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Ka and Kb Calculations

Ka and Kb Calculations. For Weak Acid Reactions:. HA + H 2 O  H 3 O + + A - K a = [H 3 O + ][A - ] K a < 1 [HA]. For Weak Base Reactions:. B + H 2 O  HB + + OH - K b = [HB + ][ OH - ] K b < 1

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Ka and Kb Calculations

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  1. Ka and Kb Calculations

  2. For Weak Acid Reactions: HA + H2O  H3O+ + A- Ka = [H3O+][A-] Ka < 1 [HA]

  3. For Weak Base Reactions: B + H2O  HB+ + OH- Kb= [HB+][OH-] Kb < 1 [B]

  4. Notice that Ka and Kb expressions look very similar. The difference is that a base produces the hydroxide ion in solution, while the acid produces the hydronium ion in solution.

  5. Another note on this point: H+ and H3O+ are both equivalent terms here. Often water is left completely out of the equation since it does not appear in the equilibrium. This has become an accepted practice. (*However, water is very important in causing the acid to dissociate.)

  6. H2O(l) + H2O(l) <=> H3O+(aq) + OH-(aq) Keq[H2O]2 = Kw = [H3O+][OH-] Kw = 1.0 x 10-14 Kw = Ka x Kb (Another useful equation) Knowing this value allows us to calculate the OH- and H+ concentration for various situations.

  7. [OH-] = [H+] : solution is neutral (in pure water, each of these is 1.0 x 10-7) [OH-] > [H+] : solution is basic [OH-] < [H+] : solution is acidic

  8. Kw = [H+] [OH-] = 1.0 X 10-14 M Calculate [H+] or [OH-] as required for each of the following solutions at 25°C, and state whether the solution is neutral, acidic, or basic. a. 1.0 X 10-5M OH- b. 1.0 X 10-7M OH- c. 10.0 M H+

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