Question

21. White precipitate of silver chloride is formed on adding a solution
of sodium chloride to silver nitrate solution.
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Answer 1

Answer:

Explanation:

This is the way to test the presence of chloride ion in a solution.

Just add silver nitrate solution to the solution (here, sodium chloride), and you will obtain a white precipitate of Silver Chloride, which is insoluble in water and nitric acid, but soluble when ammonia solution (ammonium hydroxide) is added to it.

So, reaction when silver nitrate solution is added to silver chloride:

AgNO₃ + NaCl ----> NaNO₃ + AgCl

The above reaction is already balanced.

Now, when ammonium hydroxide is added to silver chloride (the white precipitate), it dissolves, forming a colourless solution of Diammene Silver Chloride. Here's the reaction:

AgCl + 2NH₄OH ------> Ag[(NH₃)₂]Cl + 2H₂O

The above complex salt formed is diammene silver chloride, and is colourless.