SATHEE: Qualitative Analysis - Result Question 70

Qualitative Analysis - Result Question 70

####73. A mixture of two salts was treated as follows :

(i) The mixture was heated with maganese dioxide and concentrated sulphuric acid, when yellowish green gas was liberated.

(ii) The mixture on heating with sodium hydroxide solution gave a gas which turned red litmus blue.

(iii) Its solution in water gave blue precipitate with potassium ferricyanide and red colouration with ammonium thiocyanate.

(iv) The mixture was boiled with potassium hydroxide and the liberated gas was bubbled through an alkaline solution of $K_{2} H g I_{4}$ to give brown precipitate.

Identify the two salts. Give ionic equations for reactions involved in the tests (i), (ii) and (iii).

(1987, 5M)

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Solution:

(i) Mixture $+ M n O_{2} + H_{2} S O_{4} ⟶$ Yellowish green gas.

The above reaction suggest that the mixture contain $C l^{-}$ .

$\begin{aligned} 2 C l^{-} + M n O_{2} + H_{2} S O_{4} ⟶ M n S O_{4} & + C l_{2} ↑ + H_{2} O \end{aligned}$

(ii) Mixture $+ N a O H ⟶$ Gas turning red litmus blue.

The above reaction indicates that the gas is ammonia.

$N H_{4}^{+} + N a O H ⟶ N H_{3} + H_{2} O + N a^{+}$

(iii) Solution of mixture $+ K_{3} [F e (C N)_{6}] ⟶$ blue ppt.

The above reaction suggest that mixture contain a Fe(II) salt. $F e$ (II) salt react with $K_{3} [F e (C N)_{6}]$ to give blue ppt of Prussian blue complex

$3 F e^{2 +} + 2 F e (C N)_{6}^{3 -} ⟶ \underset{Blue}{F e_{3} {[F e (C N)_{6}]}_{2} ↓}$

Red colouration with $N H_{4} S C N$ suggests that some $F e (I I I)$ is also present. It is likely that a part of $F e (I I)$ is oxidised to Fe(III) by air :

$F e^{3 +} + 3 N H_{4} S C N ⟶ \underset{Red colouration}{F e (S C N)_{3}} + 3 N H_{4}^{+}$

(iv) Mixture $+ K O H ⟶$ gas $↑ \overset{K_{2} H g I_{4}}{\to}$ brown ppt.

The above reaction indicates that the gas is $N H_{3}$ , i.e. mixture contain $N H_{4}^{+}$ ion. Hence, the mixture contains $F e^{2 +}, N H_{4}^{+}$ and $C l^{-}$ ions with some impurity of $F e^{3 +}$ ion.