2. Acids and Bases Part 2

3. The pH Scale The pH scale is used to describe the concentration of an acidic or basic solution.

4. 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.

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

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

7. Water n Water ionizes; it dissociates into ions. H 2 O  H + + OH - H 2 O  H + + OH - n The reaction above is called the autoionization of water. n H + is the hydronium ion and OH - is the hydroxide ion.

8. 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.

9. Ion-Product Constant n If [H + ] > 10 -7 then [OH - ] 10 -7 then [OH - ] < 10 -7 n As [H + ] increases, [OH - ] decreases. n The solution is acidic when [H + ] > [OH - ] n If [H + ] 10 -7 n The solution is basic when [OH - ] > [H + ]

10. 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.

11. pH At 25 °C: 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 hydronium 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. 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

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

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

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

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

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

24. 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

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

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

27. 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

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

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

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

31. 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

32. 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

33. 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

34. Indicators

35. 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.

36. Indicators

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

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

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

40. Neutralization Reactions

41. 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

42. Neutralization Reactions

43. n Consider the following neutralization reaction.

44. Neutralization Reactions

46. 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 

47. 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

48. Example, cont. n Predict the products of and balance the following neutralization reaction. HNO 3 + KOH  HNO 3 + KOH  K NO3 1+1- Since the 2 oxidation numbers add to give zero, you do not need to criss-cross.

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

50. 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

51. 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

52. Neutralization

53. 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

54. 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

55. 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

56. 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

57. Titration

58. Titration Determining an Unknown

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

60. Titration

61. 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.

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

63. 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

64. 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 =

65. 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

66. 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

67. Shorcut! n For monoprotic acid titrations, the mole:mole ratio of acid:base = 1:1, therefore… M a V a = M b V b

68. 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) MaMa VaVa MbMb VbVb = (57 mL)(1.0 M)(45 mL)

69. Problem n If it takes 67.0 mL of a 0.500 M HClO 4 to neutralize 15.0 mL of KOH, what was the concentration of the KOH? (2.23 M)

70. Problem n How many milliliters of 0.275 M HCl will be needed to neutralize 25.0 mL of a 0.154 M NaOH solution? (14.0 mL)

71. Titration Curves

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

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

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

75. –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

76. –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

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

78. –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