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  • White coloured precipitate (B) is formed when an aqueous solution of salt (A) is treated with s

White coloured precipitate (B) is formed when an aqueous solution of salt (A) is treated with \mathrm{NaCl} solution. On passing \mathrm{H}_2 \mathrm{S}(\mathrm{g}) through it, the filtrate gives a black coloured precipitate (C). The compound B so formed is dissolved in hot water and on treatment of the obtained solution with sodium iodide yellow precipitate (D) is formed. It was obsessed that compound A does not liberate any gas when treated with dil.\mathrm{HCl} but when heated, it gives out a gas which is reddish brown in colour. Identify the compounds\mathrm{ A, B, C \& D} from the given options.

Option: 1
A B C D
\mathrm{PbSO}_4 \mathrm{PbBr}_2 \mathrm{PbS} \mathrm{PbCl_2}




 


Option: 2
A B C D
\mathrm{PbSO}_4 \mathrm{PbCl}_2 \mathrm{PbS} \mathrm{PbI}_2


 


Option: 3
A B C D
\mathrm{Pb}_2\left(\mathrm{NO}_3\right)_2 \mathrm{PbCl}_2 \mathrm{PbS} \mathrm{PbI}_2

 


Option: 4
A B C D
\mathrm{Pb}\left(\mathrm{NO}_3\right)_2 \mathrm{PbCl}_2 \mathrm{PbS} \mathrm{PbBr}

 


Answers (1)

best_answer

\mathrm{' B '\: is \: \mathrm{PbCl}_2} as it dissolves in hot water & forms a yellow ppt with \mathrm{\mathrm{NaI}.}

\mathrm{\mathrm{PbCl}_2(\mathrm{~B})+2 \mathrm{NaI} \rightarrow \mathrm{PbI}_2(D)+2 \mathrm{NaCl} }
\mathrm{'A'} does not give any gas with dill \mathrm{HCl} but gives a reddish brown gas on heating

\mathrm{\therefore \: A \: is \: \mathrm{Pb}\left(\mathrm{NO}_3\right)_2}

\mathrm{ 2 \mathrm{~Pb}_2\left(\mathrm{NO}_3\right)_2(\mathrm{~A}) \stackrel{\triangle}{\longrightarrow} 2 \mathrm{PbO}+2 \mathrm{NO}_2(\mathrm{~g})+\mathrm{O}_2 }

\mathrm{ \mathrm{PbCl}_2 } is sparingly soluble in water when \mathrm{ \mathrm{H}_2 \mathrm{S} } is passed it gives a black ppt. of \mathrm{ P b S }

\mathrm{\mathrm{PbCl}_2 \left.+\mathrm{H}_2 \mathrm{S} \rightarrow \mathrm{PbS(C}\right)+2 \mathrm{HCl} }

\mathrm{\therefore A \left.\rightarrow \mathrm{Pb}_2 (\mathrm{NO}_3\right)_2 }

    \mathrm{B \rightarrow \mathrm{PbCl}_2 }

    \mathrm{C \rightarrow \mathrm{PbS} }

   \mathrm{D \rightarrow \mathrm{PbI}_2}

Hence option 3 is correct.

Posted by

Gunjita

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