Qualitative Analysis - Result Question 51
####54. A white substance $A$ reacts with dilute $H _2 SO _4$ to produce a colourless gas $B$ and a colourless solution $C$. The reaction between $B$ and acidified $K _2 Cr _2 O _7$ solution produces a green solution and a slightly coloured precipitate $D$. The substance $D$ burns in air to produce a gas $E$ which reacts with $B$ to yield $D$ and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous $NH _3$ or $NaOH$ to $C$ produces first a precipitate, which dissolves in the excess of the respective reagent to produce a clear solution in each case. Identify $A, B, C, D$ and $E$. Write the equations of the reactions involved. (2001, 10M)
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Solution:
- Since, the white substance $A$ gives a colourless gas $B$ with dil. $H _2 SO _4$, such gas may be $H _2 S$. So, the substance $A$ may be a metal sulphide $(Na / K / Zn$, etc.)
When $H _2 S$ gas reacts with acidified $K _2 Cr _2 O _7$, it gives green coloured solution of $Cr _2\left(SO _4\right) _3$ alongwith slightly yellow ppt of $D$ as sulphur.
$$ \begin{array}{r} K _2 Cr _2 O _7+4 H _2 SO _4+3 H _2 S \longrightarrow K _2 SO _4+\underset{\text { Green }}{Cr _2\left(SO _4\right) _3} \\ +3 S+7 H _2 \\ D \end{array} $$
$S$ on burning in air gives $SO _2(E)$. Substance $E$ on reaction with $B\left(H _2 S\right)$ produces $D(s)$ :
$$ \underset{B}{2 H _2 S}+\underset{E}{SO _2} \longrightarrow 2 H _2 O(l)+\underset{D}{3 S \downarrow} $$
Anhydrous $CuSO _4$ produces blue colour in water.
Solution $C$ produces ppt first with $NH _3 / NaOH$ which dissolve in excess $NH _3 / NaOH$. Hence, $A$ must be $ZnS$.
$$ \begin{gathered} \underset{A}{ZnS+\text { dil. } H _2 SO _4} \longrightarrow \underset{C}{ZnSO _4(a q)+H _2 S(g)} \\ \underset{C}{ZnSO _4+2 NaOH} \longrightarrow \underset{\text { White }}{Zn(OH) _2 \downarrow}+Na _2 SO _4 \\ Zn(OH) _2+2 NaOH \longrightarrow\left[Zn(OH) _4\right]^{2-}+2 Na^{+} \end{gathered} $$
