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answer questions 2 & 3
2. Since dilute NaOH (ltd. OH) and dilute aqueous NHg produce similar results, why do the results differ between excess NaOH
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Answer #1

2. In case of NaOH, the change in dilute solution to concentrate solution only changes the concentration of OH- ion present in that solution.

in case of Ammonia, in dilute solution, it dissociates in NH4+ and OH- but in concentrate solution, the excess ammonia present in the solution can act as a ligand to form metal ammonia complex which has deferent property

3. Unknown 1 = Co2+

  Unknown 1 = Fe3+

  Unknown 1 = Cu2+

All sulfide of this three ion are black in color

Cobalt hydroxide is blue Iron hydroxide is brown   copper hydroxide is blue

cobalt form Dark blue ppt with ammonia then with ex ammonia it changes to colorless

[Co(H2O)6]2++ 2NH3 = [Co(H2O)4(OH)2] + 2NH4+   

(dark blue ppt)

[Co(H2O)6]2++6NH3   = [Co(NH3)6]2+ + 6H2O

(colorless solution)

Iron form orange-brown  ppt with ammonia then with ex ammonia more ppt form

Fe3+(aq) + 3NH3(aq) + 3H2O(l) = Fe(OH)3(s) + 3NH4+(aq)

(brown ppt)

coper from blue ppt with ammonia then with ex ammonia it changes to a royal blue solution

2Cu2++ 2SO42- + 2NH3  + 2H2O = Cu(OH)2 . CuSO4   + 2NH3

( blue ppt )

Cu(OH) 2.CuSO4 + 8NH3 = 2 [Cu(NH3)4 ]SO4 + 2OH–   + 2SO42-

(blue sol)

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