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

(a) $BaSO_4$
(b) $SrSO_4$
(c) $PbSO_4$
(d) $CdSO_4$

The white ppt. obtained with $H_2SO_4$ is that of $PbSO_4$. The white crystalline may be that of $Pb(NO_3)_2$.
$Pb(NO_3)_2+H_2S \Rightarrow 2HNO_3+PbS$
$Pb(NO_3)_2 + H_2SO_4 \longrightarrow PbSO_4 + 2HNO_3$
Answer: $PbSO_4$