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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 the concept of K w the concept of the K w circle p -functions (pH, pK a , pK w ). Today in Chem104: pH scale

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

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  1. Previously in Chem104: • plant pigments do acid/base chemistry • it’s just equilibrium • new names: Ka, Kb for same K expressions • the concept of Kw • the concept of the Kwcircle • p-functions (pH, pKa, pKw) • Today in Chem104: • pH scale • How Ka relates to Kb and pKa to pKb • More ways to use the Kw circle • Group worksheet on The Most Important Equilibrium on the Planet (Part 1)

  2. 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

  3. 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

  4. Recall how we used this picture and this relationship: Kw= [H+] x[OH-] = 10-14 [H3O+] [OH-] Kw= 10-14

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

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

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

  8. Let’s do some problems !!

  9. Example problems to be used with reaction: [Fe-OH2]2++ H2O [Fe-OH]++ H3O+ Keq = 10-10

  10. When is the conjugate base (or acid) important in acid / base equilibria? Here? HCl + H2O Cl-+ H3O+ acid base conjugate base conjugate acid

  11. Write the Ka expression for AH and the Kb expression for A- . AH + H2O A-+ H3O+ acid base conjugate base conjugate acid

  12. Alright, now we can understand why Cl- isn’t basic: We proved Kw = Kax Kb Use the Kw circle! Ka Kb Kw = 10-14

  13. If AH has a larger Ka, like 10-4 then A- must have a smaller Kb like 10-10 Kb Ka Kw Because Kw = KaxKbmust = 10-14 The stronger the acid (Ka large), the weaker the conjugate base, (Kbsmall)

  14. If A- has a larger Kb, like 10-3 then AH must have a smaller Ka like 10-11 Kb Ka Kw Because Kw = KaxKbmust = 10-14 The stronger the base (Kblarge), the weaker the conjugate acid, (Kasmall)

  15. Let’s apply P-Functions We already did this one: Kw = [H3O+][OH-] = 10-14 pKw = pH + pOH = 14 Now do the same with Kw = Kax Kb = 10-14 p of Kw = p of [Kax Kb ] = p of 10-14 -log Kw = -log (Kax Kb)= -log 10-14 -log Kw = -log Ka+ ( -log Kb) = -log 10-14 pKw = pKa + pKb = 14

  16. Now apply this equation: pKw = pKa+ pKb = 14 to this picture pKa pKb pKw

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