A white crystalline substance dissolves in wa | vedclass

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Question

(C) The correct answer is $C$.$1$. When $H_2S$ is passed through the solution,the formation of a black precipitate indicates the presence of $Pb^{2+}$

Vedclass · Accepted Answer

$PbSO_4$

Vedclass · Answer

(C) The correct answer is $C$.$1$. When $H_2S$ is passed through the solution,the formation of a black precipitate indicates the presence of $Pb^{2+}$ ions,forming $PbS$ (lead sulfide).$Pb^{2+} + H_2S \xrightarrow{	ext{acidic}} PbS \downarrow 	ext{ (black ppt.)} + 2H^+$$2$. The black precipitate $(PbS)$ dissolves in hot $HNO_3$ to form lead nitrate:$3PbS + 8HNO_3 	o 3Pb(NO_3)_2 + 2NO + 3S + 4H_2O$$3$. Upon adding conc. $H_2SO_4$ to the solution containing $Pb^{2+}$,a white precipitate of lead sulfate $(PbSO_4)$ is formed:$Pb(NO_3)_2 + H_2SO_4 	o PbSO_4 \downarrow 	ext{ (white ppt.)} + 2HNO_3$

$A$ white crystalline substance dissolves in water. On passing $H_2S$ in this solution,a black precipitate is obtained. The black precipitate dissolves completely in hot $HNO_3$. On adding a few drops of conc. $H_2SO_4$ a white precipitate is obtained. This precipitate is that of:

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