2. ACIDS and BASES (unit 11) Notes start on slide 35 ***

3. Acids and Bases An acid is a compound that produces hydrogen ion (H + ) when dissolved in water. HCl  H + + Cl - A base is a compound that produces hydroxide ion (OH - ) when dissolved in water. NaOH  Na + + OH -

4. Types of Problems Strong Acid or Strong Base in Water Strong Acid and Strong Base in Water Weak Acid or Weak Base in Water Weak Acid with a Common Ion in Water Titration Hydrolysis of a Salt

5. In Reality… H + + H 2 O  H 3 O + Hydronium Ion (Can be used interchangeably with H + )

6. Strong Acids A strong acid is one that completely dissociates in water. HCl  H + + Cl - Common Strong Acids: HClHNO 3 H 2 SO 4 HBrHIHClO 3 HClO 4

7. Strong Bases In a like manner, strong bases are those that completely dissociate in water. Common Strong Bases: LiOHNaOHKOH Sr(OH) 2 Ba(OH) 2

8. Example: Strong Acid in Water Calculate the hydrogen ion concentration when 3 moles of HCl is mixed in 6 liters water. HCl  H + + Cl - (Strong Acid)

9. Example 2: What is the [H + ] of a 2M solution of H 2 SO 4 ? H 2 SO 4  2H + + SO 4 2- [H+] = 4M

10. An acid and a base neutralize each other and form a salt and water HCl + NaOH  NaCl + H 2 O Acid Base Salt Water

11. H___ + ____OH  ____ ____ + H 2 O

12. pH = -log [H + ] AcidNeutralBase 0 7 14

13. Example: What is the pH of a.0025M solution of HNO 3 ? HNO 3  H + + NO 3 - [H+] .0025M pH = -log [H + ] pH = -log (.0025M) = -(-2.602) = +2.602

14. Is Water an Acid or a Base? H 2 O  H + + OH - Note that this is an equilibrium reaction. K w = [H + ][OH - ] =1 x 10 -14 This relationship is always true when water is present!

15. Strong Acid and a Strong Base Calculate the pH when 3 moles of HNO 3 is added to 6 liters of 2.5M NaOH solution. HNO 3  H + + NO 3 - (Strong Acid) NaOH  Na + + OH - (Strong Base)

16. [H + ] = 3moles/6liters = 0.5M [OH - ] = 2.5M But… [H + ][OH - ] = 10 -14 So, OH - is in excess

17. When H + runs out… [OH - ] 2.5M [H + ]0.5M [OH - ]2.0M -

18. pH = -log[H + ] = -log (5.0 x 10 -15 ) = 14.3

19. Weak Acids & Bases A weak acid is one that does not completely dissociate when dissolved in water. This is contrasted with strong acids, that do completely dissociate when dissolved in water.

20. Weak Acids & Bases For Example: HF  H + + F - (Weak Acid) HCl  H + + Cl - (Strong Acid) Note that the dissociation of a weak acid or base is always an equilibrium reaction.

21. Weak Acids & Bases Because a weak acid is one that does not completely dissociate when dissolved in water it is an equilibrium reaction. Both the parent acid, and the daughter ions are present in solution.

22. Weak Acids & Bases The equilibrium law expression can be written, and the constant calculated. For example… HF  H + + F -

23. Example: Calculate the pH of a 0.010M solution of HF (K a = 4 x 10 -5 ) Step 1: Write out the dissolving equation & the equilibrium law expression HF  H + + F -

24. Step 2: Let x = [H+] HF  H+ + F- Initial.01 0 0 Change –x x x Eq. (.01-x) (x) (x)

25. Step 3: Substitute equilibrium values into the equilibrium law expression. Solve for x…

26. This has the form of a quadratic equation. To simplify the solution, we make an assumption…. Assume x <<.01 Then,.01 – x .01

27. Step 4: Solve the problem using the assumed values.

28. pH = -log [H + ] pH = -log(6.3 x 10 -4 ) pH = -(-3.2) pH = 3.2 Step 5: Solve for pH

29. Another Example Calculate the Ka for a 0.10M solution of formic acid, HCHO 2 if it’s pH is 2.38 HCHO 2  H + + CHO 2 -

30. Calculate the [H + ] from the pH pH = -log [H + ] = 2.38 (given) log [H + ] = -2.38 [H + ] = 4.2 x 10 -3 M

31. HCHO 2  H + + CHO 2 - I C E 0.1M 0 0 -4.2x10 -3 4.2x10 -3 4.2x10 -3 (.1 - 4.2x10 -3 ) 4.2x10 -3 4.2x10 -3 Use the “ICE” notation to calculate equilibrium values of all the species.

