Question

On saturating aqueous solution of Cu(II), Pb(II) and Zs(II) ions with H₂​S will precipitate
(A) Only $CuS$
(B) Only $PbS$
(C) Both $CuS$and $PbS$
(D) $CuS$,$PbS$and $ZnS$

Answer 1

Hint: Sulphides are precipitated in two conditions – basic and acid medium as group $2$and group $4$cations respectively. The cations falling in the later group require more concentration of sulphide ion hence are categorised in a separate group.

Complete Step by Step Solution:
In acidic medium: on passing \[{H_2}S\;\]gas in an acidic medium like\[\;HCl\], common ion effect takes place as follows:
${H_2}S\underset {} \leftrightarrows {H^ + } + {S^{2 - }}$
$HCl\xrightarrow{{}}{H^ + } + C{l^ - }$

${H^ + }$ is the common ion that increases due to addition of acid. This suppresses the ionisation of \[{H_2}S\;\] which is a weak acid and becomes even more weaker. Group 2 ions precipitated as sulphides are $H{g^{ + 2}},P{b^{ + 2}},B{i^{ + 3}},C{u^{ + 2}},C{d^{ + 2}}$known as copper group and $S{b^{ + 3}},S{b^{ + 5}},A{s^{ + 3}},A{s^{ + 5}},S{n^{ + 2}},S{n^{ + 4}}$known as arsenic group.

In basic medium: when \[{H_2}S\;\]os passed in alkaline medium, ${H^ + }$ion combines with $O{H^ - }$from dissociation of $N{H_4}OH$to form unionised water. The reactions are as follow:
${H_2}S\underset {} \leftrightarrows {H^ + } + {S^{2 - }}$
$N{H_4}OH\underset {} \leftrightarrows N{H_4}^ + + O{H^ - }$
${H^ + } + O{H^ - } \to {H_2}0$

The removal of hydronium ions from the solution causes more dissociation of \[{H_2}S\;\]thus, increasing the concentration of sulphide ions.
Cations precipitated in group $4$are $C{o^{ + 2}},N{i^{ + 2}},Z{n^{ + 2}},M{n^{ + 2}}$
Hence all the given cations will be precipitated according to the desired condition and thus, the correct option is D.

Note: The precipitate formed can be confirmed by performing various tests. $CuS$can be confirmed by adding ferrocyanide in acidic condition to give a chocolate precipitate of copper sulphate. Adding two drops of potassium dichromate in $N{H_4}Ac$gives yellow precipitate of lead sulphide. The white precipitate of zinc sulphide can be confirmed by passing hydrogen sulphide and acetic acid in the first part division of solution used for confirming manganese, resulting in white precipitate.