Al^{3+},Fe^{3+},Zn^{2+},and Ni^{2+} ions are | vedclass

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(B) The addition of $NH_4Cl$ followed by $NH_4OH$ creates a buffer solution that suppresses the concentration of $OH^-$ ions due to the common ion eff

Vedclass · Accepted Answer

$Al(OH)_3$ and $Fe(OH)_3$

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(B) The addition of $NH_4Cl$ followed by $NH_4OH$ creates a buffer solution that suppresses the concentration of $OH^-$ ions due to the common ion effect of $NH_4^+$.In this condition,the concentration of $OH^-$ is sufficient to exceed the solubility product $(K_{sp})$ of $III^{rd}$ group cations ($Al^{3+}$,$Fe^{3+}$,$Cr^{3+}$) but insufficient to precipitate $IV^{th}$ group cations ($Zn^{2+}$,$Ni^{2+}$,$Mn^{2+}$,$Co^{2+}$).Therefore,only $Al(OH)_3$ and $Fe(OH)_3$ will precipitate as hydroxides.The correct option is $B$.

List-$I$ ($Test$/$reagent$)	List-$II$ ($Radical$ $identified$)
$A$. Lake Test	$I$. $NO_3^{-}$
$B$. Nessler's Reagent	$II$. $Fe^{3+}$
$C$. Potassium thiocyanate	$III$. $Al^{3+}$
$D$. Brown Ring Test	$IV$. $NH_4^{+}$

$Al^{3+}$,$Fe^{3+}$,$Zn^{2+}$,and $Ni^{2+}$ ions are present in an acidic solution. Excess of ammonium chloride $(NH_4Cl)$ solution is added followed by the addition of ammonium hydroxide $(NH_4OH)$ solution. The resulting precipitate will contain:

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