32. Note that 0.0042 << 0.10 So 0.10 – 0.0042  0.10

33. Percent Ionization Calculate the percent ionization of a 0.05 M solution of HCN 1. Solve for the [H + ] as before… HCN  H + + CN -

34. Let [H+] = x X = 5.48x10 -6

35. 0.011% ionization

36. ACIDS, BASES & SALTS Unit 11

37. The Arrhenius Theory of Acids and Bases

38. Arrhenius Theory of Acids and Bases: an acid contains hydrogen and ionizes in solutions to produce H + ions: HCl  H + (aq) + Cl - (aq)

39. Arrhenius Theory of Acids and Bases: a base contains an OH - group and ionizes in solutions to produce OH - ions: NaOH  Na + (aq) + OH - (aq)

40. Neutralization Neutralization: the combination of H + with OH - to form water. H + (aq) + OH - (aq)  H 2 O (l) Hydrogen ions (H + ) in solution form hydronium ions (H 3 O + )

41. In Reality… H + + H 2 O  H 3 O + Hydronium Ion (Can be used interchangeably with H + )

42. Commentary on Arrhenius Theory… One problem with the Arrhenius theory is that it’s not comprehensive enough. Some compounds act like acids and bases that don’t fit the standard definition.

43. Bronsted-Lowry Theory of Acids & Bases

44. Bronsted-Lowry Theory of Acids & Bases: An acid is a proton (H + ) donor A base is a proton (H + ) acceptor

45. for example… HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) Proton transfer Acid Base

46. NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq) BASE ACID CONJUGATE ACID CONJUGATE BASE Ammonia is a proton acceptor, and thus a base another example… Water is a proton donor, and thus an acid.

47. Conjugate acid-base pairs Conjugate acid-base pairs differ by one proton (H + ) A conjugate acid is the particle formed when a base gains a proton. A conjugate base is the particle that remains when an acid gives off a proton.

48. Examples: In the following reactions, label the conjugate acid-base pairs: H 3 PO 4 + NO 2 -  HNO 2 + H 2 PO 4 - CN - + HCO 3 -  HCN + CO 3 2- HCN + SO 3 2-  HSO 3 - + CN - H 2 O + HF  F - + H 3 O + acidbasec. acidc. base acidbasec. acidc. base acid base c. acid c. base acidbase c. acidc. base

49. Amphoteric Substances A substance that can act as both an acid and a base (depending on what it is reacting with) is termed amphoteric. Water is a prime example.

50. Properties of Acids and Bases ACIDS Have a sour taste Change the color of many indicators Are corrosive (react with metals) Neutralize bases Conduct an electric current BASES Have a bitter taste Change the color of many indicators Have a slippery feeling Neutralize acids Conduct an electric current

51. Strength of Acids and Bases A strong acid dissociates completely in sol’n: HCl  H + (aq) + Cl - (aq) A weak acid dissociates only partly in sol’n: HNO 2  H + (aq) + NO 2 - (aq) A strong base dissociates completely in sol’n: NaOH  Na + (aq) + OH - (aq) A weak base dissociates only partly in sol’n: NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq)

52. The Lewis Theory of Acids and Bases

53. The Lewis Theory of Acids & Bases Lewis acid: a substance that can accept an electron pair to form a covalent bond (electron pair acceptor). Lewis base: a substance that can donate an electron pair to form a covalent bond (electron pair donor).

54. Neutralization (using Lewis) Neutralization: the formation of a coordinate covalent bond in which both electrons originated on the same (donor) atom.

