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Equilibrium constants

Equilibrium constants. Equilibrium constants. P C c P D d. K =. P A a P B b. [C] c [D] d. K =. [A] a [B] b. Equilibrium constants. P C c P D d. K =. P A a P B b. [C] c [D] d. K =. [A] a [B] b. [solid] = 1. Equilibrium of solids and solutions.

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Equilibrium constants

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  1. Equilibrium constants

  2. Equilibrium constants PCcPDd K = PAaPBb [C]c[D]d K = [A]a[B]b

  3. Equilibrium constants PCcPDd K = PAaPBb [C]c[D]d K = [A]a[B]b [solid] = 1

  4. Equilibrium of solids and solutions

  5. Equilibrium of solids and solutions NaCl(s) + H2O(l)

  6. Equilibrium of solids and solutions NaCl(s) + H2O(l) Na+(aq) + Cl-(aq) To remove an ion from the crystal lattice, the solvating interactions must be stronger than the lattice interactions.

  7. Equilibrium of solids and solutions NaCl(s) + H2O(l) Na+(aq) + Cl-(aq) Solution must be saturated for this equilibrium to take place.

  8. Dynamic equilibrium

  9. Equilibrium for dissolution-precipitation reaction:

  10. Equilibrium for dissolution-precipitation reaction: I2(s) + CCl4(l) I2(CCl4)

  11. Equilibrium for dissolution-precipitation reaction: I2(s) + CCl4(l) I2(CCl4) K [I2]CCl = 4

  12. Equilibrium for dissolution-precipitation reaction: I2(s) + CCl4(l) I2(CCl4) [I2]CCl K = 4 Molarity of saturated solution = K

  13. [I2]CCl K = 4 K will vary when conditions are changed based on Le Chatelier’s Principle.

  14. [I2]CCl K = 4 If solution of a material is exothermic, increasing the temperature will decrease K. A(s) A(sol)

  15. [I2]CCl K = 4 If solution of a material is endothermic, increasing the temperature will increase K. A(s) A(sol)

  16. Kinetics (reaction rates) must be considered.

  17. The value of K is not an indicator of howlong it takes to attain equilibrium.

  18. A K = 5 does not guarantee a 5 M solution in a few minutes.

  19. Not all solutions are ‘ideal’

  20. Not all solutions are ‘ideal’ Ideal Solution: Widely separated species (ions or molecules) that do not interact.

  21. CsCl(s) Cs+(aq) + Cl-(aq)

  22. CsCl(s) Cs+(aq) + Cl-(aq) As the concentration of ions increases, Cs+to Cl- distances decrease.

  23. CsCl(s) Cs+(aq) + Cl-(aq) As the concentration of ions increases, Cs+ to Cl- distances decrease. Cs+…Cl- Cs+…Cl- Cl-…Cs+

  24. CsCl(s) Cs+(aq) + Cl-(aq) Cs+…Cl- Cs+…Cl- Cl-…Cs+ Ion pairing may occur before equilibrium.

  25. CsCl(s) Cs+(aq) + Cl-(aq) Cs+…Cl- Cs+…Cl- Cl-…Cs+ Ion pairing may occur before equilibrium. These solutions are non-ideal.

  26. CsCl(s) Cs+(aq) + Cl-(aq) Salts of low solubilities allow the study of solutions that are essentially ideal.

  27. CsCl(s) Cs+(aq) + Cl-(aq) Salts of low solubilities allow the study of solutions that are essentially ideal. A saturated solution of 0.1 or less is a sign of low solubility in a salt.

  28. Solubility product

  29. Solubility product AgCl(s) Ag+(aq) + Cl-(aq)

  30. Solubility product Same form as equilibrium constant AgCl(s) Ag+(aq) + Cl-(aq)

  31. Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] [AgCl]

  32. Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp 1

  33. Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp

  34. Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp 25oC Ksp = 1.6 x 10-10

  35. Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp 25oC Ksp = 1.6 x 10-10 [Ag+] = [Cl-] = y

  36. Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp 25oC Ksp = 1.6 x 10-10 [Ag+] = [Cl-] = y y2 = Ksp = 1.6 x 10-10

  37. Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp y2 = Ksp = 1.6 x 10-10 [Ag+] = [Cl-] = 1.26 x 10-5 M

  38. Solubility product AgCl(s) Ag+(aq) + Cl-(aq) [Ag+][Cl-] = Ksp y2 = Ksp = 1.6 x 10-10 1.8 x 10-3 g/L [Ag+] = [Cl-] = 1.26 x 10-5 M

  39. BaF2(s) Ba2+(aq) + 2 F-(aq)

  40. BaF2(s) Ba2+(aq) + 2 F-(aq) [Ba2+][F-]2 = Ksp

  41. Fe3+(aq) + 3 OH-(aq) Fe(OH)3

  42. Fe3+(aq) + 3 OH-(aq) Fe(OH)3 [Fe3+][OH-]3 = 1.1 x 10-36

  43. Fe3+(aq) + 3 OH-(aq) Fe(OH)3 [Fe3+][OH-]3 = 1.1 x 10-36 For every Fe3+ that goes into solution, 3 OH- go into solution.

  44. Fe3+(aq) + 3 OH-(aq) Fe(OH)3 [Fe3+][OH-]3 = 1.1 x 10-36 [y][3y]3 = 1.1 x 10-36

  45. Fe3+(aq) + 3 OH-(aq) Fe(OH)3 [Fe3+][OH-]3 = 1.1 x 10-36 [y][3y]3 = 1.1 x 10-36 If there is another source of OH- (NaOH) that provides a higher [OH-] then that is the value of [OH-] to be used.

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