Question

In presence of $HCl$, ${{H}_{2}}S$ result the precipitation of group-2 elements but not group-4 elements during qualitative analysis .It is due to :
  (A) higher concentration of ${{S}^{2-}}$
  (B) higher concentration of ${{H}^{+}}$
  (C) lower concentration of ${{S}^{2-}}$
  (D) lower concentration of ${{H}^{+}}$

Answer 1

Hint: This question is a direct application of common ion effect. The answer I related to solubility products of radicals and hence by understanding the ionization of ions we would be able to determine the common ion and its effect on the given example of qualitative analysis.


Complete step by step solution:
- Let's start with the concept of common ion effect. The effect in which the degree of ionization of an electrolyte is being suppressed by the addition of a strong electrolyte which contains a common ion is called the common ion effect. Or it also can be described as the relative lowering of degree of ionization of a weak electrolyte by the addition of a strong electrolyte which has a common ion.
- We are asked about the qualificative analysis of group-2 and group-4 cation. So, let's have a look at each of these analyses. In group-2 analysis when ${{H}_{2}}S$ gas is passed through acidified solution of $HCl$, sulphides of basic radicals of group-2 are precipitated and when ${{H}_{2}}S$ gas is passed through the solution in presence of $N{{H}_{4}}OH$,cations of group-4 are precipitated as sulphides .
- The dissociation of ${{H}_{2}}S$ and $HCl$ can be written as follows
${{H}_{2}}S\rightleftharpoons 2{{H}^{+}}+{{S}^{2-}}$
$HCl\rightleftharpoons {{H}^{+}}+C{{l}^{-}}$
- Here ${{H}^{+}}$acts as the common ion and it shifts the equilibrium to the left by Le Chatleier's principle. The ionization of ${{H}_{2}}S$ is suppressed by the addition of $HCl$and lowers the concentration of ${{S}^{2-}}$ions, just exceeds the solubility product of group-2 sulphides. As a result, only group-2 cations are precipitated and the precipitation of group-4 sulphides are prevented because they have higher solubility product values as compared to the sulphides of group-2.
- Let's summarize the points we have discussed. The dissociation of ${{H}_{2}}S$ is suppressed in the presence of $HCl$due to common ion effect and this decreases the ${{S}^{2-}}$ion concentration. Thus, only group 2 radicals are precipitated as they have low solubility products.



Therefore, the answer is option(C) lower concentration of ${{S}^{2-}}$


Note: It should be noted that the common ion effect is very useful in analytical chemistry and it is often applied in qualitative analysis as we have seen above. An electrolyte will precipitate only if the concentration of its ions exceeds the solubility product which means that the precipitation is obtained only if the concentration of any one of the ions is increased. Hence, adding a common ion, will increase the solubility product