Qualitative Analysis - Result Question 53
####56. An aqueous blue coloured solution of a transition metal sulphate reacts with $H _2 S$ in acidic medium to give a black precipitate $A$, which is insoluble in warm aqueous solution of $KOH$. The blue solution on treatment with KI in weakly acidic medium, turns yellow and produces a white precipitate $B$. Identify the transition metal ion. Write the chemical reactions involved in the formation of $A$ and $B$.
(2000, 4M)
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Solution:
- The transition metal is $Cu^{2+}$. The compound is $CuSO _4 \cdot 5 H _2 O$. It dissolves in water to give blue coloured solution due to presence of $Cu^{2+}$. On passing $H _2 S(g)$ in acid medium of salt solution black precipitate of $CuS$ is obtained which is not soluble in aqueous $KOH$ solution.
$$ CuSO _4+H _2 S \xrightarrow{H^{+}} \underset{\text { black }}{CuS \downarrow}+H _2 SO _4 $$
On adding KI solution to aqueous solution of $CuSO _4$, yellow solution of $CuI _2$ is formed in the beginning which decompose into white ppt of CuI.
$$ \begin{aligned} & CuSO _4+2 KI \longrightarrow CuI _2+K _2 SO _4 \\ & 2 CuI _2 \longrightarrow \underset{\text { White }}{2 CuI \downarrow}+I _2 \end{aligned} $$