55. Example 1: Ionization of NH 3 : NH 3 + H 2 O  NH 4 + + OH - N N H H H OHH+  H H H HH + O.. + - acid base Acid = electron pair acceptor, base = electron pair donor (to form the covalent bond)

56. Example 2: Auto-ionization of water: H 2 O + H 2 O  H 3 O + + OH - O O HH OHH+  H HHH + O.. + - acid base.. Acid = electron pair acceptor, base = electron pair donor (to form the covalent bond)

57. Example 3: Reaction of NH 3 with HBr (a Lewis AND a Bronsted-Lowry acid-base reaction): NH 3 + HBr  NH 4 + + Br - N N H H H BrH+  H H H H +.. + - acid base

58. SUMMARY OF ACID-BASE THEORIES TheoryAcid DefinitionBase Definition Arrhenius Theory Any substance which releases H + ions in water solution. Any substance which releases OH - ions in water solution Brǿnsted- Lowry Theory Any substance which donates a proton. Any substance which accepts a proton. Lewis Theory Any substance which can accept an electron pair. Any substance which can donate an electron pair.

59. Acid-Base Reactions Neutralization reactions: reactions between acids and metal hydroxide bases which produce a salt and water. H + ions and OH - ions combine to form water molecules: H + (aq) + OH - (aq)  H 2 O (l)

60. Example 1: the reaction of HCl and NaOH (there are 3 ways to write the chemical equation): Balanced formula unit equation: HCl (aq) + NaOH (aq)  H 2 O (l) + NaCl (aq) Total ionic equation: H + + Cl - + Na + + OH -  H 2 O + Na + + Cl - Net ionic equation: H + (aq) + OH - (aq)  H 2 O (l)

61. Example 2: Write the 3 types of equations for the reaction of hydrobromic acid, HBr, with potassium hydroxide, KOH. Balanced formula unit equation: HBr (aq) + KOH (aq)  H 2 O (l) + KBr (aq) Total ionic equation: H + + Br - + K + + OH -  H 2 O + K + + Br - Net ionic equation: H + (aq) + OH - (aq)  H 2 O (l)

62. Example 3: Write the 3 types of equations for the reaction of nitric acid, HNO 3, with calcium hydroxide, Ca(OH) 2. Balanced formula unit equation: 2HNO 3 (aq) + Ca(OH) 2 (aq)  2H 2 O (l) + Ca(NO 3 ) 2 (aq) Total ionic equation: 2H + + 2NO 3 - + Ca 2+ + 2OH -  2H 2 O + Ca 2+ + 2NO 3 - Net ionic equation: H + (aq) + OH - (aq)  H 2 O (l)

63. Demos…

64. DEMO: Sponge

65. How does that work?... The sponge is soaked in Congo red. Congo red is a dye, a biological stain, and a pH indicator. It has been used as a direct fabric dye for cotton to produce a bright red color. Scientists use Congo red as a pH indicator (a substance that will change color in the presence of different ion concentrations, [H + ])

66. Variety of pH indicators… There are many different types of pH indicators, such as universal indicator and litmus paper. Litmus paper comes in red Litmus paper and blue Litmus paper. 

67. Red litmus paper in an acids turns… Blue litmus paper in a base turns … BLUE RED

68. Demo: tap water vs. dH 2 O Both waters have Universal indicator in them (= pH indicator (changes color in the presence of ions), which is a type of weak acids) The water will change pH, and therefore COLOR (which helps us determine if a solution is acidic or basic) with the addition of HCl (acid) and NaOH (base)

69. Universal Indicator Color Chart pH scale 0 7 14 AcidNeutral Base

70. Why does it take more drops of acid or base to make the tap water change color than it does for the distilled water? What is distilled water made of? What is tap water made of?

71. Buffered Solutions A solution of a weak acid and a common ion is called a buffered solution.

72. Consider the following buffered solution… HAc  H + + Ac - H 2 O  H + + OH - Add additional acid…(H + ) The H + will combine with the Ac - producing HAc. There is an excess of Ac - from the common ion salt. HAc  H + + Ac -

73. Now, add additional base (OH - ) The OH - will combine with the H + to produce water… H 2 O  H + + OH - The H+ comes from the HAc HAc  H + + Ac -

74. Thus, the solution maintains it’s pH in spite of added acid or base.

75. pH and pOH

76. Ionization of water Experiments have shown that pure water ionizes very slightly: 2H 2 O  H 3 O + + OH - Measurements show that: [H 3 O + ] = [OH - ]=1 x 10 -7 M Pure water contains equal concentrations of H 3 O + + OH -, so it is neutral.

