2. Acids and Bases Part 2

3. The pH Scale

4. Water n Water ionizes; it falls apart into ions. H 2 O  H + + OH - H 2 O  H + + OH - n The reaction above is called the self-ionization of water.

5. Water n [H + ] = [OH - ] = 1 x 10 -7 M n When [H + ] = [OH - ], the solution is neutral. n At 25 °C K w = [H + ] x [OH - ] = 1 x 10 -14 n K w is called the ion-product constant.

6. Ion-Product Constant n If [H + ] > [OH - ], the solution is acidic. n The solution is basic when [OH - ] > [H + ].

7. pH n In most applications, the observed range of possible hydronium or hydroxide ion concentrations spans 10 –14 M to 1M. n To make this range of possible concentrations easier to work with, the pH scale was developed.

8. pH n pH is a mathematical scale in which the concentration of hydronium ions (H 3 O + ) in a solution is expressed as a number from 0 to 14.

9. pH n pH meters are instruments that measure the exact pH of a solution.

10. pH n Indicators register different colors at different pH’s.

11. pH n pH = - log [H + ] n In neutral solution, pH = 7. n In an acidic solution, pH < 7. n In a basic solution, pH > 7.

12. pH n As the pH drops from 7, the solution becomes more acidic. n As pH increases from 7, the solution becomes more basic. alkaline (basic)

13. pH n The pH of a solution equals the negative logarithm of the hydrogen or hydronium ion concentration.

14. pH n pH “goes with” the terms hydrogen and hydronium.

15. pOH n The pOH of a solution equals the negative logarithm of the hydroxide ion concentration.

16. pOH n pOH “goes with” the term hydroxide.

17. pH and pOH n If either pH or pOH is known, the other may be determined by using the following relationship.

18. Problem 16a) Find the pH of the following solution. The hydronium ion concentration equals: 10 –2 M (which = 1 x 10 -2 M) pH = - log[H+](1 x 10 -2 ) pH = 2

19. Problem 16b) Find the pH of the following solution. The hydrogen ion concentration equals: 10 –11 M. pH = - log[H+](1 x 10 -11 ) pH = 11

20. Problem 16c) Find the pH of the following solution. The hydronium ion concentration equals: 1 x 10 –6 M. pH = - log[H+](1 x 10 -6 ) pH = 6

21. Problem 16d) Find the pH of the following solution. The hydroxide ion concentration equals: 10 –8 M. pOH = - log[OH-](1 x 10 -8 ) pOH = 8

22. Problem, cont. 16d cont.) Find the pH of the following solution. The hydroxide ion concentration equals: 10 –8 M. pH + = 14pOH8 pH = 6

23. Problem 16e) Find the pH of the following solution. The hydroxide ion concentration equals: 10 –5 M. (pH = 9)

24. Problem 16f) Find the pH of the following solution. The hydroxide ion concentration equals: 10 –3 M. (pH = 11)

25. Problem 17) If a certain carbonated soft drink has a hydrogen ion concentration of 1.0 x 10 –4 M, what are the pH and pOH of the soft drink? (pH = 4) (pOH = 10)

26. Problem 18) Find the pH if the hydrogen ion concentration equals: 3.25 x 10 –3 M. (pH = 2.49)

27. Problem 19) Find the pH if the hydroxide ion concentration equals: 7.36 x 10 –5 M. (pH = 9.87)

28. Problem 20) Find the pOH if the hydroxide ion concentration equals: 8.34 x 10 –9 M. (pOH = 8.08)

29. Problem 21) Find the pOH if the hydronium ion concentration equals: 1.45 x 10 –4 M. (pOH = 10.16)

30. Calculating Ion Concentrations From pH n If either pH or pOH is known, the hydrogen/hydronium ion or hydroxide ion can be found. [H + ] = 10 -pH [OH - ] = 10 -pOH

