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Previously in Chem104: the concept of K w the concept of the K w circle

Previously 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. Today in Chem104: How K a relates to K b and pK a to pK b More ways to use the K w circle

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Previously in Chem104: the concept of K w the concept of the K w circle

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  1. Previously in Chem104: • the concept of Kw • the concept of the Kw circle • p-functions (pH, pKa, pKw) • pH scale • P-functions in calcs • Today in Chem104: • 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) What I had to use to find logarithms

  2. Lauren said: “Cl- is a very weak base because HCl is a strong acid” How did she know that? AH + H2O A-+ H3O+ acid base conjugate base conjugate acid Now write the Ka expression for AH and the Kb expression for A- .

  3. Alright, now we’re can understand what Lauren said: We proved Kw = Ka x Kb Use the Kw circle! Ka Kb Kw = 10-14

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

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

  6. 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 = Ka x Kb = 10-14 p of Kw = p of [Ka x Kb ] = p of 10-14 -log Kw = -log (Ka x Kb)= -log 10-14 -log Kw = -log Ka+ ( -log Kb) = -log 10-14 pKw = pKa + pKb = 14

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

  8. Let’s do some problems !!

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