1. Working with [H 3 O + ], [OH – ], pH, and pOH Introduction Here, we’ll introduce some useful relationships that exist among hydronium concentration, hydroxide concentration, pH, and pOH

2. From Math: Here’s something that we should be aware of from Math, concerning logs

3. From Math: If: a × b = c If A TIMES B is equal to c

4. From Math: If: a × b = c Then: log(a) + log(b) = log(c) Then the log of (a) PLUS the log of (b) is equal to the log of (c).

5. We already know that in any aqueous solution the concentration of hydronium times the concentration of hydroxide is equal to Kw.

6. If we take the log of everything, it follows that the log of H3O+ concentration plus the log of OH minus concentration is equal to the log of Kw.

7. Now, we’ll multiply everything by negative 1 –1

8. And we get that the negative log of hydronium ion concentration plus the negative log of hydroxide ion concentration is equal to the negative log of Kw.

9. The negative log of the hydronium ion concentration is the pH,

10. the negative log of the hydroxide ion concentration is pOH,

11. And the negative log of Kw is equal to something called pKw.

12. So we have two important equations: the concentration of hydronium times the concentration of hydroxide is equal to Kw.

13. And the pH PLUS the pOH is equal to pKw.

14. Where the pK w is defined as the –logK w Where pK w = –logK w

15. Both of these equations are true for ANY temperature at which water is a liquid. True for any temperature at which water is a liquid

16. Now, we’ll zoom into a temperature of 25°C. 25°C

17. At 25°C, K w = 1.0 × 10 -14 At 25°C K w = 1.0 × 10 -14 pK w = –logK w = –log(1.0 × 10 –14 ) = 14.00

18. So at 25°C, the pKw… At 25°C K w = 1.0 × 10 -14 pK w = –logK w = –log(1.0 × 10 –14 ) = 14.00

19. Which is the negative log of Kw At 25°C K w = 1.0 × 10 -14 pK w = –logK w = –log(1.0 × 10 –14 ) = 14.00

20. Is the –log of 1.0 × 10 –14 … At 25°C K w = 1.0 × 10 -14 pK w = –logK w = –log(1.0 × 10 –14 ) = 14.00

21. Which is equal to 14.00 At 25°C K w = 1.0 × 10 -14 pK w = –logK w = –log(1.0 × 10 –14 ) = 14.00

22. So we can say that specifically at 25°C At 25°C pK w = 14.00

23. The pKw is equal to 14.00 At 25°C pK w = 14.00

24. Remember, we had recently determined that pKw is equal to pH + pOH At 25°C pK w = 14.00 pK w = pH + pOH = 14.00

25. And at 25° pkw = 14.00 At 25°C pK w = 14.00 pK w = pH + pOH = 14.00

26. Therefore we can say that at 25°C, pH + pOH = 14.00 At 25°C pH + pOH = 14.00

27. You’ll be using this equation a lot. Just make sure you use caution. Remember, this is true ONLY at 25°C. At 25°C pH + pOH = 14.00 This is true ONLY at 25°C!

28. Remember that if temperature is not mentioned in a problem, we can assume that it is 25°C At 25°C If temperature is not mentioned, assume that it is 25°C

29. And we can assume that pH + pOH is equal to 14. At 25°C If temperature is not mentioned, assume that pH + pOH = 14.00

30. Here’s an example. We’re told that the pOH of a solution is 3.49 and we’re asked what the pH is? The pOH of a solution is 3.49. What is the pH? pH + pOH = 14.00

31. We are not given the temperature, so we can assume its 25°C and that pH + pOH is equal to 14 The pOH of a solution is 3.49. What is the pH? pH + pOH = 14.00

32. We want to find the pH, so we rearrange the blue equation to solve for pH, and we get the yellow equation: pH = 14 minus pOH. The pOH of a solution is 3.49. What is the pH? pH + pOH = 14.00

33. Which is 14 minus 3.49 The pOH of a solution is 3.49. What is the pH? pH + pOH = 14.00

34. And that equals 10.51. So the pH is 10.51. The pOH of a solution is 3.49. What is the pH? pH + pOH = 14.00

35. Now we’ll review the things we know are true at any temperature and things we know are true ONLY at 25°C. At ANY TemperatureONLY at 25°C

36. We’ll start with equations that are true at ANY temperature At ANY TemperatureONLY at 25°C

37. [H + ][OH – ] = K w At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w

38. pH + pOH = pK w At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w pH + pOH = pK w

39. pH = –log[H 3 O + ] At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w pH + pOH = pK w pH = –log[H 3 O + ]

40. [H 3 O + ] = 10 –pH At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w pH + pOH = pK w pH = –log[H 3 O + ] [H 3 O + ] = 10 –pH

41. pOH = –log[OH – ] At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w pH + pOH = pK w pH = –log[H 3 O + ] [H 3 O + ] = 10 –pH pOH = –log[OH – ]

42. [OH – ] = 10 –pOH At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w pH + pOH = pK w pH = –log[H 3 O + ] [H 3 O + ] = 10 –pH pOH = –log[OH – ] [OH – ] = 10 –pOH

43. pK w = –log(K w ) At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w pH + pOH = pK w pH = –log[H 3 O + ] [H 3 O + ] = 10 –pH pOH = –log[OH – ] [OH – ] = 10 –pOH pK w = –log(K w )

44. We can solve the previous equation for Kw, we get K w = 10 –pKw At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w pH + pOH = pK w pH = –log[H 3 O + ] [H 3 O + ] = 10 –pH pOH = –log[OH – ] [OH – ] = 10 –pOH pK w = –log(K w ) K w = 10 –pKw

