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Chapter 6 Problems

Chapter 6 Problems. 6.6, 6.9, 6.15, 6.16, 6.19, 6.21, 6.24 Comments on “Lab Report & Pop Rocks”. 6.6. From the equations HOCl H + + OCl - K = 3.0 x 10 -8 HOCl + OBr -  HOBr + OCl - K= 15 Find K for HOBr  H + + OBr —. Flip. HOBr + OCl -  HOCl + OBr -. K’ = 1/K= 1/15. K= ?.

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Chapter 6 Problems

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  1. Chapter 6 Problems • 6.6, 6.9, 6.15, 6.16, • 6.19, 6.21, 6.24 Comments on • “Lab Report & Pop Rocks”

  2. 6.6 • From the equations HOCl H+ + OCl-K = 3.0 x 10-8 HOCl + OBr-  HOBr + OCl-K= 15 Find K for HOBr  H+ + OBr— Flip HOBr + OCl-  HOCl + OBr- K’ = 1/K= 1/15 K= ? K= 3.0 x 10-8/15 K= 0.2 x 10-8

  3. 6.9 • The formation of Tetrafluorethlene from its elements is highly exothermic. • 2 F2 (g) + 2 C (s)  F2C=CF2 (g) • (a) if a mixture of F2, graphite, and C2F4 is at equilibrium in a closed container, will the reaction go to the right or to the left if F2 is added? • (b) Rare bacteria … eat C2F4 and make teflon for cell walls. Will the reaction go to the right or to the left if these bacteria are added?

  4. 6.9 • The formation of Tetrafluorethlene from its elements is highly exothermic. • 2 F2 (g) + 2 C (s)  F2C=CF2 (g) • (C) will the reaction go to the right or to the left if graphite is added? • (d) will reaction go left or right if container is crushed to one-eighths of original volume? • (e) Does “Q” get larger or smaller if vessel is Heated?

  5. 6-15. • What concentration of Fe(CN)64- is in equilibrium with 1.0 uM Ag+ and Ag4Fe(CN)6 (s). Ag4Fe(CN)6  4Ag+ + Fe(CN)64- Ksp = [Ag+]4 [Fe(CN)64-] 8.5 x 10-45 = [1.0 x 10-6]4 [Fe(CN)64-] [Fe(CN)64-] = 8.5 x 10-21 M = 8.5 zM

  6. 6-16. Cu4(OH)6(SO4)  4 Cu2+ + 6OH- + SO42- I’d first set up an ICE table: Cu4(OH)6(SO4)  4 Cu2+ + 6 OH + SO42- I Some - 1.0 x 10-6 M - C -x + 4x Fixed at 1.0 x 10-6 M + x E Some –x + 4x Fixed at 1.0 x 10-6 M + x Ksp = [Cu2+]4 [OH-]6 [SO42-] = 2.3 x 10-69 Ksp = [4x]4 [1.0 x 10-6]6 [x] = 2.3 x 10-69 X = 9.75 x 10-8 M Cu2+ = 4x = 3.90 x 10-7 M

  7. Chapter 6 Chemical Equilibrium

  8. Chemical Equilibrium • Equilibrium Constant • Solubility product (Ksp) • Common Ion Effect • Separation by precipitation • Complex formation

  9. Separation by Precipitation

  10. Separation by Precipitation Completeseparation can mean a lot … we should define complete. Complete means that the concentration of the less soluble material has decreased to 1 X 10-6M or lower before the more soluble material begins to precipitate

  11. Separation by Precipitation EXAMPLE:Can Fe+3 and Mg+2 be separated quantitatively as hydroxides from a solution that is 0.10 M in each cation? If the separation is possible, what range of OH- concentrations is permissible.

  12. Add OH- Mg2+ Mg2+ Fe3+ Fe3+ Fe3+ Fe3+ Mg2+ Mg2+ Mg2+ Mg2+ Fe3+ Fe3+ Mg2+ Fe3+ Mg2+ Fe3+ Fe3+ Fe3+ Mg2+ Mg2+ Mg2+ Fe3+ Fe3+

  13. Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ @ equilibrium Mg2+ Fe3+ What is the [OH-] when this happens ^ Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ Fe(OH)3(s)

  14. EXAMPLE: Separate Iron and Magnesium? Ksp = [Fe+3][OH-]3 = 2 X 10-39 Ksp = [Mg+2][OH-]2 = 7.1 X 10-12 Assume [Fe+3]eq = 1.0 X 10-6M when “completely” precipitated. What will be the [OH-] @ equilibrium required to reduce the [Fe+3] to [Fe+3] = 1.0 X 10-6M ? Ksp = [Fe+3][OH-]3 = 2 X 10-39

  15. EXAMPLE: Separate Iron and Magnesium? Ksp = [Fe+3][OH-]3 = 2 X 10-39 (1.0 X 10-6M)*[OH-]3 = 2 X 10-39

  16. Dealing with Mg2+ • Find [OH-] to start precipitating Mg2+ • Conceptually – • Will assume a minimal amount of Mg2+ will precipitate and determine the respective concentration of OH- • Evaluate Q • If • Q>K • Q<K • Q=K “Left” “Right” “Equilibrium”

