salt Hydrolysis

94. 94 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sodium ethanoate (CH 3 COONa) is the salt of a weak acid and a strong base. In solution, the salt is completely ionized. CH 3 COONa(aq) → CH 3 COO − (aq) + Na + (aq) Sodium ethanoate Ethanoate ion Sodium ion Salt Hydrolysis

95. 95 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. The ethanoate ion is a Brønsted-Lowry base, which means it is a hydrogen acceptor. It reacts with water to form ethanoic acid and hydroxide ions. At equilibrium, the reactants are favored. Salt Hydrolysis CH 3 COO – (aq) + H 2 O(l) CH 3 COOH(aq) + OH – (aq) H + donor Brønsted-Lowry acid H + acceptor Brønsted-Lowry base (makes the solution basic)

96. 96 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. This process is called hydrolysis because a hydrogen ion is split off a water molecule. The suffix -lysis comes from a Greek word meaning to “separate” or “loosen.” In the solution, the hydroxide-ion concentration is greater than the hydrogen-ion concentration. Thus, the solution is basic. Salt Hydrolysis CH 3 COO – (aq) + H 2 O(l) CH 3 COOH(aq) + OH – (aq) H + donor Brønsted-Lowry acid H + acceptor Brønsted-Lowry base (makes the solution basic)

97. 97 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Ammonium chloride (NH 4 Cl) is the salt of the strong acid hydrochloric acid (HCl) and the weak base ammonia (NH 3 ). It is completely ionized in solution. NH 4 Cl(aq) → NH 4 + (aq) + Cl − (aq) Salt Hydrolysis

98. 98 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. The ammonium ion (NH 4 + ) is a strong enough acid to donate a hydrogen ion to a water molecule. The products are ammonia molecules and hydronium ions. The reactants are favored at equilibrium, as shown by the relative sizes of the arrows. Salt Hydrolysis NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) H + donor Brønsted-Lowry acid H + acceptor Brønsted-Lowry base (makes the solution acidic)

99. 99 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. This process is another example of hydrolysis. At equilibrium the [H 3 O + ] is greater than the [OH – ]. Thus, a solution of ammonium chloride is acidic. Salt Hydrolysis NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) H + donor Brønsted-Lowry acid H + acceptor Brønsted-Lowry base (makes the solution acidic)

100. 100 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. To determine if a salt will form an acidic or basic solution, remember the following rules: Strong acid + Strong acid + Strong base → Neutral solution Strong acid + Acidic Strong acid + Weak base → Acidic solution Weak acid + Weak acid + Strong base → Basic solution Strong wins!!! Salt Hydrolysis