1. You’ll find out what buffer solutions are and how they are prepared. Buffer Solutions Definition and Preparation

2. A buffer solution can be defined as a solution that Minimizes Changes in pH when small amounts of acid or base are added to it A Buffer Solution is a solution that minimizes changes in pH when small amounts of acid or base are added to it.

3. Or it can also be defined as… A Buffer Solution is a solution that minimizes changes in pH when small amounts of acid or base are added to it. or

4. a solution that maintains a relatively constant pH when small amounts of acid or base are added to it. A Buffer Solution is a solution that minimizes changes in pH when small amounts of acid or base are added to it. A Buffer Solution is a solution that maintains a relatively constant pH when small amounts of acid or base are added to it. or

5. To get an idea of what a buffer solution does, we’ll start with 1 litre of pure water. Water is unbuffered, and it has an initial pH of 7. 7.00 unbuffered Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 Initial pH

6. Now, we’ll add 0.10 mol of the strong acid, HCl to the water. 7.00 0.10 mol HCl Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09

7. Watch the pH meter. 7.00 0.10 mol HCl 1.00 Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09

8. 1.00 We’ll note here that the final pH is 1 1.00 0.10 mol HCl Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 Final pH

9. 1.00 The pH went from 7 all the way down to 1, so we can see that it has decreased by 6 whole units. 1.00 0.10 mol HCl Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09

10. Now, we’ll go back again and start with 1 Litre of pure water. Again, it’s neutral pH is 7. And remember, water is unbuffered 7.00 unbuffered Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 Initial pH

11. This time, we’ll add 0.1 mol of the strong base, NaOH. 7.00 0.10 mol NaOH Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09

12. Watch the pH meter 7.00 0.10 mol NaOH 13.00 Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09

13. 1.00 We’ll make a note here that the final pH is 13. 13.00 0.10 mol NaOH Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 Final pH

14. 1.00 The pH went from 7 all the way up to 13, so that’s an increase of 6 whole units. 13.00 0.10 mol NaOH Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09

15. What we’ll do now is replace the water with a buffer solution. This particular solution contains 1M acetic acid and 1M sodium acetate. 4.74 [CH 3 COOH] = 1.0 M [CH 3 COO – ] = 1.0 M Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09

16. We see the initial pH is 4.74 4.74 [CH 3 COOH] = 1.0 M [CH 3 COO – ] = 1.0 M Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 Initial pH

17. Now, we’ll add 0.1 mol of the strong acid HCl to this buffer solution and see what happens… 4.74 0.10 mol HCl Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09

18. 4.66 Watch the pH meter 4.74 0.10 mol HCl Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 4.66

19. 1.00 We see that the pH has gone down, but only down to 4.66. 0.10 mol HCl Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 4.66 Final pH

20. 1.00 In going from 4.74 down to 4.66, the pH has dropped only by 0.08. This is a very small change in pH. Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 4.66 0.10 mol HCl

21. 1.00 Compare this with the very large drop of 6 pH units when 0.1 mol of HCl was added to unbuffered pure water. Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 4.66 0.10 mol HCl

22. Now, we’ll go back and start again with our buffer solution 4.74 Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 [CH 3 COOH] = 1.0 M [CH 3 COO – ] = 1.0 M

23. that has an initial pH of 4.74. 4.74 Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 [CH 3 COOH] = 1.0 M [CH 3 COO – ] = 1.0 M Initial pH

24. This time, we’ll add 0.1 mol of the strong base NaOH to 1 Litre of this buffer solution and see what happens. Make a prediction. 4.74 0.10 mol NaOH Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09

25. 4.66 Watch the pH meter 4.74 0.10 mol NaOH Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 4.83

26. 1.00 As a result of adding the base, the pH rose slightly to a final value of 4.83. 0.10 mol NaOH Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 4.83 Final pH

27. 1.00 The pH started at 4.74 and rose to 4.83, so that is an increase of only 0.09, which is a very small increase. 0.10 mol NaOH Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 4.83

28. 1.00 Compare this with an increase of 6 whole units when NaOH was added to pure, unbuffered water. 0.10 mol NaOH Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 4.83

29. We’ll summarize our results. When a small amount of acid is added to pure, unbuffered water, the pH drops dramatically Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 When a small amount of acid is added to pure, unbuffered water, the pH drops dramatically

30. And when a small amount of base is added to pure, unbuffered water, the pH Rises dramatically Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 When a small amount of base is added to pure, unbuffered water, the pH rises dramatically

31. But when a small amount of acid is added to a buffer solution, the pH drops very slightly Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 When a small amount of acid is added to a buffer solution, the pH drops very slightly

