Question

Which of the following radicals will not be precipitated by passing\[{H_2}S\] in concentrated acid solution?
A. Copper
B. Antimony
C. Arsenic
D. Cadmium

Answer 1

Hint:Qualitative inorganic analysis of cations is based on the solubility products of their precipitates. The concentration of the acid present in the solution affects the concentrations of various ions which, in turn, affects the solubility product.
Complete step-by-step answer:
The qualitative inorganic analysis is a method in analytical chemistry that aims to determine the elemental composition of a given inorganic compound (or a mixture of compounds). This form of analysis mostly focuses on the detection of the ions of the given inorganic compound in an aqueous solution.
Anions and cations are tested separately in this method.
The cations are divided into six analytical groups. Each group has a common reagent which is used to separate the cations of that group from the aqueous solution of the given inorganic compound.
In the qualitative inorganic analysis of cations, those belonging to the 2^nd and 4^th analytical groups precipitate as sulphides. Thus, hydrogen sulphide gas (\[{H_2}S\]) is the group reagent for both 2^nd and 4^th analytical groups.
All the cations given in the options belong to the 2^nd analytical group. Cations of this group precipitate as sulphides which are insoluble in acidic conditions.
Analysis of cations is based on their solubility products. When the cation gains the optimum concentration required for its precipitation, it precipitates allowing us to detect it.
Among the sulphides of the 2^nd group, cadmium sulphide (\[CdS\]) has the highest solubility product. \[CdS\] exists in the following solubility equilibrium:
\[CdS(s) \rightleftharpoons C{d^{2 + }}(aq) + {S^{2 - }}(aq)\]
Its solubility product (\[{K_{sp}}\] ) is defined as \[{K_{sp}} = [C{d^{2 + }}][{S^{2 - }}]\] where \[[C{d^{2 + }}]\] and\[[{S^{2 - }}]\] are the concentrations of the cadmium and sulphide ions respectively.
A high \[{K_{sp}}\] value of \[CdS\] means that the optimum concentration of \[C{d^{2 + }}\] and\[{S^{2 - }}\] ions required for its precipitation is high.
\[CdS\] precipitates when \[{H_2}S\] gas is passed through the solution of \[C{d^{2 + }}\] ions. \[{H_2}S\] gas exists in the following equilibrium in an aqueous medium:
\[{H_2}S(aq) \rightleftharpoons 2{H^ + }(aq) + {S^{2 - }}(aq)\]
We can see that the sulphide ions necessary for the precipitation of \[CdS\] are supplied by the dissociation of \[{H_2}S\].
Since the solution is acidified, there are \[{H^ + }\] ions floating around in the solution already. The presence of \[{H^ + }\] ions in the solution inhibits the dissociation of \[{H_2}S\] in the aqueous solution. This phenomenon is called the common ion effect. This will also cause the \[{H_2}S\] to furnish a lesser number of sulphide ions and the concentration of sulphide ions will decrease.
If the solution is acidified using a concentrated acid, there will be a higher amount of \[{H^ + }\] ions present in the solution beforehand. This will inhibit the dissociation of \[{H_2}S\] and lower the concentration of \[{S^{2 - }}\] ions even further.
Since the \[{K_{sp}}\] of \[CdS\] is high, the lower concentration of \[{S^{2 - }}\] ions will not be enough to cause its precipitation.
Thus, option D is correct.
Note:The concept of solubility product, solubility equilibrium, ionic product and their relationship with the process of precipitation must be clear to the student when it comes to solving questions about the qualitative inorganic analysis.