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The Ionic Product of Water

The Ionic Product of Water. K w. Ionic Product of water, K w. Just because a solution contains [H + ] it doesn’t necessarily mean it’s acidic. All aqueous solutions will contain H + and OH - ions For an acid solution [H + ] > [OH - ]

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The Ionic Product of Water

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  1. The Ionic Product of Water Kw

  2. Ionic Product of water, Kw • Just because a solution contains [H+] it doesn’t necessarily mean it’s acidic. • All aqueous solutions will contain H+ and OH- ions • For an acid solution [H+] > [OH-] • For an alkaline solution [H+] <[OH-] • For an neutral solution [H+] =[OH-]

  3. Ionic Product of water, Kw H2O(l)+ H2O (l)H3O+(aq)+ OH-(aq) H2O(l)H+(aq)+ OH-(aq) Kc = [H+][OH-] [H2O]

  4. Ionic Product of water, Kw H2O(l)H+(aq)+ OH-(aq) Kc = [H+][OH-] [H2O] Since [H2O] is effectively constant and in large excess Kw = [H+][OH-] mol2dm-6

  5. Ionic Product of water, Kw At 298K the value of Kw is 1 x 10-14mol2dm-6 Kw = [H+][OH-] mol2dm-6 In pure water [H+] = [OH-] So Kw = [H+]2 Hence [H+] = √Kw At 298K [H+] = √(1 x 10-14) = 1 x 10-7 So at 289K the pH of pure water is 7

  6. Effect of temperature on Kw H2O(l)H+(aq)+ OH-(aq) Dissociation of water is endothermic Increasing temp increases Kw Increasing temp increases [H+] At higher tempspH of water will decrease

  7. At 321K the value of Kw is 4 x 10-14 mol2dm-6 calculate the pH of water at this temp Kw = [H+][OH-] mol2dm-6 In pure water [H+] = [OH-] So Kw = [H+]2 Hence [H+] = √Kw At 321K [H+] = √(4 x 10-14) = 2 x 10-7 pH= -log10[H+] So at 321K the pH of pure water is 6.70

  8. The pH of Strong Bases Kw gives us a method of calculating the pH of strong bases Kw = [H+][OH-] mol2dm-6 So [H+] = Kw [OH-]

  9. Calculate the pH of 0.1M KOH at 298K Since KOH is a strong base [OH-] = 0.1 Kw = [H+][OH-] = 1 x 10-14 mol2dm-6 [H+] = Kw [OH-] [H+] = 1 x 10-14 0.1 = 1 x 10-13 pH= -log10[H+] pH= 13

  10. At 338K the pH of pure water is 6.5. Calculate Kw at this temp. Hence calculate the pH of 0.01M NaOH at this temp pH= -log10[H+] [H+] = 3.162 x 10-7 Kw = [H+][OH-] [H+] = [OH-] for pure water Kw = [H+]2 for pure water

  11. At 338K the pH of pure water is 6.5. Calculate Kw at this temp. Hence calculate the pH of 0.01M NaOH at this temp Kw = [H+]2 for pure water Kw = (3.162 x 10-7)2 Kw = 1x 10-13 Kw = [H+][OH-] [H+] = Kw [OH-] = 1 x 10 -130.01 = 1 x 10 -11

  12. At 338K the pH of pure water is 6.5. Calculate Kw at this temp. Hence calculate the pH of 0.01M NaOH at this temp [H+] = 1 x 10 -11 pH= -log10 [H+] pH = 11

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