32. And when a small amount of base is added to a buffer solution, the pH rises very slightly Start with Initial pHAdd Final pH Change in pH water7.00 0.10 mol HCl 1.00– 6.00 water7.00 0.10 mol NaOH 13.00+ 6.00 buffer4.74 0.10 mol HCl 4.66– 0.08 buffer4.74 0.10 mol NaOH 4.83+ 0.09 When a small amount of base is added to a buffer solution, the pH rises very slightly

33. So now we know what a buffer solution does. It minimizes changes in pH when a small amount of acid or base is added to it. A Buffer Solution is a solution that minimizes changes in pH when small amounts of acid or base are added to it.

34. What we’ll do now is take a look at how buffer solutions are prepared. How Buffer Solutions are Prepared

35. To be able to minimize changes in pH, A Buffer Solution is a solution that minimizes changes in pH when small amounts of acid or base are added to it.

36. A Buffer Solution must be able to partially neutralize BOTH acids and bases that are added. A Buffer Solution is a solution that minimizes changes in pH when small amounts of acid or base are added to it. A Buffer Solution must be able to partially neutralize BOTH acids and bases that are added.

37. (click) A Buffer Solution must be able to partially neutralize BOTH acids and bases that are added.

38. In order to do this, it must contain relatively high amounts of BOTH a base and an acid. A Buffer Solution must be able to partially neutralize BOTH acids and bases that are added. In order to do this, it must contain relatively high amounts of BOTH a base and an acid.

39. (click) In order to do this, it must contain relatively high amounts of BOTH a base and an acid.

40. This can ONLY occur if the base and acid are both WEAK. In order to do this, it must contain relatively high amounts of BOTH a base and an acid. This can ONLY occur if the base and acid are both WEAK.

41. (click) This can ONLY occur if the base and acid are both WEAK.

42. A buffer solution consists of a weak conjugate acid-base pair in which both the acid and the base have relatively high concentrations. This can ONLY occur if the base and acid are both WEAK. A buffer solution consists of a weak conjugate acid-base pair in which both the acid and the base have relatively high concentrations.

43. (click) A buffer solution consists of a weak conjugate acid-base pair in which both the acid and the base have relatively high concentrations.

44. An example is a solution that contains 1 molar ethanoic or acetic acid, which is a weak acid and 1 molar ethanoate or acetate ion, which is a weak base. A buffer solution consists of a weak conjugate acid-base pair in which both the acid and the base have relatively high concentrations. An example is a solution with 1 M CH 3 COOH and 1 M CH 3 COO –. Weak Acid Weak Base

45. We’ll use the more familiar names acetic acid, and acetate ion here. A buffer solution consists of a weak conjugate acid-base pair in which both the acid and the base have relatively high concentrations. An example is a solution with 1 M CH 3 COOH and 1 M CH 3 COO –. Weak Acid Weak Base

46. (click after animation) An example is a solution with 1 M CH 3 COOH and 1 M CH 3 COO –. Weak Acid Weak Base

47. An example is a solution that contains 1 molar acetic acid, which is a weak acid and 1 molar acetate ion, which is a weak base. An example is a solution with 1 M CH 3 COOH and 1 M CH 3 COO –. An Equilibrium is established in which CH 3 COOH and CH 3 COO – are both 1 M:

48. In this solution, an equilibrium is established in which the concentrations of acetic acid and the acetate ion are both 1 molar. An Equilibrium is established in which CH 3 COOH and CH 3 COO – are both 1 M: CH 3 COOH + H 2 O  H 3 O + + CH 3 COO – 1 M

49. And the hydronium ion concentration is quite low. An Equilibrium is established in which CH 3 COOH and CH 3 COO – are both 1 M: CH 3 COOH + H 2 O  H 3 O + + CH 3 COO – 1 M low

50. The 1 molar acetic acid is available to neutralize small amounts of strong base that might be added to the solution. An Equilibrium is established in which CH 3 COOH and CH 3 COO – are both 1 M: CH 3 COOH + H 2 O  H 3 O + + CH 3 COO – 1 M low Is able to neutralize added strong base

51. And the 1 molar acetate ion is available to neutralize small amounts of strong acid that might be added to the solution. An Equilibrium is established in which CH 3 COOH and CH 3 COO – are both 1 M: CH 3 COOH + H 2 O  H 3 O + + CH 3 COO – 1 M low Is able to neutralize added strong base Is able to neutralize added strong acid

