During chemical transformations of copper to series of copper containing product

Question

During chemical transformations of copper to series of copper containing products and then back to copper experiment:

What would the result be if the solution was not basic when it was heated to form CuO? What would be the result if a slight blue solution was poured off in the last step (part V) of the reaction? What would be the result if, in the final step, all of the water boiled away and the Cu was exposed to the Bunsen burner flame for a long period of time?Why couldn't you substitute 3M H2SO4 for concentrated HNO3 in the transformation from Cu(s) to [Cu(H2O)6]2+? Why couldn't you substitute concentrated HNO3 solution for 3M H2SO4 in Parts IV (dissolving CuO with H2SO4) and V (reducing with zinc)?

Explanation / Answer

1)What would the result be if the solution was not basic when it was heated to form CuO?

Solution :- If the solution is not basic while converting the Cu^2+ ions to Cu then some of the Cu^2+ will not convert to the Cu(OH)2 therefore all of the Cu^2+ cannot be transformed in the CuO.

This will lead to the lower recovery of the Cu

2)What would be the result if a slight blue solution was poured off in the last step (part V) of the reaction?

Solution :- If some of the slight blue solution is poured off then some of the complex of the Cu(H2O)6^2+ will not get reduced to the Cu(s) therefore it will lead to the lower recovery of the Cu(s)

3)What would be the result if, in the final step, all of the water boiled away and the Cu was exposed to the Bunsen burner flame for a long period of time?

Solution :- If the Cu is overheated and exposed to the Bunsen burner then it will react with the Oxygen in the air to form the CuO. This CuO will lead to the higher mass of the Cu recovered. Therefore it will given higher percent of the recovery of the Cu(s) than actual.

4)Why couldn't you substitute 3M H2SO4 for concentrated HNO3 in the transformation from Cu(s) to [Cu(H2O)6]2+?

Solution :- H2SO4   used to dissolve the Cu(s) to given the Cu(H2)6^2+ complex while the HNO3 is used to oxidize the Copper from Cu(s) to Cu^2+

Therefore they cannot be exchanged.

5)Why couldn't you substitute concentrated HNO3 solution for 3M H2SO4 in Parts IV (dissolving CuO with H2SO4) and V (reducing with zinc)?

Solution :- Because the HNO3 does the oxidation of the copper but we need to convert the CuO to the complex therefore it is important to use the H2SO4

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