Question

$Fe(OH{)}_3$ can be separated from $Al(OH{)}_3$ by addition of:

• A. $H_{2}S{O}_4$ solution
• B. dil. $HCl$ solution
• C. $NaOH$ solution
• D. $N{H}_{4}Cl$ and $N{H}_{4}OH$

Answer 1

The correct option is C $NaOH$ solution

$F{e}^{3+}$ and $A{l}^{3+}$ both belong to group III of inorganic analysis and both are precipitated as their hydroxides on adding group reagent, i.e., $N{H}_{4}OH$ in presence of $N{H}_{4}Cl$.

$F{e}^{3+}+3O{H}^{-}\to Fe(OH{)}_3$
$A{l}^{3+}+3O{H}^{-}\to Al(OH{)}_3$

only the precipitate of $Al(OH{)}_3$ is soluble in $NaOH$ solution. Hence, $Fe(OH{)}_3$ and $Al(OH{)}_3$ are separated by using $NaOH$ solution.