52. To prepare a buffer solution, molecular acids can be added directly. CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO – (aq) 1 M Molecular acids can be added directly

53. To prepare the buffer in this example, concentrated acetic or ethanoic acid can be diluted to 1 molar. CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO – (aq) 1 M Molecular acids can be added directly Concentrated acetic acid (ethanoic acid) can be diluted to 1 M

54. We can use soluble ionic salts as sources of ions in a buffer solution.. CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO – (aq) 1 M Soluble Ionic Salts are Sources of Ions

55. To make this buffer, we could add 1 mole of solid sodium acetate (or sodium ethanoate) to enough 1 M acetic acid, to make 1 L of solution. CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO – (aq) 1 M Soluble Ionic Salts are Sources of Ions 1 mol of solid sodium acetate (sodium ethanoate) (NaCH 3 COO) can be added

56. The salt, sodium acetate dissociates into sodium and acetate ions. CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO – (aq) Soluble Ionic Salts are Sources of Ions 1 mol of solid sodium acetate (sodium ethanoate) (NaCH 3 COO) can be added

57. The ion in the salt that’s not part of the buffer solution is a spectator ion. In this case it’s the sodium ion… CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO – (aq) Soluble Ionic Salts are Sources of Ions 1 mol of solid sodium acetate (sodium ethanoate) (NaCH 3 COO) can be added

58. And we’ll discard it, and we’re left with the acetate ion. CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO – (aq) Soluble Ionic Salts are Sources of Ions 1 mol of solid sodium acetate (sodium ethanoate) (NaCH 3 COO) can be added

59. Which is part of the buffer solution. CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO – (aq)

60. This buffer is prepared using the Weak Acid, CH 3 COOH, and a salt of its conjugate base, NaCH 3 COO … CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO – (aq) 1 M CH 3 COOH (Weak Acid) 1 M NaCH 3 COO ( Salt of Its Conjugate Base ) 1 M

61. Buffers prepared using a weak acid and a salt of its conjugate base are often acidic buffers, like this one. This is true only for combinations that produce a pH below 7. CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO – (aq) 1 M CH 3 COOH (Weak Acid) 1 M NaCH 3 COO ( Salt of Its Conjugate Base ) Called an ACIDIC Buffer 1 M

62. Another example of a buffer solution is the weak base ammonia (NH 3 ), and it’s conjugate acid the ammonium ion (NH 4 + ) NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH – (aq) Another example of a buffer solution is the weak base ammonia (NH 3 ), and it’s conjugate acid the ammonium ion (NH 4 + )

63. Here’s the equilibrium equation for this buffer solution. NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH – (aq) Another example of a buffer solution is the weak base ammonia (NH 3 ), and it’s conjugate acid the ammonium ion (NH 4 + ) 1 M

64. Molecular bases like NH 3 can be added directly NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH – (aq) 1 M Molecular bases like NH 3 can be added directly

65. Again, soluble ionic salts are sources we can use for ions in buffer solutions. NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH – (aq) 1 M Soluble Ionic Salts are Sources of Ions

66. NH 4 Cl will dissociate into NH 4 + and Cl minus ions NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH – (aq) 1 M Soluble Ionic Salts are Sources of Ions 1 mol of solid ammonium chloride (NH 4 Cl) can be added

67. In this salt, the chloride ion is not part of the buffer, so it is acting as a spectator ion… NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH – (aq) 1 M Soluble Ionic Salts are Sources of Ions 1 mol of solid ammonium chloride (NH 4 Cl) can be added

68. and can be discarded… NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH – (aq) 1 M Soluble Ionic Salts are Sources of Ions 1 mol of solid ammonium chloride (NH 4 Cl) can be added

69. Leaving us with the ammonium ion NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH – (aq) 1 M Soluble Ionic Salts are Sources of Ions 1 mol of solid ammonium chloride (NH 4 Cl) can be added

70. Which is part of this buffer solution. NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH – (aq) 1 M

71. So this buffer is prepared using the weak base, NH3, and a salt of its conjugate acid NH 4 Cl. NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH – (aq) 1 M 1 M NH 3 (Weak Base) 1 M NH 4 Cl ( Salt of Its Conjugate Acid )

72. Buffers prepared with a weak base and a salt of its conjugate acid are called basic buffers. NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH – (aq) 1 M 1 M NH 3 (Weak Base) 1 M NH 4 Cl ( Salt of Its Conjugate Acid ) Called a BASIC Buffer

