Dreamerish*~
Love Addict - Nakashima
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- Jan 16, 2005
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- HSC
- 2005
This's from last year's trials paper of my school.
In the laboratory a test solution was analysed. It formed a pale yellow precipitate with silver nitrate solution. When sodium hydroxide was added to the solution it formed a blue precipitate which dissolved when excess sodium hydroxide was added. Deduce the ions present in the test solution and write suitable equations.
What I thought was:
PO43-(aq) + 3AgNO3(aq) → Ag3PO4(s) + 3NO3- ... which forms a yellow precipitate.
Cu2+(aq) + 2NaOH(aq) → Cu(OH)2(s) + 2Na2+(aq) ... which forms a blue precipitate.
But Cu3(PO4)2 is a precipitate itself, so how can it exist as a solution?
Also, why does the blue precipitate disappear after adding excess NaOH?
In the laboratory a test solution was analysed. It formed a pale yellow precipitate with silver nitrate solution. When sodium hydroxide was added to the solution it formed a blue precipitate which dissolved when excess sodium hydroxide was added. Deduce the ions present in the test solution and write suitable equations.
What I thought was:
PO43-(aq) + 3AgNO3(aq) → Ag3PO4(s) + 3NO3- ... which forms a yellow precipitate.
Cu2+(aq) + 2NaOH(aq) → Cu(OH)2(s) + 2Na2+(aq) ... which forms a blue precipitate.
But Cu3(PO4)2 is a precipitate itself, so how can it exist as a solution?
Also, why does the blue precipitate disappear after adding excess NaOH?