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(C) To form a white precipitate with both dilute $NaOH$ and dilute $HCl$,the cations must satisfy the following conditions:$1$. With dilute $HCl$,the

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$A, B$

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(C) To form a white precipitate with both dilute $NaOH$ and dilute $HCl$,the cations must satisfy the following conditions:$1$. With dilute $HCl$,the cation must form a white precipitate (e.g.,$Pb^{2+}$ or $Ag^+$).$2$. With dilute $NaOH$,the cation must form a white precipitate (e.g.,$Pb^{2+}$,$Zn^{2+}$,or $Bi^{3+}$).Analyzing the options:- $Pb(NO_3)_2$: Forms white $PbCl_2$ with $HCl$ and white $Pb(OH)_2$ with $NaOH$.- $Zn(NO_3)_2$: Forms soluble $ZnCl_2$ with $HCl$ and white $Zn(OH)_2$ with $NaOH$.- $Bi(NO_3)_3$: Forms soluble $BiCl_3$ with $HCl$ and white $Bi(OH)_3$ with $NaOH$.- $AgNO_3$: Forms white $AgCl$ with $HCl$ but forms brown $Ag_2O$ with $NaOH$.- $Hg(NO_3)_2$: Forms soluble $HgCl_2$ with $HCl$ and yellow $HgO$ with $NaOH$.For the mixture to give white precipitates only with both reagents:- Mixture $A$ ($Pb(NO_3)_2$ and $Zn(NO_3)_2$): $Pb^{2+}$ gives white $PbCl_2$ and $Pb(OH)_2$; $Zn^{2+}$ gives white $Zn(OH)_2$. This works.- Mixture $B$ ($Pb(NO_3)_2$ and $Bi(NO_3)_3$): $Pb^{2+}$ gives white $PbCl_2$ and $Pb(OH)_2$; $Bi^{3+}$ gives white $Bi(OH)_3$. This works.- Mixture $C$ ($AgNO_3$ and $Bi(NO_3)_3$): $Ag^+$ gives brown $Ag_2O$ with $NaOH$. Incorrect.- Mixture $D$ ($Pb(NO_3)_2$ and $Hg(NO_3)_2$): $Hg^{2+}$ gives yellow $HgO$ with $NaOH$. Incorrect.Thus,mixtures $A$ and $B$ are correct.

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