Question

Step 1)

Cu (s) + 4 HNO3 -> Cu^(2+) + 2 NO3^(-) + 2 NO2 + 2 H2O

Cu^(2+) + 6 H2O -> Cu(H2O)6 &^(2+)

Description: Weight .49 of Cu wire and record . Under the heat add roughly 5 ML HNO3(nitric acid). When CU is dissolved add 10 mL of DI H2O.

Step2)

Cu(H2O)6 &^(2+) + 2 NaOH -> Cu(OH)3 + 6 H2O

Description: Add 6 MNaOH drop wise while stirring. Stop when litmus paper turns blue.

Step3)

Cu(OH)3 ->(heat) CuO + H2O

Description: Add 80 90 mL DI H2O Ba gently white string for 4 min. Allow to cool and vacuum filter precipitate. Wash with 5 mL hot water 3 times. Use 250 mL vacuum flask. Discard filtrate.

Step4)

CuO + H2SO4 -> Cu(H2O)^2+

Description: Dissolve CuO with 10 mL 3 MH2SO4. If not dissolved wash with filtrate again. Wash filter paper with DI H2O and collect in filtrate.

Step5)

Cu(H2O)6 ^(2+) + Zn -> Cu + Zn ^(2+)+ + 6 H2O

Description: Add .5 g of Zn and stir occasionally. Reaction is complete when all Zn is dissolved and solution is colorless.

Above is a description of an Experiment involving copper.

Question 1:

What haoppens to Copper(the yeild) is going to be affected if Step 3 (Copper CU(OH)2) is basic and NOT acidic?

Question 2:

In part 5, In the solution that was being filtered (copper vs solution) still light blue -> how does it affect yield?

Question 3:

What happens if in Step 1, you are going to use sulfuric acied instead of Nitric Acid?

Questions 4:

Why couldnt you substitute concentraed HNO3 solution for 3 mol H2SO4 in part 4 and 5?

Answer 1

Hi, some of the equations you typed, contain errors. However, here are the answers:

Question 1:

copper hydroxide that we heat in step 3 is formed in step 2 like this:

this is the hydrated form of copper hydroxide and this is what is actually converted into CuO upon heating.

If the solution is basic, the hydroxide precipitate will dissolve forming the following complex and the above reaction will not take place.

This is tetrahydroxocuprate(II) complex ion.

Question 2:

The blue colour of the solution is due to the ion, if the solution is still blue, this means the reaction with Zn is not complete and all of the is not reduced to Cu. You have to either wait for the reaction to complete or add more zinc to the solution.

Question 3:

If you are using dilute acids, then no reaction will take place with dilute sulphuric acid as copper will not react with it. Nitrate ions are necessary to oxidize copper to copper (II) ions. The reaction that happens is already written in your part 1.

Question 4:

You cannot substitute nitric acid with sulphuric acid since, nitrate ions oxidize copper you will not get copper metal back in step 5. Since, nitrate ions will be present in the solution, once you get copper from this reaction:

it will react with nitrate ions in the solution like this:

Hope that helps. If you need more info, please do not hesitate to ask.