31. Calculating Ion Concentrations From pH n On a graphing calculator, hit   and then the number.2nd LOG ( )

32. Calculating Ion Concentrations From pH n On a scientific calculator, hit  the number  shift LOG +/-

33. Problem 22) Find the [H + ] of a solution that has a pH equal to 6. [H + ] = 1 x 10 -6 M [H + ] =2nd LOG ( ) 6

34. Problem 23) Find the [H + ] of a solution that has a pH equal to 12. [H + ] = 1 x 10 -12 M

35. Problem 24) Find the [H + ] of a solution that has a pH equal to 5. [H + ] = 1 x 10 -5 M

36. Example 25) Find the [H + ] of a solution that has a pOH equal to 6. pH + = 14pOH 6 pH = 8

37. Example, cont 25) Find the [H + ] of a solution that has a pOH equal to 6. [H + ] = 1 x 10 -8 M [H + ] =2nd LOG ( ) 8

38. Example, cont 26) Find the [OH - ] of a solution that has a pOH equal to 6. [H + ] = 1 x 10 -6 M [OH - ] =2nd LOG ( ) 6

39. Problem 27) Find the [H + ] of a solution that has a pOH equal to 2. [H + ] = 1 x 10 -12 M

40. Problem 28) Find the [H + ] of a solution that has a pOH equal to 4. [H + ] = 1 x 10 -10 M

41. Problem 29) Find the [OH - ] of a solution that has a pH equal to 10. [OH - ] = 1 x 10 -4 M

42. Problem 30) Find the [H + ] of a solution that has a pH equal to 4.23. [H + ] = 5.89 x 10 -5 M

43. Problem 31) Find the [H + ] of a solution that has a pOH equal to 6.34. [H + ] = 2.19 x 10 -8 M

44. Problem 32) Find the [OH - ] of a solution that has a pH equal to 10.5. [OH - ] = 3.16 x 10 -4 M

45. Problem 33) Find the [OH - ] of a solution that has a pOH equal to 13.5. [OH - ] = 3.17 x 10 -14 M

46. Calculating Ion Concentration From Ion Concentration n If either [H + ] or [OH - ] is known, the hydrogen ion or hydroxide ion can be found. [H + ] [OH - ] = 1 x 10 -14

47. Problem 34) Find the hydrogen ion concentration if the hydroxide ion concentration equals: 1 x 10 –8 M. [H + ] = 1 x 10 -14 [OH - ]1 x 10 -8 [H + ] = 1 x 10 -6 M

48. Problem 35) Find the hydrogen ion concentration if the hydroxide ion concentration equals: 1 x 10 –2 M. [H + ] = 1 x 10 -14 [OH - ]1 x 10 -2 [H + ] = 1 x 10 -12 M

49. Problem 36) Find the hydroxide ion concentration if the hydrogen ion concentration equals: 1 x 10 –4 M. [OH - ] = 1 x 10 -14 [H + ]1 x 10 -4 [OH - ] = 1 x 10 -10 M

50. Problem 37) Find the hydroxide ion concentration if the hydrogen ion concentration equals: 1 x 10 –9 M. [OH - ] = 1 x 10 -14 [H + ]1 x 10 -9 [OH - ] = 1 x 10 -5 M

51. Problem 38) Find the hydrogen ion concentration if the hydroxide ion concentration equals: 3.25 x 10 –3 M. [H + ] = 1 x 10 -14 [OH - ] [H + ] 3.25 x 10 -3 [H + ] = 3.08 x 10 -12 M

52. Problem 39) Find the hydroxide ion concentration if the hydrogen ion concentration equals: 6.44 x 10 –6 M. [OH - ] = 1 x 10 -14 [H + ] 6.44 x 10 -6 [OH - ] = 1.55 x 10 -9 M

53. Indicators

54. Indicators n Chemical dyes whose colors are affected by acidic and basic solutions are called indicators. n n Many indicators do not have a sharp color change as a function of pH. n n Most indicators tend to be red in more acidic solutions.

55. Indicators

56. Indicators 40) Which indicator is best to show an equivalence point pH of 4? Methyl orange

57. Indicators 41) Which indicator is best to show an equivalence point pH of 11? Alizarin yellow R

58. Indicators 42) Which indicator is best to show an equivalence point pH of 2? Thymol blue

59. Neutralization Reactions

60. n The reaction of an acid and a base is called a neutralization reaction. Acid + Base  Salt + water Acid + Base  Salt + water n Salt = an ionic compound

61. Neutralization Reactions

62. n Consider the following neutralization reaction.

