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:

Similar Questions

$A$ $0.3 \, M$ $HCl$ solution contains the following ions: $Hg^{2+}, Cd^{2+}, Sr^{2+}, Fe^{2+}, Cu^{2+}$. The addition of $H_2S$ to the above solution will precipitate which of the following?

Which sulphide is soluble in $(NH_4)_2CO_3$?

The reagent used in the qualitative analysis of the $IV^{th}$ group basic radicals is:

In the fifth group of qualitative analysis,$(NH_4)_2CO_3$ is added to precipitate out the carbonates. We do not add $Na_2CO_3$ because

The pair$(s)$ of ions where $BOTH$ the ions are precipitated upon passing $H_2S$ gas in the presence of dilute $HCl$,is(are):
$(A) Ba^{2+}, Zn^{2+}$
$(B) Bi^{3+}, Fe^{3+}$
$(C) Cu^{2+}, Pb^{2+}$
$(D) Hg^{2+}, Bi^{3+}$

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