1 / 15

Previously in Chem104: plant pigments do acid/base chemistry it’s just equilibrium

Previously in Chem104: plant pigments do acid/base chemistry it’s just equilibrium new names: K a , K b for same K expressions. Today in Chem104: the concept of K w the concept of the K w circle p-functions (pH, pK a , pK w ) pH scale P-functions in calcs.

xandra-everett
Télécharger la présentation

Previously in Chem104: plant pigments do acid/base chemistry it’s just equilibrium

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. Previously in Chem104: • plant pigments do acid/base chemistry • it’s just equilibrium • new names: Ka, Kb for same K expressions • Today in Chem104: • the concept of Kw • the concept of the Kw circle • p-functions (pH, pKa, pKw) • pH scale • P-functions in calcs

  2. Dual Personality of Water: both an acid and a base Amphoteric (or amphiprotic) The Autoionization Reaction of water. 2 H20 H3O+ + OH- Keq =

  3. Because Kw = [H3O+][OH-] = 10-14 and Kw is a constant: then the solution is neutral when [H3O+] = [OH-] = 10-7 M [H3O+] [OH-] Kw

  4. But, if [H3O+] > [OH-], [H3O+] > 10-7 M, the solution is acidic [H3O+] > 10-7 M [OH-] < 10-7 M Kw

  5. Or if [OH-] > [H3O+], [OH-] > 10-7 M, the solution is basic [H3O+] < 10-7 M [OH-] > 10-7 M Kw Because Kw = [H3O+] x [OH-] must be 10-14

  6. Because all these exponential numbers are a pain: i.e. Kw = [H3O+][OH-] = 10-14 [H3O+] = [OH-] = 10-7 M (in neutral water) Ka = 10-10 we will use: The P-Function p of (a number) = -log10(a number) so “pH” means p of [H3O+], or -log10[H3O+] so the pH of neutral water, where [H3O+] = 10-7 M is pH 7

  7. The P-Function and K’s p of (Keq) = -log10Keq = pKeq p of (Ka) = -log10Ka = pKa p of (Kw) = -log10Kw = pKw Examples: Keq = 10-10 the pKeq = 10 Ka = 1.20 x 10-5 the pKeq = 4.92 Kw = 10-14 the pKw = 14

  8. The P-Function simplifies exponential numbers Ka = 1.20 x 10-5 , pKeq = 4.92 [H3O+] = 4.20 x 10-5M, pH = 4.376 [OH-] = 6.66 x 10-6M, pOH = 5.176 (note number of sig figs) Examples converting from p-function: if pH = 7.47, [H3O+] = 10-7.47 M= 3.39 x 10-8M, if pKa = 4.92, Ka = 10-4.92 = 1.20 x 10-5

  9. P-Function simplifies a large range of numbers: graphically 10 1 10-110-310-5 10-7 10-9 10-11 10-13 10-14 [H3O+], M converts to a simpler scale -1 0 1 3 5 7 9 11 13 pH Note that on a p-scale, the smaller the p-number, the larger the actual number

  10. Working in P-Functions can simplify problems Recall Kw = [H3O+][OH-] = 10-14 Apply the P-function to each side p of Kw = p of [H3O+][OH-] = p of 10-14 -log Kw = -log ( [H3O+][OH-] )= -log 10-14 -log Kw = -log [H3O+] + ( -log [OH-] ) = -log 10-14 pKw = pH + pOH = 14

  11. Now apply this equation: pKw = pH + pOH = 14 to this picture [H3O+] [OH-] Kw = 10-14

  12. Now apply this equation: pKw = pH + pOH = 14 to this picture pH pOH pKw = 14

  13. When the solution is acidic [H3O+] > 10-7 M, pH < 7 : pH is small pH < 7 pOH > 7 pKw Because pKw = pH + pOH must be 14

  14. Fill in the blanks! pOH is _______ pKw pH is _______ When the solution is ____________ [H3O+] __10-7 M, pH ___ 7 : pH is ________

  15. Let’s do some problems !!

More Related