63. Neutralization Reactions

65. Example 43) Predict the products of and balance the following neutralization reaction. (Remember to check the oxidation numbers of the ions in the salt produced.) a) HNO 3 + KOH 

66. Example, cont. n Predict the products of and balance the following neutralization reaction. a) HNO 3 + KOH  The salt is composed of the ___________ ion and the _______ ion. potassium nitrate

67. Example, cont. n Predict the products of and balance the following neutralization reaction. a) HNO 3 + KOH  K NO 3 1+1- Since the 2 oxidation numbers add to give zero, you do not need to criss-cross.

68. Example, cont. n Predict the products of and balance the following neutralization reaction. a) HNO 3 + KOH  KNO 3 + H 2 O

69. Problem n Predict the products of and balance the following neutralization reaction. (Remember to check the oxidation numbers of the ions in the salt produced.) b) HCl + Mg(OH) 2  MgCl 2 + 2H 2 O2

70. Problem n Predict the products of the following neutralization reaction. (Remember to check the oxidation numbers of the ions in the salt produced.) c) H 2 SO 4 + NaOH  Na 2 SO 4 + 2H 2 O2

71. Neutralization

72. Problem 44) How many moles of HNO 3 are needed to neutralize 0.86 moles of KOH? 0.86 moles KOH 1 mole KOH 1 mole HNO 3 = 0.86 moles KOH HNO 3 + KOH  KNO 3 + H 2 O

73. Problem 45) How many moles of HCl are needed to neutralize 3.5 moles of Mg(OH) 2 ? 3.5 moles Mg(OH) 2 1 mol Mg(OH) 2 2 mol HCl = 7.0 moles HCl 2HCl + Mg(OH) 2  MgCl 2 + 2H 2 O

74. Problem 46) How many moles of H 3 PO 4 are needed to neutralize 3.5 moles of Mg(OH) 2 ? 2.3 moles H 3 PO 4 2H 3 PO 4 + 3Mg(OH) 2  Mg 3 (PO 4 ) 2 + 6H 2 O

75. Problem 47) How many moles of HC 2 H 3 O 2 are needed to neutralize 3.5 moles of Cr(OH) 3 ? 11 moles HC 2 H 3 O 2 3HC 2 H 3 O 2 + Cr(OH) 3  Cr(C 3 H 3 O 2 ) 3 + 3H 2 O

76. Problem 48) If it takes 87 mL of an HCl solution to neutralize 0.67 moles of Mg(OH) 2 what is the concentration of the HCl solution? 0.67 moles Mg(OH) 2 1 mol Mg(OH) 2 2 mol HCl = 1.3 moles HCl 2HCl + Mg(OH) 2  2KCl + 2H 2 O

77. Problem, cont. 48) If it takes 87 mL of an HCl solution to neutralize 0.67 moles of Mg(OH) 2 what is the concentration of the HCl solution? M = 0.087 L moles liters 1.3 mol M = 15 M

78. Problem 49) If it takes 58 mL of an H 2 SO 4 solution to neutralize 0.34 moles of NaOH what is the concentration of the H 2 SO 4 solution? 0.34 moles NaOH 2 mol NaOH 1 mol H 2 SO 4 = 0.17 moles H 2 SO 4 H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O

79. Problem, cont. 49) If it takes 58 mL of an H 2 SO 4 solution to neutralize 0.34 moles of NaOH what is the concentration of the H 2 SO 4 solution? M = 0.058 L moles liters 0.17 mol M = 2.9 M

80. Problem 50) If it takes 85 mL of an HNO 3 solution to neutralize 0.54 moles of Mg(OH) 2 what is the concentration of the HNO 3 solution? M = 13 M 2HNO 3 + Mg(OH) 2  Mg(NO 3 ) 2 + 2H 2 O

