Alkaline solution of ${N H}_{4} C l$ gives brown ppt. with $K_{2} {H g I}_{4}$

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Solution

Nessler's reagent can be prepared by using $K I, H g l_{2}, a n d N a O H$
or $K I (e x c e s s), H g C l_{2} a n d N a O H$

Mercuric chloride reacts with potassium iodide to form an orange ppt of mercuric iodide.

$H g C l_{2} + 2 K I ⟶ H g I_{2} o r a n g e ↓$

Mercuric iodide reacts with KI to form Nessler's reagent which is soluble in water.

$H g I_{2} + 2 K I ⟶ K_{2} H g I_{4} s o l u b l e$

Nessler's reagent is used to identify $N H_{4}^{+}$ ions.

$2 K_{2} H g I_{4} + 4 N a O H + N H_{4} C l \to H g_{2} N H_{2} O I b r o w n p r e c i p i t a t e + 7 N a I + K C l + 3 H_{2} O$

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