The qualitative distinction of ZnSO_4 and Al_ | vedclass

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Question

(A) Both $Zn^{2+}$ and $Al^{3+}$ ions form white precipitates with $NaOH$ and $NH_4OH$. However,$Zn(OH)_2$ is soluble in excess of both $NaOH$ and $NH

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$NH_4OH$

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(A) Both $Zn^{2+}$ and $Al^{3+}$ ions form white precipitates with $NaOH$ and $NH_4OH$. However,$Zn(OH)_2$ is soluble in excess of both $NaOH$ and $NH_4OH$ due to the formation of soluble complexes like $[Zn(OH)_4]^{2-}$ and $[Zn(NH_3)_4]^{2+}$. $Al(OH)_3$ is soluble in excess $NaOH$ to form $[Al(OH)_4]^-$,but it is insoluble in excess $NH_4OH$. Therefore,$NH_4OH$ can be used to distinguish between them because $Al(OH)_3$ will remain as a precipitate while $Zn(OH)_2$ will dissolve. $NaOH$ is less effective for distinction because both hydroxides dissolve in excess $NaOH$.

The qualitative distinction of $ZnSO_4$ and $Al_2(SO_4)_3$ can be done by using the reagent

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