81. Problem 51) If it takes 150. mL of an Ca(OH) 2 solution to neutralize 0.800 moles of HCl what is the concentration of the Ca(OH) 2 solution? M = 2.67 M 2HCl + Ca(OH) 2  CaCl 2 + 2H 2 O

82. Acid Rain

83. n Acid Rain is any rain with a pH less than 5.6. n Pure rain is naturally acidic because of dissolved CO 2. n It is caused by the man-made oxides of sulfur and nitrogen. n SO 3 + H 2 O  H 2 SO 4

84. Acid Rain n Research shows acid rain is associated with parts of a country where heavy industries are situated & also down-wind from such sites. n Analysis of acid rain indicates that especially sulfur oxides, SO x and nitrogen oxides, NO x are mostly responsible from rain acidity.

85. Acid Rain n Snow fog, sleet, hail and drizzle all become contaminated with acids when SO x and NO x are present as pollutants.

86. Acid Rain n Acid Rain damage caused by all the fires and ash in the air.

87. Acid Rain n More damage from Acid Rain.

88. Titration

89. Titration Determining an Unknown

90. Titration n The general process of determining the molarity of an acid or a base through the use of an acid-base reaction is called an acid-base titration.

91. Titration

92. Titration n The known reactant molarity is used to find the unknown molarity of the other solution. n Solutions of known molarity that are used in this fashion are called standard solutions.

93. Titration n In a titration, the molarity of one of the reactants, acid or base, is known, but the other is unknown.

94. Problem 52) A 15.0-mL sample of a solution of H 2 SO 4 with an unknown molarity is titrated with 32.4 mL of 0.145M NaOH to the bromothymol blue endpoint. Based upon this titration, what is the molarity of the sulfuric acid solution? H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O

95. Problem, cont. n n First find the number of moles of the solution for which you know the molarity and volume. n n … is titrated with 32.4 mL of 0.145M NaOH … M = 0.0324 L moles liters x x = 4.70 x 10 -3 moles NaOH 0.145 M =

96. Problem, cont. n n Next, use the mole-mole ratio to determine the moles of the unknown. 4.70 x 10 -3 mol NaOH 2 mol NaOH 1 mol H 2 SO 4 = 2.35 x 10 -3 moles H 2 SO 4 H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O

97. Problem, cont. n Finally, determine the molarity of the unknown solution. n n A 15.0-mL sample of a solution of H 2 SO 4 … M = 0.015 L moles liters 2.35 x 10 -3 mol M = 0.157 M

98. Problem 53) If it takes 45 mL of a 1.0 M NaOH solution to neutralize 57 mL of HCl, what is the concentration of the HCl ? HCl + NaOH  NaCl + H 2 O (0.79 M)

99. Problem 54) If it takes 67.0 mL of 0.500 M H 2 SO 4 to neutralize 15.0 mL of Al(OH) 3 what was the concentration of the Al(OH) 3 ? 3H 2 SO 4 + 2Al(OH) 3  Al 2 (SO 4 ) 3 + 6H 2 O (1.49 M)

100. Problem 55) How many moles of 0.275 M HCl will be needed to neutralize 25.0 mL of 0.154 M NaOH? HCl + NaOH  NaCl + H 2 O (0.00385 moles)

101. Titration Curves

102. A plot of pH versus volume of acid (or base) added is called a titration curve. Titration Curves

103. Strong Base-Strong Acid Titration Curve Titration Curves

104. Consider adding a strong base (e.g. NaOH) to a solution of a strong acid (e.g. HCl). Titration Curves M M HCl

105. –Before any base is added, the pH is given by the strong acid solution. Therefore, pH < 7. Titration Curves M M HCl

106. –When base is added, before the equivalence point, the pH is given by the amount of strong acid in excess. Therefore, pH < 7. Titration Curves M M HCl equivalence point

107. –At equivalence point, the amount of base added is stoichiometrically equivalent to the amount of acid originally present. Therefore, pH = 7. Titration Curves M M HCl

108. To detect the equivalent point, we use an indicator that changes color somewhere near 7.00. Titration Curves

109. –Past the equivalence point all acid has been consumed. Thus one need only account for excess base. Therefore, pH > 7. Titration Curves M M HCl