When the NaOH solution is added, Cu(OH)_2 does not precipitate immediately. What
ID: 520121 • Letter: W
Question
When the NaOH solution is added, Cu(OH)_2 does not precipitate immediately. What else present in the reaction mixture from Part A reacts with the NaOH before the copper(II) ion? Explain. The sample in Part B was not centrifuged. Why? Perhaps the student chemist had to be across campus for another appointment. Because of the student's "other priorities" the percent recovery of copper in the experiment will decrease. Explain why. 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. Sulfuric acid has a dual role in the chemistry of this experiment. What are its two roles in the recovery of the copper metal? Part E.2. Jacob couldn't find the 6M H_2SO_4s, so instead substituted the 6 M HNO_3 that was available. What change was most likely observed as a result of this decision? Explain. 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.Explanation / Answer
2. When NaOH is added, it initially reacts with the acid present already in solution and once that has been neutralized, it reacts with Cu2+ ions to form the hydroxide precipitate.
3. If the solution is not centrifuged, the percentage recovery of copper will decrease as soem of the formed precipitate would get redissolved in solution which would thus reduce the total amount of precipitate and hence the copper precipitated out.
4. If all of CuO does not react with H2SO4, the the recorded percent recovery of copper would be lower than theoretical and some of CuO would be in unreacted and not accounted for in the calculation.
5. H2SO4 acts as both the reagent to form sulfate and dehydrating agent to remove formed water in the reaction.
6. Using HNO3 (a monoprotic acid) instead of H2SO4 (a diprotic acid), would change the reaction results, here we would get Cu(NO3)2 formed and the water formed would not be removed by HNO3 unlike H2SO4. We would need twice the moles of HNO3 to react as compared to 1 : 1 mole ratio for CuO and H2SO4.
7. If two large excess of H2SO4 is used, the calculated percent copper would be too high. To avoid that we must do this reaction very slowly. The concentration of acid used must be accurately measured and to be standarized before use to avoid error due to this reaction.
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