1 / 8

The Ion Product Constant for Water ( Kw )

The Ion Product Constant for Water ( Kw ). Pure water dissociates according to the following reaction: H 2 O (l)  H + ( aq ) + OH - ( aq ) There is an equal amount of H + and OH - ions in solution (neutral, pH = 7) at 25°C [H + ] = [OH - ] = 1x10 -7 mol/L

velma
Download Presentation

The Ion Product Constant for Water ( Kw )

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The Ion Product Constant for Water (Kw)

  2. Pure water dissociates according to the following reaction: H2O(l) H+(aq) + OH-(aq) There is an equal amount of H+ and OH- ions in solution (neutral, pH = 7) at 25°C [H+] = [OH-] = 1x10-7 mol/L equilibrium constant for the dissociation of water: Kw Kw = [H+][OH-] = (1x10-7)(1x10-7) = 1x10-14 * small k, reactants are favoured (does not go to completion)

  3. Since strong acids and bases dissociate completely in water, [H+] = [acid] @ 25°C acids: [H+] > [OH-] [H+] > 1x10-7 [OH-] < 1x10-7 bases: [OH-] > [H+] [H+] < 1x10-7 [OH-] > 1x10-7 We can use Kw to calculate [H+] and [OH-] in solutions

  4. Ex 1) Find the [H+] and [OH-] in: (a) 2.5 M nitric acid (b) 0.16 M Barium hydroxide (a) HNO3 H+ + NO3- C 2.5 M 2.5 M Kw = [H+][OH-] [OH-] = Kw / [H+] = (1x10-14)/(2.5) = 4x10-15 M (b) Ba(OH)2 Ba2+ + 2 OH- C 0.16 M 0.32 M Kw = [H+][OH-] [H+] = Kw / [OH-] = (1x10-14)/(0.32) = 3.1x10-14 M

  5. pH and pOH pH: The Power of the Hydronium Ion A measure of the amount of H+ ions in a solution Convenient way to represent acidity since [H3O+] is usually a very small number 2 factors determine pH ionization concentration because they both contribute to the number of H+ or OH- molecules in a solution. The practical scale goes from 0  14

  6. pH = -log [H3O+] pOH = -log [OH-] In neutral water, pH = -log [H3O+] = -(log(1 x 10-7) = 7 pOH = -log [OH-] = -(log(1 x 10-7) = 7 Note: pH + pOH = 14, always, regardless of solution! Another way to calculate [H3O+] & [OH-] in solution: [H3O+] = 10-pH [OH-] = 10-pOH

  7. Ex) A liquid shampoo has a [OH-] of 6.8x10-5 mol/L (a) Is the shampoo acid, basic or neutral? (b) What is [H3O+]? (c) What is the pH and pOH of the shampoo? (a) [OH-] = 6.8x10-5 > 1.0x10-7, basic (b) [H3O+] = Kw / [OH-] = (1.0x10-14)/(6.8x10-5) = 1.5x10-10 mol/L (c) pH = -log [H3O+] pOH = -log [OH-] = -log [1.5x10-10] = -log [6.8x10-5] = 9.83 = 4.17 check: 9.83 + 4.17 = 14 !

  8. HOMEWORK P382 #1-4 p390 #9-12 p392 #13-18

More Related