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Question:
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Concentrated sodium hydroxide can separate a mixture of:
A. $\mathrm{Al}^{3+}$ and $\mathrm{Cr}^{3+}$
B. $\mathrm{Zn}^{2+}$ and $\mathrm{Pb}^{2+}$
C. $\mathrm{Cr}^{3+}$ and $\mathrm{Fe}^{3+}$
D. $\mathrm{Al}^{3+}$ and $\mathrm{Zn}^{3+}$
Options:
A. $\mathrm{Al}^{3+}$ and $\mathrm{Cr}^{3+}$
B. $\mathrm{Zn}^{2+}$ and $\mathrm{Pb}^{2+}$
C. $\mathrm{Cr}^{3+}$ and $\mathrm{Fe}^{3+}$
D. $\mathrm{Al}^{3+}$ and $\mathrm{Zn}^{3+}$
Solution:
2707 Upvotes
Verified Answer
The correct answer is:
$\mathrm{Cr}^{3+}$ and $\mathrm{Fe}^{3+}$
When the $\mathrm{Cr}^{3+}$ reacts with the conc. $\mathrm{NaOH}$, then following reaction, takes place,
$\mathrm{Cr}^{3+}+3 \mathrm{OH}^{-} \rightarrow \mathrm{Cr}(\mathrm{OH})_3$
Which on further reactions gives $\mathrm{Cr}(\mathrm{OH})_4^{-}$(green coloured solution)
$\mathrm{Cr}(\mathrm{OH})_3+\mathrm{OH}^{-} \rightarrow \mathrm{Cr}(\mathrm{OH})_4^{-}$(green coloured solution)
Whereas $\mathrm{Fe}^{3+}$ reaction with the conc. $\mathrm{NaOH}$ gives the $\mathrm{Fe}(\mathrm{OH})_3$ precipitate
$\mathrm{Fe}^{3+}+3 \mathrm{OH}^{-} \rightarrow \mathrm{Fe}(\mathrm{OH})_3$ (precipitate)
So, as the $\mathrm{Cr}(\mathrm{OH})^{4-}$ is in solution and $\mathrm{Fe}(\mathrm{OH})_3$ (precipitate)
formed is not in solution so, it can be separated out from the solution.
$\mathrm{Cr}^{3+}+3 \mathrm{OH}^{-} \rightarrow \mathrm{Cr}(\mathrm{OH})_3$
Which on further reactions gives $\mathrm{Cr}(\mathrm{OH})_4^{-}$(green coloured solution)
$\mathrm{Cr}(\mathrm{OH})_3+\mathrm{OH}^{-} \rightarrow \mathrm{Cr}(\mathrm{OH})_4^{-}$(green coloured solution)
Whereas $\mathrm{Fe}^{3+}$ reaction with the conc. $\mathrm{NaOH}$ gives the $\mathrm{Fe}(\mathrm{OH})_3$ precipitate
$\mathrm{Fe}^{3+}+3 \mathrm{OH}^{-} \rightarrow \mathrm{Fe}(\mathrm{OH})_3$ (precipitate)
So, as the $\mathrm{Cr}(\mathrm{OH})^{4-}$ is in solution and $\mathrm{Fe}(\mathrm{OH})_3$ (precipitate)
formed is not in solution so, it can be separated out from the solution.
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