Question

Briefly explain what you would observe if you used the test described in (a) on a solution that needed more NaOH solution

Answer 1

If we add more NaOH to the solution in test(a), the Cu2+ turns into Cu(OH)(s) blue precipitate only, the extra OH- ions will remain in the solution idle.

But they are useful in giving information that all the Cu2+ ions are precipitated as Cu(OH)2, by turning red litmus into blue.

Explanation:

Transform [Cu(H2O)6]2+ (aq) to Cu(OH)2(s)

    Be careful in handling NaOH, for it is a strong base which will sting if it contacts the skin. Add the NaOH solution dropwise to the copper solution.
    After a blue precipitate is formed, periodically test the acidity of the solution by dipping your stirring rod into the solution and touching it to red litmus paper. Try not to transfer the blue precipitate onto the litmus paper: that will result in some loss of copper, and a possibly false blue on the litmus paper. The solution starts out acidic because of excess nitric acid from the previous step, so the first OH- added goes into neutralizing the acid; once the acid is neutralized, the next OH- added goes to forming the blue Cu(OH)2 precipitate.
Only after that is finished does added OH- hang around idle, and only at that time will it turn red litmus paper blue.
We want to make sure all the copper present is turned to Cu(OH)2, so we add OH- until the solution turns the litmus paper blue.