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\mathrm{Fe}(\mathrm{OH})_{3} can be separated from \mathrm{Al}(\mathrm{OH})_{3} by addition of

Option: 1

dil.\ \mathrm{HCl}


Option: 2

\mathrm{NaCl} \; \text{Solution}


Option: 3

\mathrm{NaOH}\; \text{Solution}


Option: 4

\mathrm{NH}_{4} \mathrm{Cl}\; \text{ and }\mathrm{NH}_{4} \mathrm{OH}


Answers (1)

best_answer

As we have learnt, 

\text{Al}(\text{OH})_{3}  is amphoteric and it is because of this reason it is soluble in \text{NaOH}. It dissolves in the excess of the reagent to form Sodium Aluminate. 

\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{NaOH} \longrightarrow \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_{4}\right]

\text{Fe}(\text{OH})_{3} is basic and it does not dissolve in \mathrm{NaOH} and hence \mathrm{NaOH} can be used to separate \text{Al}(\text{OH})_{3} from \text{Fe}(\text{OH})_{3}.

Hence, the correct answer is Option (3)

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mansi

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