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Warmup

Ms Jonson: prepare warm and cold water baths. Do not write the questions! ANSWER them!. Warmup. Co(H 2 O) 6 2+ + 4Cl - + heat  CoCl 4 2+ + 6H 2 O. Based on your knowledge of Le Chatelier’s Principle, answer the questions below using the equation above:

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Warmup

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  1. Ms Jonson: prepare warm and cold water baths Do not write the questions! ANSWER them! Warmup Co(H2O)62+ + 4Cl- + heat  CoCl42+ + 6H2O Based on your knowledge of Le Chatelier’s Principle, answer the questions below using the equation above: 1. What will be the effect of removingCoCl42+from the system? 2. What if more chloride ions are added to the reaction? 3. What will be the effect of removingheat (cooling) the reaction above?

  2. Co(H2O)62+ + 4Cl- + heat CoCl42+ + 6H2Ohydrated cobalt ioncobalt chloride ion Demonstrating Le Chatelier’s Principle Flask P 1. 10 ml CoCl2 2. Add about 10 ml HCl in 1 ml increments. Observe! 3. Add10 ml distilled water slowly. Observe! 4. Place in the hot water bath. Prediction? (start on Flask B) 5. Immerse in the ice water bath. Observe! Flask B 1. 10 ml CoCl2 2. Add 10ml HCl. Observe! 3. Add 5ml AgNO3. Observe!

  3. 1. Given the reaction: Co(H2O)62+(aq) + 4Cl-(aq) + heat  CoCl42+(aq) + 6H2O(l) a. Write equilibrium expression for this reaction. Keq = [CoCl42+]1 [Cl-]4[Co(H2O)62+]1 b. Given equilibrium concentrations, the Keq is calculated to be 1.95 x 10-4. What does this MEAN? The equilibrium lies to the left.

  4. Co(H2O)62+(aq) + 4Cl-(aq) + heat  CoCl42+(aq) + 6H2O(l) Keq = 1.95 x 10-4 c. Some Cl- are precipitated by Ag+. Once equilibrium is reestablished, the concentrations are: [CoCl42+] = 8.9 x 104 M, [Cl-]= 6.7 x 102 M. What is the new [Co(H2O)62+]? Keq = 1.95 x 10-4 = [CoCl42+]1 [Cl-]4[Co(H2O)62+]1 1.95 x 10-4 = (8.9 x 104)1. . (6.7 x 102)4 [Co(H2O)62+]1 (3.929468 x 107) [Co(H2O)62+] = 8.9 x 104 [Co(H2O)62+] = 2.3 x 10-3 M

  5. (we’ll do this on the board) 2. Ca(OH)2(s)  Ca2+(aq) + 2OH-(aq) a. Write the Ksp expression b. If [Ca2+] = 0.0021M and [OH-] = 0.025M, what is the Ksp for this reaction? c. According to the Ksp, does this reaction favor the product or reactant side? d. If Ca2+ ions were added, how would this affect the equilibrium position?

  6. 3. HNO2(aq)  H+(aq) + NO2-(aq) a)Write the Ka expression for the dissociation of HNO2 Ka = [H+][NO2-] [HNO2] b) The Ka for this acid is 7.2 x 10-4 . If the [HNO2] at equilibrium is 0.35 M, what should be the [H+] and [NO2-] at equilibrium? 0.016 M for EACH ion! c) At equilibrium, the [H+] for a 0.35M HNO3 solution is 2.90M. What is the Ka of nitric acid? Compare it to the Ka of HNO2. What does this mean? Ka = 2.4 x 101. Nitric acid is a stronger acid.

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