45. Now we’ll review what is true ONLY at 25°C At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w pH + pOH = pK w pH = –log[H 3 O + ] [H 3 O + ] = 10 –pH pOH = –log[OH – ] [OH – ] = 10 –pOH pK w = –log(K w ) K w = 10 –pKw

46. [H + ][OH – ] = 1.0 × 10 –14 At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w [H 3 O + ][OH – ] =1.0 × 10 –14 pH + pOH = pK w pH = –log[H 3 O + ] [H 3 O + ] = 10 –pH pOH = –log[OH – ] [OH – ] = 10 –pOH pK w = –log(K w ) K w = 10 –pKw

47. pH + pOH = 14.00 At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w [H 3 O + ][OH – ] =1.0 × 10 –14 pH + pOH = pK w pH + pOH = 14.00 pH = –log[H 3 O + ] [H 3 O + ] = 10 –pH pOH = –log[OH – ] [OH – ] = 10 –pOH pK w = –log(K w ) K w = 10 –pKw

48. K w = 1.0 × 10 –14 At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w [H 3 O + ][OH – ] =1.0 × 10 –14 pH + pOH = pK w pH + pOH = 14.00 pH = –log[H 3 O + ]K w = 1.0 × 10 –14 [H 3 O + ] = 10 –pH pOH = –log[OH – ] [OH – ] = 10 –pOH pK w = –log(K w ) K w = 10 –pKw

49. pK w = 14.00. At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w [H 3 O + ][OH – ] =1.0 × 10 –14 pH + pOH = pK w pH + pOH = 14.00 pH = –log[H 3 O + ]K w = 1.0 × 10 –14 [H 3 O + ] = 10 –pH pK w = 14.00 pOH = –log[OH – ] [OH – ] = 10 –pOH pK w = –log(K w ) K w = 10 –pKw

50. So we see that any equations that contain the number 14, are ONLY true at 25°C. At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w [H 3 O + ][OH – ] =1.0 × 10 –14 pH + pOH = pK w pH + pOH = 14.00 pH = –log[H 3 O + ]K w = 1.0 × 10 –14 [H 3 O + ] = 10 –pH pK w = 14.00 pOH = –log[OH – ] [OH – ] = 10 –pOH pK w = –log(K w ) K w = 10 –pKw

51. In order to succeed in the rest of this unit You REALLY need to KNOW all these equations! Pause and make a screen capture of this, save it, and go over it periodically. At ANY TemperatureONLY at 25°C [H 3 O + ][OH – ] = K w [H 3 O + ][OH – ] =1.0 × 10 –14 pH + pOH = pK w pH + pOH = 14.00 pH = –log[H 3 O + ]K w = 1.0 × 10 –14 [H 3 O + ] = 10 –pH pK w = 14.00 pOH = –log[OH – ] [OH – ] = 10 –pOH pK w = –log(K w ) K w = 10 –pKw You REALLY need to KNOW all these equations!

52. Here’s the “square” at 25°C. It shows all the formulas you can use to make one step or two step conversions among [H3O+], [OH-], pH, and pOH. [H 3 O + ] [OH – ] pH pOH pH = –log[H 3 O + ] [OH – ] = 10 –pOH pOH = –log[OH – ] [H 3 O + ] = 10 –pH [H 3 O + ][OH–] = 1.00 × 10 –14 pH + pOH = 14.00

53. It would be good if you could draw something similar to this from memory. It will help you with the calculations you’ll be required to do. [H 3 O + ] [OH – ] pH pOH pH = –log[H 3 O + ] [OH – ] = 10 –pOH pOH = –log[OH – ] [H 3 O + ] = 10 –pH [H 3 O + ][OH–] = 1.00 × 10 –14 pH + pOH = 14.00

54. For now, you may want to pause the video, take a screen shot and print yourself a copy of this to work with. [H 3 O + ] [OH – ] pH pOH pH = –log[H 3 O + ] [OH – ] = 10 –pOH pOH = –log[OH – ] [H 3 O + ] = 10 –pH [H 3 O + ][OH–] = 1.00 × 10 –14 pH + pOH = 14.00

55. Here’s a simpler version we can use to help us come up with plans for calculations [H 3 O + ] [OH – ] pH pOH

56. For example, Let’s say we’re given the hydronium ion concentration and we want to find the pOH [H 3 O + ] [OH – ] pH pOH ? Given

57. We can do this in two steps. We could start (click) by converting hydronium ion concentration of pH… [H 3 O + ] [OH – ] pH pOH Given ?

58. And in the second step (click), we’ll convert pH to pOH. [H 3 O + ] [OH – ] pH pOH Given ?

59. Alternately, we could have started by converting (click) hydronium concentration to hydroxide concentration [H 3 O + ] [OH – ] pH pOH Given ?

60. And then (click) hydroxide ion concentration to pOH. This would give us the same answer as the other method. [H 3 O + ] [OH – ] pH pOH Given ?

61. In another example, let’s say we’re given the pH and we want to find hydroxide ion concentation. [H 3 O + ] [OH – ] pH pOH ? Given

62. We could start (click) by converting pH to pOH [H 3 O + ] [OH – ] pH pOH ? Given

63. And then (click) pOH to hydroxide concentration [H 3 O + ] [OH – ] pH pOH ? Given

64. Or alternately, we could have started with pH (click) and converted to hydronium concentration [H 3 O + ] [OH – ] pH pOH ? Given

65. And then from (click) hydronium concentration to hydroxide concentration. [H 3 O + ] [OH – ] pH pOH ? Given