Asked by Ksenija
Hi. I have a little problem with my chemistry task. I have 4 unknown solutions: 1 light blue (A),1 green (B), 2 without colour (C & D).
As I understood that the blue one is CuSO4, green one is NiCl2, but 1 uncoloured is NH3, then...
A - CuSO4 [couldn't ir be Cu(OH)2?]
B - NiCl2
C - x
D - NH3 [couldn't it be NaOH?]
It's said that when combine A+B+C then nothing happens. A+B -> light blue sediments, after adding D the colour becomes intensive blue/purple. B+D -> purple solution. C+D -> brown sediments, but after adding D, sediments dissolve.
Did I understand it correctly, if so, than what is C?
As I understood that the blue one is CuSO4, green one is NiCl2, but 1 uncoloured is NH3, then...
A - CuSO4 [couldn't ir be Cu(OH)2?]
B - NiCl2
C - x
D - NH3 [couldn't it be NaOH?]
It's said that when combine A+B+C then nothing happens. A+B -> light blue sediments, after adding D the colour becomes intensive blue/purple. B+D -> purple solution. C+D -> brown sediments, but after adding D, sediments dissolve.
Did I understand it correctly, if so, than what is C?
Answers
Answered by
GK
Here are some tests that may help:
CuSO4 + NH3 ----> deep purplish blue color when an excess of NH3 is used. That would confirm NH3. If no clues were given for the other colorless solution, it would not be easy to identify it.
Cu(OH)2, unlike CuSO4, has low solubility in water. NaOH would form gel like precipitates with CuSO4 and NiCl2.
CuSO4 + NH3 ----> deep purplish blue color when an excess of NH3 is used. That would confirm NH3. If no clues were given for the other colorless solution, it would not be easy to identify it.
Cu(OH)2, unlike CuSO4, has low solubility in water. NaOH would form gel like precipitates with CuSO4 and NiCl2.
Answered by
Ksenija
thanks :) it brought in a bit of clearness in my head :)
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