7 notecards = 2 pages (4 cards per page)
Part A.2 What is the formula and the color of the gas that is evolved?
Part B.1. When the NaOH solution is added, Cu(OH)2 does not precipitate immediately. What else present in the retain mixture from Part A reacts with the NaOH before the copper(II) ion? Explain.
H2O. Cu(No3)2 is aqueous and has Cu(2+) and NO3(-) ions. When NaOH is added, Na+ ions react with the No3- ions to form the NaNo3, and the Cu(2+) ions react with the OH- ions to form Cu(Oh)2 which is light blue precipitate. It is possible that these H+ and OH- ions that form water started to react with NaOH before Cu(No3)2 did. But the reaction, other than that, needed time.
Part C.1. The sample in Part B was not centrifuged. Why? Perhaps the student chemist had to be across campus for another appointment. Because of that student's "other priorities" the percent recovery of the copper in the experiment will decrease. Explain why.
Initially, a blue precipitate is formed. But because it was not centrifuged, the reaction did not fully complete in the tube, and reversed somewhat from sitting still, so some copper will remain within the the Cu (II) hydroxide and not react with the NaOH, lessening the percent recovery of copper from the experiment. Failure to form all of the Cu(Oh)2 will do this if not centrifuged.
Part D.1. All of the CuO does not react with the sulfuric acid. Will the reported percent recovery of copper in the experiment be too high or too low? Explain.
The reported recovery will be too low. If 1 mole is present of CuO, and only half of it reacts for example, you only get a part of the Cu that has reacted. 1 mole, .78 moles reacts, then percent collected will be 78%.
Part. E. Sulfuric acid has a dual role in the chemistry. What are its two roles in the recovery of the copper metal?
It speeds up the reaction in the copper cycle, and also acts as a dehydrating agent. It takes away the water from the Cu (II) hydroxide, allowing you to eventually recover Copper itself.
Part E.2. Jacob couldn't find the 6MH2SO4, so instead substituted the 6M HNO3 that was available. What change was most likely observed as a result of this decision. Explain?
HNO3 cannot be use in the redox titration because it oxidizes itself. Unlike HCl, it isn’t volatile, and doesn’t take part in the reaction.
Part E. Errors in experimental technique can lead to the percent recovery of copper being too high-one such error may occur in Part E.2 and another in Part E.3. Cite those two errors and explain what should be done to ensure that those errors do not occur in the recovery,
E1. Failure to centrifuge. - Don’t let it sit idle, balance centrifuge, and avoid contaminants.