73. In some buffer solutions both the weak acid and the weak base are ions:

74. An example is the solution containing 1 molar dihydrogen phosphate, H2PO4 minus, In some buffer solutions both the weak acid and the weak base are ions: + +  1 M

75. And 1 molar monohydrogen phosphate HPO4 2 minus. + +  1 M In some buffer solutions both the weak acid and the weak base are ions:

76. Here is the equilibrium equation for this buffer. The dihydrogen phosphate ion is the weak acid and its conjugate base, the monohydrogen phosphate ion, is the weak base. + +  Weak Acid Weak Base 1 M

77. Any soluble ionic compound which contains the dihydrogen phosphate ion would be a good source of this ion. An example is the soluble salt potassium dihydrogen phosphate. + +  Possible Source is 1 M KH 2 PO 4 1 M

78. It dissociates into K+ and H2PO4 minus ions + +  Possible Source is 1 M KH 2 PO 4 1 M

79. K+ is a spectator ion. + +  Possible Source is 1 M KH 2 PO 4 1 M

80. So it doesn’t appear in the overall reaction and we’re left with H2PO4 minus + +  Possible Source is 1 M KH 2 PO 4 1 M

81. Any soluble ionic compound which contains the monohydrogen phosphate ion would be a good source of this ion. An example is the soluble salt potassium monohydrogen phosphate, K2HPO4 + +  Possible Source is 1 M KH 2 PO 4 1 M Possible Source is 1 M K 2 HPO 4

82. K2HPO4 dissociates into 2K+ ions and one monohydrogen phosphate ion, HPO4 2 minus. + +  Possible Source is 1 M KH 2 PO 4 1 M Possible Source is 1 M K 2 HPO 4

83. K+ is a spectator and can be discarded… + +  Possible Source is 1 M KH 2 PO 4 1 M Possible Source is 1 M K 2 HPO 4

84. leaving us with monohydrogen phosphate ion, HPO4 2 minus. + +  Possible Source is 1 M KH 2 PO 4 1 M Possible Source is 1 M K 2 HPO 4

85. It is important for you to know that Strong acids and their conjugate bases, Strong acids and their conjugate bases, Or Strong bases and their conjugate acids Are NOT used to prepare buffer solutions.

86. Or Strong bases and their conjugate acids, Strong acids and their conjugate bases, Or Strong bases and their conjugate acids Are NOT used to prepare buffer solutions.

87. are NOT used to prepare buffer solutions. Strong acids and their conjugate bases, Or Strong bases and their conjugate acids Are NOT used to prepare buffer solutions.

88. Let’s say we have a solution made by mixing solid NaCl with HCl, in which they both have a resulting concentration of 1 molar. Strong acids and their conjugate bases, or Strong bases and their conjugate acids are NOT used to prepare buffer solutions. + +  1 M 1 M HCl (aq) 1 M NaCl (aq)

89. Our solution would initially contain 1M HCl and 1M Cl minus. Strong acids and their conjugate bases, or Strong bases and their conjugate acids are NOT used to prepare buffer solutions. + +  1 M 1 M HCl (aq) 1 M NaCl (aq)

90. Because HCl is a strong acid… Strong acids and their conjugate bases, or Strong bases and their conjugate acids are NOT used to prepare buffer solutions. + +  1 M Strong Acid

91. it will ionize 100% + +  1 M Strong Acid 1 0 0 % 0 M 2 M

92. When ionization is complete + +  0 M 2 M 1 0 0 %

93. There is no HCl left + +  0 M 2 M 1 0 0 %

94. And the concentrations of chloride and hydronium ions have increased + +  0 M 2 M 1 0 0 %

95. Hydronium ions are available to neutralize any added bases. +  2 M Available to neutralize bases

96. But the chloride ion is a neutral spectator, so it is unable to neutralize any added acids. +  2 M Neutral Spectator

97. The single arrow pointing to the right indicates that this reaction will ONLY move forward. The reverse reaction cannot occur. Cl minus ions do not react with H3O+ ions. +  2 M Neutral Spectator Single arrow pointing to the right

98. A buffer solution has to be able to neutralize both acids and bases. What we have remaining after 100% ionization can neutralize bases, but not acids. Therefore, it will NOT function as a buffer solution. +  2 M Will NOT function as a Buffer

99. Now, let’s consider strong bases. NaOH is an example of a strong base. It dissociates to form… Strong Base

100. Na + ions Strong Base

101. And OH minus ions. Strong Base

102. OH minus is considered to be a strong base. Strong Base

103. Will neutralize added acids. A strong base is able the neutralize added acids. Strong Base

104. The conjugate acid of OH minus is just water. So there is no such thing as the salt of the conjugate acid of a strong base. The Conjugate Acid of OH – is H 2 O

105. Strong bases are hydroxides of Group 1 or alkali metals, and Group 2 Alkaline Earth’s. The cation of a strong base will always be a neutral spectator ion. In this base it is the Na+ ion. Spectator Ion

106. Since they are spectators, cations of strong bases are not able to neutralize any added bases. Spectator Ion Will Not neutralize added bases.

