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Hydrochloric Acid

HgCl 2 gives ...

Question

${H g C l}_{2}$ gives orange ppt with $K I$ , which dissolves in excess of it forming $K_{2} {H g I}_{4}$ called Nessler's reagent.

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Solution

${H g C l}_{2}$ gives orange ppt with. $K I$ which dissolves in excess of it forming $K_{2} {H g I}_{4}$ called.Nessler's reagent.
Mercuric chloride reacts with potassium iodide to form 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 idntify $N H_{4}^{+}$ 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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