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 + ] > 10 -7 then [OH - ] 10 -7 then [OH - ] < 10 -7 n The solution is acidic when [H + ] > [OH - ] n If [H + ] 10 -7 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 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.

13. pH and pOH n The pH of a solution equals the negative logarithm of the hydrogen ion concentration.

14. pOH n Chemists have also defined a pOH scale to express the basicity of a solution.

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

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

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

18. Problem n n 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

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

20. Example, cont. n n Find the pH of the following solution. The hydroxide ion concentration equals: 10 –8 M. pH + = 14pOH8 pH = 6

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

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

23. Problem n n 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)

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

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

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

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

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

29. Example n n Find the [H + ] of a solution that has a pOH equal to 6. pH + = 14pOH6 pH = 8

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

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

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

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

34. 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

35. Example n n 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

36. Problem n n 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

37. Problem n n 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

38. Problem n n 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

39. Indicators

40. 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.

41. Indicators

42. Indicators n Which indicator is best to show an equivalence point pH of 4? Methyl orange

43. Indicators n Which indicator is best to show an equivalence point pH of 11? Alizarin yellow R

44. Indicators n Which indicator is best to show an equivalence point pH of 2? Thymol blue

45. Neutralization Reactions

46. 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

47. Neutralization Reactions

48. n Consider the following neutralization reaction.

49. Neutralization Reactions

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

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

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

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

55. 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.) HCl + Mg(OH) 2  HCl + Mg(OH) 2  MgCl 2 + 2H 2 O 2

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

57. Neutralization

58. Example n How many moles of HNO 3 are need 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

59. Problem n 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

60. Problem n 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

61. Problem n 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

62. Example n 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

63. Example, cont. n 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

64. Problem n 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

65. Problem, cont. n 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

66. Problem n 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

67. Problem n 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

68. Acid Rain

69. 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

70. 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.

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

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

73. Acid Rain n More damage from Acid Rain.

74. Titration

75. Titration Determining an Unknown

76. 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.

77. Titration

78. 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.

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

80. Example n n 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

81. Example, 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 =

82. Example, 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

83. Example, 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

84. Problem n 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)

85. Problem n 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)

86. Problem n 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 (0.0140 moles)

87. Titration Curves

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

89. Strong Base-Strong Acid Titration Curve Titration Curves

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

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

92. –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

93. –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

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

95. –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