The solution of ‘X’ having excess caustic potash is used to detect ammonia. Which of the following is ‘X’?
A. ${{K}_{2}}[Hg{{I}_{4}}]$
B. $[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}$
C. ${{K}_{3}}[Fe{{(NCS)}_{6}}]$
D. $[Co{{(N{{H}_{3}})}_{5}}S{{O}_{4}}]Br$
Answer
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Hint: Recall which reagent is used to detect the presence of ammonia during qualitative salt analysis and what kind of precipitate is formed.
Complete answer:
The reagent to detect the presence of ammonia in inorganic, qualitative analysis is the Nessler’s reagent. When Nessler’s reagent is added to ammonia, it reacts with the mercury (Hg) and iodine (I) to form a complex called Iodide of Million’s base which precipitates with a brown colour. This brown coloured precipitate signifies the presence of ammonia in the solution. Nessler’s reagent is a solution of ${{K}_{2}}[Hg{{I}_{4}}]$ and an excess of $KOH$ which is also known as caustic potash. The reaction that occurs is:
\[2{{K}_{2}}Hg{{I}_{4}}+N{{H}_{3}}+KOH\to HgO\cdot Hg(N{{H}_{2}})I+7KI+2{{H}_{2}}O\]
This reaction is done regularly to detect the presence of ammonia or ammonium ion in labs. The easier method of adding dilute sodium hydroxide and then checking for fumes is not used since ammonia can cause poisoning and will harm your health.
Hence, option A is the correct answer.
Additional information:
The IUPAC name for Nessler’s reagent is potassium tetraiodomercurate (II). Nessler’s reagent can be prepared using anhydrous mercuric iodide and anhydrous potassium iodide. It is mainly used to detect the presence of ammonium ions in various labs. It forms a brown precipitate when it reacts with ammonium ion.
Note: While handling Nessler’s reagent, it is natural to find some brown precipitate already present in the bottle if it hasn’t been used for a long time or is very old. The reagent will still be usable.
Complete answer:
The reagent to detect the presence of ammonia in inorganic, qualitative analysis is the Nessler’s reagent. When Nessler’s reagent is added to ammonia, it reacts with the mercury (Hg) and iodine (I) to form a complex called Iodide of Million’s base which precipitates with a brown colour. This brown coloured precipitate signifies the presence of ammonia in the solution. Nessler’s reagent is a solution of ${{K}_{2}}[Hg{{I}_{4}}]$ and an excess of $KOH$ which is also known as caustic potash. The reaction that occurs is:
\[2{{K}_{2}}Hg{{I}_{4}}+N{{H}_{3}}+KOH\to HgO\cdot Hg(N{{H}_{2}})I+7KI+2{{H}_{2}}O\]
This reaction is done regularly to detect the presence of ammonia or ammonium ion in labs. The easier method of adding dilute sodium hydroxide and then checking for fumes is not used since ammonia can cause poisoning and will harm your health.
Hence, option A is the correct answer.
Additional information:
The IUPAC name for Nessler’s reagent is potassium tetraiodomercurate (II). Nessler’s reagent can be prepared using anhydrous mercuric iodide and anhydrous potassium iodide. It is mainly used to detect the presence of ammonium ions in various labs. It forms a brown precipitate when it reacts with ammonium ion.
Note: While handling Nessler’s reagent, it is natural to find some brown precipitate already present in the bottle if it hasn’t been used for a long time or is very old. The reagent will still be usable.
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