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4.8-4.9: Ka, Kb and the Conjugate Pair

4.8-4.9: Ka, Kb and the Conjugate Pair. Chemistry 12. 4.8 – 4.9. Keq = Ksp = Kw = Ka = Kb =. equilibrium constant solubility product eqb exp for ionization of water acid ionization constant base ionization constant. Concept 1.

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4.8-4.9: Ka, Kb and the Conjugate Pair

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  1. 4.8-4.9: Ka, Kb and the Conjugate Pair Chemistry 12

  2. 4.8 – 4.9 • Keq = • Ksp = • Kw = • Ka = • Kb = equilibrium constant solubility product eqb exp for ionization of water acid ionization constant base ionization constant

  3. Concept 1 Using Kw to calculate unknown [H3O+] and [OH-] from a strong acid or a strong base.

  4. Eg.) At 250C an HClsolution has a concentration of 0.0100 M. What is the [OH-]? HCl + H20 ↔ H3O++ Cl- [H3O+] = 0.0100 M [H3O+][OH-] = Kw 0.0100 M [OH-] = 1.00 x 10-14 [OH-] = 1.00 x 10-12 M

  5. Concept 2: Equilibrium equations and Ka and Kb Expressions What is the Ka expression for oxalic acid? (Hint: use the table) H2O(l) + H2C2O4(aq)↔ H3O+(aq) + HC2O4-(aq) Ka = [H3O+][HC2O4-] [H2C2O4]

  6. What is the equilibrium equation for HC2O4- acting as a base and the corresponding Kb expression? HC2O4-(aq) + H2O(l) ↔H2C2O4(aq) +OH-(aq) Kb = [H2C2O4][OH-] [HC2O4-]

  7. Concept 3: Finding Ka and calculating Kb What is Ka value for the following? (simply use the table) H2C2O4 Ka = 5.9 x 10-2 HC2O4- Ka = 6.4 x 10 -5

  8. …but solving for Kb is not that easy.

  9. What is the Kb value of HPO42-? … get out your B-L Table and follow along.

  10. We must use the Ka value of H2PO4- to determine Kb of HPO42- HPO42- is acting as a base HPO42- is acting as an acid Ka value of HPO42- so we cannot use this to calculate Kb

  11. Calculation You should notice that this is the Ka value of the conjugate acid for HPO42- Kb of HPO42- = Kw/ Ka(H2PO4-) = 10-14/6.2 x 10-8 = 1.6 x10-7

  12. Now you try Actually Ka of H2C2O4 Kb for HC2O4-? Find HC2O4- onright side of table Ka = 5.9 x 10 -2 Kw = Ka x Kb 1.0 x 10-14 = 5.9 x 10 -2 x Kb Kb = 1.7 x 10-13

  13. Concept 4: Explaining the connection between Kw, Ka and Kb. This can be explained using HC2O4- but other weak acids could also be used Why do we use the formula Kw = Ka x Kb? H2O(l) + HC2O4-(aq)↔ H3O+(aq) + C2O42-(aq) Acid dissociation Base dissociation

  14. 4.8 – 4.9 To get Kb we must figure out the base dissociation equation of C2O42-.

  15. 4.8 – 4.9 H2O(l) + HC2O4-(aq)↔ H3O+(aq) + C2O42-(aq) Invert H3O+(aq) + C2O42-(aq) ↔ H2O(l) + HC2O4-(aq) + OH- (aq)↔ OH-(aq) 2 H2O(l) H2O(l) + C2O42-(aq) ↔ OH-(aq) + HC2O4-(aq)

  16. 4.8 – 4.9 H2O(l) + C2O42-(aq) ↔ OH-(aq) + HC2O4-(aq) H2O(l) + HC2O4-(aq)↔ H3O+(aq) + C2O42-(aq) Kb = [HC2O4-][OH-] [C2O42-] Ka = [H3O+][C2O42-] [HC2O4-]

  17. 4.8 – 4.9  if Kb = [HC2O4-][OH-] Ka = [H3O+][C2O42-] [C2O42-] [HC2O4-] Then Ka x Kb [HC2O4-][OH-] x [H3O+][C2O42-] [C2O42-] [HC2O4-]  Ka x Kb = [OH-][H3O+] = Kw

  18. Learning Check Can you calculate [H3O+] and [OH-] of a strong acid or base using Kw? Can you write equilibrium equations for weak acids and bases? Can you write out Ka and Kb expressions? Can you find Ka on the B-L table? Can you calculate Kb using the B-L table?

  19. Read & Highlight Self Notes p. 3 - 5 Hebden # 31-37

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