  17. Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ @ equilibrium Mg2+ Fe3+ = 1.3 x 10-11 [OH-] ^ Mg2+ Mg2+ Is this [OH-] (that is in solution) great enough to start precipitating Mg2+? Mg2+ Mg2+ Mg2+ Fe(OH)3(s)

  18. Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ @ equilibrium Mg2+ Fe3+ = 1.3 x 10-11 [OH-] ^ Mg2+ Mg2+ Is this [OH-] (that is in solution) great enough to start precipitating Mg2+? Mg2+ Mg2+ Mg2+ Fe(OH)3(s)

  19. EXAMPLE: Separate Iron and Magnesium? What [OH-] is required to begin the precipitation of Mg(OH)2? [Mg+2] = 0.10 M Really, Really close to 0.1 M [Mg2+]eq = 0.09999999999999999 M Ksp = (0.10 M)[OH-]2 = 7.1 X 10-12 [OH-] = 8.4 X 10-6M

  20. EXAMPLE: Separate Iron and Magnesium? @ equilibrium [OH-] to ‘completely’ remove Fe3+ = 1.3 X 10-11 M ^ [OH-] to start removing Mg2+ = 8.4 X 10-6M “All” of the Iron will be precipitated b/f any of the magnesium starts to precipitate!!

  21. EXAMPLE: Separate Iron and Magnesium? Q vs. K Ksp = [Mg2+][OH-]2 = 7.1 X 10-12 Q = [0.10 M][1.3 x 10-11]2 = 1.69 x 10-23 Mg(OH)2(s) Mg2+ + 2OH- Q<K Reaction will proceed to “Right”

  22. Dealing with Mg2+ • Find [OH-] to start precipitating Mg2+ • Conceptually – • Will assume a minimal amount of Mg2+ will precipitate and determine the respective concentration of OH- • Evaluate Q • If • Q>K • Q<K • Q=K “Left” “Right” “Equilibrium” NO PPT

  23. “Real Example” • Consider a 1 liter solution that contains 0.3 M Ca2+ and 0.5 M Ba2+. • Can you separate the ions by adding • Sodium Carbonate? • Sodium Chromate ? • Sodium Fluoride? • Sodium Hydroxide? • Sodium Iodate? • Sodium Oxylate?

  24. An example • Consider Lead Iodide PbI2 (s) Pb2+ + 2I- Ksp= 7.9 x 10-9 What should happen if I- is added to a solution? Should the solubility go up or down?

  25. Complex Ion Formation

  26. Complex Formation complex ions (also called coordination ions) Lewis Acids and Bases acid => electron pair acceptor (metal) base => electron pair donor (ligand)

  27. I- I- Pb2+ Pb2+

  28. I- I- Pb2+ Pb2+

  29. I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+

  30. I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+

  31. I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+

  32. I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- I- Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+ Pb2+

  33. Effects of Complex Ion Formation on Solubility Consider the addition of I- to a solution of Pb+2 ions

  34. Overall constants are designated with b This one is b2 Effects of Complex Ion Formation on Solubility Consider the addition of I- to a solution of Pb+2 ions

  35. Protic Acids and Bases Section 6-7

  36. Question • Can you think of a salt that when dissolved in water is not an acid nor a base? • Can you think of a salt that when dissolved in water IS an acid or base?

  37. Protic Acids and Bases - Salts • Consider Ammonium chloride • Can ‘generally be thought of as the product of an acid-base reaction. NH4+Cl- (s) NH4+ + Cl- From general chemistry – single positive and single negative charges are STRONG ELECTROLYTES – they dissolve completely into ions in dilute aqueous solution

  38. Protic Acids and Bases Conjugate Acids and Bases in the B-L concept CH3COOH + H2O  CH3COO- + H3O+ acid + base <=> conjugate base + conjugate acid conjugate base => what remains after a B-L acid donates its proton conjugate acid => what is formed when a B-L base accepts a proton

  39. Question: Calculate the Concentration of H+ and OH- in Pure water at 250C.

  40. EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C. H2O H+ + OH- Kw = [H+][OH-] = 1.01 X 10-14 KW=(X)(X) = 1.01 X 10-14 (X) = 1.00 X 10-7

  41. Example • Concentration of OH- if [H+] is 1.0 x 10-3 M @ 25 oC? “From now on, assume the temperature to be 25oC unless otherwise stated.” Kw = [H+][OH-] 1 x 10-14 = [1 x 10-3][OH-] 1 x 10-11 = [OH-]

  42. pH ~ -3 -----> ~ +16 pH + pOH = - log Kw = pKw = 14.00

  43. Is there such a thing as Pure Water? • In most labs the answer is NO • Why? • A century ago, Kohlrausch and his students found it required to 42 consecutive distillations to reduce the conductivity to a limiting value. CO2 + H2O HCO3- + H+

  44. 6-9 Strengths of Acids and Bases

  45. Strong Bronsted-Lowry Acid • A strong Bronsted-Lowry Acid is one that donates all of its acidic protons to water molecules in aqueous solution. (Water is base – electron donor or the proton acceptor). • HCl as example

  46. Strong Bronsted-Lowry Base • Accepts protons from water molecules to form an amount of hydroxide ion, OH-, equivalent to the amount of base added. • Example: NH2- (the amide ion)

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