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Three separate samples of a solution of a single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dil. HCl solution and one formed a black precipitate with $\mathrm{H}_{2} \mathrm{~S}$. The salt could be
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Verified Answer
The correct answer is:
$\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$
$\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NH}_{4} \mathrm{OH}$
$$
\begin{array}{r}
\rightarrow \mathrm{Pb}(\mathrm{OH})_{2} \downarrow+2 \mathrm{NH}_{4} \mathrm{NO}_{3}(\text { white ppt }) \\
\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{HCl} \rightarrow \mathrm{PbCl}_{2} \downarrow+2 \mathrm{HNO}_{3}(\text { whiteppt }) \\
\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{H}_{2} \mathrm{~S} \rightarrow \mathrm{PbS} \downarrow+2 \mathrm{HNO}_{3} \text { (black) }
\end{array}
$$
$$
\begin{array}{r}
\rightarrow \mathrm{Pb}(\mathrm{OH})_{2} \downarrow+2 \mathrm{NH}_{4} \mathrm{NO}_{3}(\text { white ppt }) \\
\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{HCl} \rightarrow \mathrm{PbCl}_{2} \downarrow+2 \mathrm{HNO}_{3}(\text { whiteppt }) \\
\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{H}_{2} \mathrm{~S} \rightarrow \mathrm{PbS} \downarrow+2 \mathrm{HNO}_{3} \text { (black) }
\end{array}
$$
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