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CHM 2730 – SP24 Name________________________ Homework Assignment #6 (out of 64 pts) Due: 8:00am, Wednesday, February 28, 2024 I certify that the work below is my own, I have not copied the work from someone else, nor have I searched for these answers on the internet. I have completed all of the work on this assignment. Signature________________________________________________________ For full credit, show all work (each step) in detail, including cancellation of units. You also need to express your answer to the proper number of significant figures and show units. Copper (II) hydroxide is considered essentially insoluble in water due to its very low πΎπΎπ π π π value. πΆπΆπΆπΆ(ππππ)2 (π π ) β πΆπΆπ’π’2+ (ππππ) + 2 πππ»π» − (ππππ) πΎπΎπ π π π = 4.8 × 10−20 Recall that compounds considered “insoluble” in water will dissolve to some (albeit sometimes an undetectable) extent. However, if higher concentrations of the counterion (in this case πππ»π» − ) cause formation of soluble complex ions with πΆπΆπ’π’2+ , more πΆπΆπ’π’2+ may dissolve than would be calculated in pure water for the above equilibrium. Consider the following equilibria and cumulative formation constants (from Harris 9th ed. Appendix F) for the reaction of πΆπΆπ’π’2+ (ππππ) and πππ»π» − (ππππ): πΆπΆπ’π’2+ (ππππ) + πππ»π» − (ππππ) β πΆπΆπΆπΆπΆπΆπ»π» + (ππππ) πΆπΆπ’π’2+ (ππππ) + 2 πππ»π» − (ππππ) β πΆπΆπΆπΆ(ππππ)2 (ππππ) πΆπΆπ’π’2+ (ππππ) + 3 πππ»π» − (ππππ) β πΆπΆπΆπΆ(ππππ)− 3 (ππππ) πΆπΆπ’π’2+ (ππππ) + 4 πππ»π» − (ππππ) β πΆπΆπΆπΆ(ππππ)2− 4 (ππππ) πΎπΎπ π π π = 3.16 × 106 π½π½2 = 6.31 × 1011 π½π½3 = 3.16 × 1014 π½π½4 = 3.98 × 1015 Thus, the total πΆπΆπΆπΆ present in the presence of πππ»π» − can be written as: ππππππππππ ππππππππππππππππππ πΆπΆπΆπΆ 2− = [πΆπΆπ’π’2+ ] + [πΆπΆπΆπΆ(ππππ)+ ] + [πΆπΆπΆπΆ(ππππ)2 ] + [πΆπΆπΆπΆ(ππππ)− 3 ] + [πΆπΆπΆπΆ(ππππ)4 ] 1 1) (20 pts) Using the example covered in lecture as a model (ππππ 2+ complexes with iodide πΌπΌ − ), which is also discussed in the text, [πΆπΆπ’π’2+ ], [πΆπΆπΆπΆ(ππππ)+ ], 2− [πΆπΆπΆπΆ(ππππ)2 ](ππππ), [πΆπΆπΆπΆ(ππππ)− 3 ], [πΆπΆπΆπΆ(ππππ)4 ], and Total dissolved πΆπΆπΆπΆ for a fixed concentration of πππ»π» − = 0.0010 M. Show your work. 2 2) (20 pts) Repeat the calculations from part a) but this time for a fixed concentration of πππ»π» − = 1.00 M. Show your work. 3 3) (11 pts) Complete the following table using the results from parts a) and b), again, using the example in lecture as a guide. [πΆπΆπ―π―− ] 0.0010 [πͺπͺππππ+ ] [πͺπͺπͺπͺ(πΆπΆπΆπΆ)+ ] ππ− [πͺπͺπͺπͺ(πΆπΆπΆπΆ)ππ ] [πͺπͺπͺπͺ(πΆπΆπΆπΆ)− ππ ] οΏ½πͺπͺπͺπͺ(πΆπΆπΆπΆ)ππ οΏ½ Total dissolved Cu 1.00 4) (3 pts) What is(are) the major form(s) of dissolved Cu when in the presence of 0.0010 M πππ»π» − (ππππ) ? 5) (3 pts) What is(are) the major form(s) of dissolved Cu when in the presence of 1.00 M πππ»π» − (ππππ) ? 6) (4 pts) Is there a significant difference in the major forms of dissolved Cu for 0.0010 M vs 1.00 M πππ»π» − (ππππ)? If there is, explain why there would be a difference. 7) (3 pts) Compare the Total dissolved Cu in the presence of the two different concentrations of πππ»π» − (ππππ). Does what you observe make sense? Why or why not? 4
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