Question

Alex was given a solution of an unknown salt Y for analysis. He performed the following tests and recorded his observations:
• To a part of the solution Y, he added silver nitrate solution and obtained a white precipitate which was soluble in ammonium hydroxide solution.
• To the remaining solution he added a few drops of sodium hydroxide solution and obtained a pale blue precipitate.
Based on the observations made by Alex, identify:
(a) the anion and
(b) the cation present in salt Y.

Hint:

Salt Y is Copper(II) Chloride (CuCl_2). Remember that while AgCl is soluble in NH_4OH, it is insoluble in dilute nitric acid!

Answer 1

From the observations, we can identify both the anion and the cation present in salt Y by using standard qualitative analysis.

Step 1: Identify the anion
When silver nitrate solution was added to solution Y, a white precipitate was formed.
This white precipitate was soluble in ammonium hydroxide solution.

This is the characteristic test for the chloride ion.
The reaction is:

AgNO₃ + Cl^- → AgCl + NO₃^-

The white precipitate is silver chloride (AgCl).
Silver chloride dissolves in ammonium hydroxide, which confirms the presence of chloride ion, Cl^-.

Step 2: Identify the cation
When sodium hydroxide solution was added to the remaining solution, a pale blue precipitate was obtained.

A pale blue precipitate with sodium hydroxide indicates the formation of copper(II) hydroxide, Cu(OH)₂.
This confirms the presence of copper(II) ion, Cu²⁺.

The reaction is:

Cu²⁺ + 2OH^- → Cu(OH)₂

Therefore:
The anion present is chloride ion (Cl^-).
The cation present is copper(II) ion (Cu²⁺).

Hence, salt Y is most likely copper chloride solution.

Final Answer:
(a) Anion = Chloride ion (Cl^-)
(b) Cation = Copper(II) ion (Cu²⁺)