Question

Which one of the following sulphides is only completely precipitated when the acidic solution is made dilute
(A) \[HgS\;\]
(B) \[\;PbS\;\]
(C) \[CdS\;\]
(D) \[CuS\]

Answer 1

Hint: The given sulphides are precipitated when ${H_2}S$ is passed in presence of $HCl$ or any acidic medium. But when the medium is diluted the one which requires least concentration of sulphide ion will be precipitated only.

Complete Step by Step Solution:
The ionic product when exceeds the solubility product then precipitation takes place. Ionic product is defined as the product or multiplication of the concentration of the ions of the compound raised to their respective stoichiometric terms.

Solubility product is similar to ionic product but defined for sparingly soluble salts.
On dilution of the acid, the acid becomes weaker and supplies less ${H^ + }$ for the common ion effect. Thus, the suppression of ${H_2}S$ does not take place to a large extent. Hence, the concentration of sulphide increases. Therefore, this increases the value of Ksp of the sulphide compound.

In other words, on dilution that sulphide compound will be precipitated which has greater value of K_sp.
Values of k_sp are given below:
\[HgS\;\]$ = {10^{ - 54}}$
\[\;PbS\;\]$ = {10^{ - 29}}$
\[CdS\;\]$ = {10^{ - 28}}$
 \[CuS\]$ = {10^{ - 44}}$
\[CdS\;\] has the largest value of solubility product so, cadmium sulphide will be precipitated completely.
Thus, the correct answer is C.

Note: $P{b^{ + 2}}$ is placed in both group \[{\mathbf{1}}\] and group \[{\mathbf{2}}\]due to intermediate value of solubility product of its sulphide and chloride as compared to chlorides and sulphides of cations of respective groups. The value of solubility product for lead chloride is ${10^{ - 5}}$ and for silver chloride is ${10^{ - 10}}$. The value of solubility product for lead sulphide is ${10^{ - 29}}$ and for copper sulphide is ${10^{ - 44}}$.