The sodium fusion extract $(N a X)$ when acidified with nitric acid $(H N O_{3})$ and then treated with silver nitrate $(A g N O_{3})$ gives a white curdy precipitate. Then choose the correct statement(s) :

X

is chlorine

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X

is bromine.

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Precipitate formed is soluble in

N H_{4} O H

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Precipitate formed is insoluble in

N H_{4} O H

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Solution

The correct option is C Precipitate formed is soluble in $N H_{4} O H$
Test for halogens :
The sodium fusion extract having $N a X$ is acidified with nitric acid $(H N O_{3})$ and then treated with silver nitrate $(A g N O_{3})$
Case 1 :
White curdy precipitate soluble in $N H_{4} O H$ indicates presence of $C l$
$N a C l + A g N O_{3} \to A g C l White ppt. ↓ + N a N O_{3}$
$A g C l$ dissolves in ammonium hydroxide $(N H_{4} O H)$

Case 2:
Pale yellow precipitate sparingly soluble in $N H_{4} O H$ indicates presence of $B r$
$N a B r + A g N O_{3} \to A g B r Pale yellow ppt. ↓ + N a N O_{3}$
$A g B r$ is sparingly soluble in ammonium hydroxide $(N H_{4} O H)$ .

Case 3 :
Yellow precipitate insoluble in $N H_{4} O H$ solution indicates presence of $I$
$N a I + A g N O_{3} \to A g I Yellow ppt. ↓ + N a N O_{3}$
$A g I$ is insoluble in ammonium hydroxide $(N H_{4} O H)$

The presence of nitrogen or sulphur interferes in the detection of halogens.These atoms should be removed by boiling the sodium fusion extract with concentrated nitric acid to decompose the cyanide or sulphide.

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Similar questions

(N a X)

(H N O_{3})

(A g N O_{3})

N H_{4} O H

(A g N O_{3})

Which two solutions will give a white precipitate when treated with dilute nitric acid followed by silver nitrate solution?

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