77. pH pH is a measure of the concentration of hydronium ions in a solution. pH = -log [H 3 O + ] or pH = -log [H + ]

78. Example: What is the pH of a solution where [H 3 O + ] = 1 x 10 -7 M? pH = -log [H 3 O + ] pH = -log(1 x 10 -7 ) pH = 7

79. Example: What is the pH of a solution where [H 3 O + ] = 1 x 10 -5 M? pH = -log [H 3 O + ] pH = -log(1 x 10 -5 ) pH = 5 When acid is added to water, the [H 3 O + ] increases, and the pH decreases.

80. Example: What is the pH of a solution where [H 3 O + ] = 1 x 10 -10 M? pH = -log [H 3 O + ] pH = -log(1 x 10 -10 ) pH = 10 When base is added to water, the [H 3 O + ] decreases, and the pH increases.

81. The pH Scale AcidNeutral Base 0 7 14

82. pOH pOH is a measure of the concentration of hydroxide ions in a solution. pOH = -log [OH - ]

83. Example: What is the pOH of a solution where [OH - ] = 1 x 10 -5 M? pOH = -log [OH - ] pOH = -log(1 x 10 -5 ) pOH = 5

84. How are pH and pOH related? At every pH, the following relationships hold true: [H 3 O + ] [OH - ] = 1 x 10 -14 M pH + pOH = 14

85. Example 1: What is the pH of a solution where [H + ] = 3.4 x 10 -5 M? pH = -log [H + ] pH = -log(3.4 x 10 -5 M) pH = 4.5

86. Example 2: The pH of a solution is measured to be 8.86. What is the [H + ] in this solution? pH = -log [H + ] 8.86 = -log [H + ] -8.86 = log [H + ] [H + ] = antilog (-8.86) [H + ] = 10 -8.86 [H + ] = 1.38 x 10 -9 M ***you may have to put your calculator into sci mode to get the decimals

87. Example 3: What is the pH of a solution where [H + ] = 5.4 x 10 -6 M? pH = -log [H + ] pH = -log(5.4 x 10 -6 ) pH = 5.3

88. Example 4: What is the [OH - ] and pOH for the solution in example #3? [H 3 O + ][OH - ]= 1 x 10 -14 (5.4 x 10 -6 )[OH - ] = 1 x 10 -14 [OH - ] = 1.9 x 10 -9 M ***you may have to put your calculator into sci mode to get the decimals pH + pOH = 14 pOH = 14 – 5.3 = 8.7

89. Extra Practice Classify each solution as acidic, basic, or neutral ***MUST SOLVE FOR pH (write down this #) and use the pH scale a. [H+] = 6.0 x 10 -10 M b. [OH-] = 3.0 x 10 -2 M c. [H+] = 2.0 x 10 -7 M d. [OH-] = 1.0 x 10 -7 M pH = 9.2, basic pOH = 1.5, pH = 12.5, basic pH = 6.7, acidic pOH = 7 = pH, neutral

90. Example #5 Which is the MOST basic from question #4? B, pH = 12.7

91. Weak Acid with a Common Ion What is the pH of a solution of 1.5M HAc and 0.5M NaAc? Step 1: Write out the dissolving equations HAc  H + + Ac - NaAc  Na + + Ac -

92. Step 2: Write out the equilibrium law expression.

93. Step 3: Use the “ICE” notation to determine EQ values HAc  H + + Ac - I C E 1.500.5 -xxx (1.5-x)x(0.5+x)

94. Step 4: Insert values into the Eq Law expression…

95. Make the usual assumptions… Assume x << 0.5 Then (0.5 + X)  0.5 (1.5 – X)  1.5 Substitute these values in the eq. expression

97. Finally, Calculate pH pH = -log [H + ] pH = -log (5.4 x 10 -5 ) pH = -(-4.27) pH = 4.27

98. Acids and bases: Titrations The amount of acid or base in a solution is determined by carrying out a neutralization reaction; an appropriate acid-base indicator (changes color in specific pH range) must be used to show when the neutralization is completed.