107. The single arrow pointing to the right indicates that this reaction will ONLY move forward. The reverse reaction cannot occur. Na + ions do not react with OH – ions. Single arrow pointing to the right

108. A buffer solution has to be able to neutralize both acids and bases. What we have remaining after dissociation can neutralize acids, but not bases. Therefore, it will NOT function as a buffer solution. Will NOT function as a Buffer

109. Just a couple more things about buffer solutions. In order to be an effective buffer solution, a solution must contain both a weak acid and a weak base. In order to be an effective buffer, a solution must contain both a weak acid and a weak base.

110. The [weak acid] and the [weak base] should be relatively high. In order to be an effective buffer, a solution must contain both a weak acid and a weak base. The [weak acid] and the [weak base] should be relatively high.

111. For example, a solution with 1 M HF and 1 M NaF has a high capacity to buffer, or partially neutralize any added acid or base In order to be an effective buffer, a solution must contain both a weak acid and a weak base. The [weak acid] and the [weak base] should be relatively high. 1 M HF and 1 M NaF — high capacity to buffer (partially neutralize added acid or base)

112. A solution containing 0.1 M HF and 0.1 M NaF has a medium capacity to buffer added acids and bases. In order to be an effective buffer, a solution must contain both a weak acid and a weak base. The [weak acid] and the [weak base] should be relatively high. 1 M HF and 1 M NaF — high capacity to buffer (partially neutralize added acid or base) 0.1 M HF and 0.1 M NaF — medium capacity to buffer

113. But a solution with 0.001 M HF and 0.001 M NaF has a very low capacity to buffer. There is very little weak acid available to neutralize added bases and very little weak base available to neutralize added acids. In order to be an effective buffer, a solution must contain both a weak acid and a weak base. The [weak acid] and the [weak base] should be relatively high. 1 M HF and 1 M NaF — high capacity to buffer (partially neutralize added acid or base) 0.1 M HF and 0.1 M NaF — medium capacity to buffer 0.001 M HF and 0.001 M NaF — very low capacity to buffer

114. One more requirement. As well as containing both a weak acid and a weak base In order to be an effective buffer, a solution must contain both a weak acid and a weak base.

115. The [weak acid] and the [weak base] should either be equal or close to each other in value. In order to be an effective buffer, a solution must contain both a weak acid and a weak base. The [weak acid] and the [weak base] should be equal or close to each other in value.

116. For example, a solution in which the [HNO 2 ] is 0.80 M and the [NaNO 2 ] is 1.0 M is an effective buffer, because these values are close to each other. In order to be an effective buffer, a solution must contain both a weak acid and a weak base. The [weak acid] and the [weak base] should be equal or close to each other in value. 0.80 M HNO 2 and 1.0 M NaNO 2 is an effective buffer.

117. But a solution in which [ HNO 2 ] is 0.80 M and the [NaNO 2 ] is 0.002 M NOT an effective buffer. The values 0.80 M and 0.002 M are too different. In order to be an effective buffer, a solution must contain both a weak acid and a weak base. The [weak acid] and the [weak base] should be equal or close to each other in value. 0.80 M HNO 2 and 1.0 M NaNO 2 is an effective buffer. 0.80 M HNO 2 and 0.002 M NaNO 2 is NOT an effective buffer.

118. There is enough HNO2 available to partially neutralize added strong bases In order to be an effective buffer, a solution must contain both a weak acid and a weak base. The [weak acid] and the [weak base] should be equal or close to each other in value. 0.80 M HNO 2 and 1.0 M NaNO 2 is an effective buffer. 0.80 M HNO 2 and 0.002 M NaNO 2 is NOT an effective buffer.

119. But not much NO2 minus available to partially neutralize added strong acids. In order to be an effective buffer, a solution must contain both a weak acid and a weak base. The [weak acid] and the [weak base] should be equal or close to each other in value. 0.80 M HNO 2 and 1.0 M NaNO 2 is an effective buffer. 0.80 M HNO 2 and 0.002 M NaNO 2 is NOT an effective buffer.