white precipitate of silver chloride is formed on adding a solution of sodium chloride to Silver Nitrate solution balance equation
pravallika7:
AgNo3+Nacl- NaNo3+Agcl
Answers
Answered by
54
Hello SIS!!
Rajdeep here....
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.
Hope my answer is satisfactory...
Thanks!!
Rajdeep here....
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.
Hope my answer is satisfactory...
Thanks!!
Answered by
29
NaCl+AgNo3------>AgCl +NaNO3
sry by mistake this is right ans
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