99. Buret Solution with Indicator DEMO of lab… Read the buret correctly = 2 decimal places

100. Acid Base Titration: the addition of a known amount of solution to determine the volume or concentration of another solution A very accurate method to measure concentration. Acid + Base  Salt + Water H + + OH -  H 2 O Moles H + = Moles OH -

101. 3 Steps: (SHOW DEMO OF TITRATION) 1. Add a measured amount of an acid of unknown concentration to a flask. 2. Add an appropriate indicator to the flask 3. Add measured amounts of a base of known concentration using a buret. Continue until the indicator shows that neutralization has occurred. This is called the end point of the titration (*** look at page 620 fig. 19-20)

102. Example: A 25 mL solution of H 2 SO 4 is neutralized by 18 mL of 1.0 M NaOH using phenolphthalein as an indicator. What is the concentration of the H 2 SO 4 solution? Equation: H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O  (steps)

103. Steps 1) How many mol of NaOH are needed for neutraliztion? 2) How many moles of H 2 SO 4 were neutralized? 3) Calculate the concentration of the acid: 

104. 1) How many mol of NaOH are needed for neutraliztion? Mol NaOH = 1.0 mol x 0.018 L = 0.018 mol 1 L

105. 2) How many moles of H 2 SO 4 were neutralized? Mol H 2 SO 4 = 0.018 mol NaOH x 1 mol H 2 SO 4 = 2 mol NaOH 9.0 x 10 -3 mol H 2 SO 4

106. 3) Calculate the concentration of the acid: M = mol = 9.0 x 10-3 mol = 0.36 M L 0.025 L

107. Titration Curve PAGE 619 fig 19-17 A graph showing how the pH changes as a function of the amount of added titrant in a titration. Data for the graph is obtained by titrating a solution and measuring the pH after EVERY drop of added titrant.

108. Equivalence point = The point on the curve where the moles of acid equal the moles of base; the midpoint of the steepest part of the curve is a good approximation of the equivalence point. PAGE 619 fig 19-17

109. PAGE 620 fig. 19-19 Knowledge of the equivalence point can then be used to choose a suitable indicator for a given titration; the indicator must change color at a pH that corresponds to the equivalence point.

110. Calculations of Titrations The mole method and molarity Calculate the molarity of a sulfuric acid solution if 23.2 mL of it reacts with 0.212 g of Na 2 CO 3. Na 2 CO 3 + H 2 SO 4  Na 2 SO 4 + CO 2 + H 2 O ****use steps 1-3 that we just did.  answer

111. ANSWER Molarity = of a sulfuric acid solution if 23.2 mL of it reacts with 0.212 g of Na 2 CO 3. Na 2 CO 3 + H 2 SO 4  Na 2 SO 4 + CO 2 + H 2 O

112. Normality The Normality (N) of a solution is defined as the molarity x the total positive oxidation number of the solute. EXAMPLES HCl  H+ + Cl- H 2 SO 4  2 H + + OH- NaOH  Na + + OH- Ba(OH) 2  Ba 2+ + 2 OH-

113. Example Calculate the molarity and normality of a solution that contains 34.2 g of Ba(OH) 2 in 8.00 L of solution. M = ? N = ?  answers

114. Answer (molarity) M = 34.2 g of Ba(OH) 2 x 1 mol = 0.0249 M 8.00 L 171.35g  answer for normality

115. Answer (normality) N = M x +2 (for Ba 2+ ) = 0.0249 M x 2 = 0.0499 N

116. Using Normality 1 equivalent of an acid reacts with 1 equivalent of a base, so in titration problems, you can use this equation N a V a = N b V b

117. Example 30.0 mL of 0.0750 N HNO 3 required 22.5 mL of Ca(OH) 2 for neutralization. Calculate the normality and molarity of the Ca(OH) 2 solution. 2 HNO 3 + Ca(OH) 2  Ca(NO 3 ) 2 + 2 H 2 O  answers

118. ANSWER (normality) N of Ca(OH) 2 = ? N a V a = N b V b (0.0750 N)(30.0 mL) = N b (22.5 mL) = 0.100 N

119. Answer (molarity) N = M x (+ oxidation #) M = N / (+ oxidation #) M = 0.100 N / (2) because Ca 2+ = 0.0500 M