Question: Why only \[A{s^{3 + }}\] gets precipitated as \[A{s_2}{S_3}\] and not \[Z{n^{ + 2}}\] as ZnS when \[...

Question

Why only $A{s^{3 + }}$ gets precipitated as $A{s_2}{S_3}$ and not $Z{n^{ + 2}}$ as ZnS when ${H_2}S$ is passed through an acidic solution containing $A{s^{3 + }}$ and $Z{n^{ + 2}}$ ?
(A) Solubility product of $A{s_2}{S_3}$ is less than that of ZnS
(B) Enough $A{s^{3 + }}$ are present in acidic medium
(C) Zinc salt does not ionise in acidic medium
(D) Solubility product changes in presence of an acid

Answer 1

Solution

The species having the minimum value of solubility product will get precipitated first because its ionic product will be more than the solubility product. Another factor is the common ion effect due to hydrogen sulphide, which decreases the solubility product.

Complete step-by-step answer:
In the presence of an acid such as HCl which is also a strong electrolyte leads to suppression of dissociation of ${H_2}S$ due to presence of common ion ${H^ + }$ and thus less ${S^{2 - }}$ ions are produced. This is called the common ion effect.
${H_2}S \rightleftharpoons 2{H^ + } + {S^{2 - }}$
The solubility product is the equilibrium constant for the dissolution of a solid substance into an aqueous solution, denoted by the symbol $K_{sp}$ . Its value depends on temperature i.e. it usually increases with an increase in temperature due to increased solubility as mobility of ions rises.
There are some factors that affect the solubility product. The first one is the common-ion effect. Due to the presence of a common ion, $K_{sp}$ value decreases. The second one is the diverse-ion effect, in which ions of the solutes are not common and thus the value of $K_{sp}$ will increase.
If a solubility product is greater than the ionic product then we say that the solution is unsaturated and no precipitate will form by the addition of more solute. When it is less than the ionic product, then the solution is supersaturated and the excess of solute will precipitate immediately.
As per our problem, $A{s^{3 + }}$ gets precipitated as $A{s_2}{S_3}$ and not $Z{n^{ + 2}}$ as ZnS when ${H_2}S$ is passed through an acidic solution containing $A{s^{3 + }}$ and $Z{n^{ + 2}}$ which means for $A{s^{3 + }}$ , solubility product is less than the ionic product. Therefore, only $A{s^{3 + }}$ forms a precipitate.

Hence, the correct option is (A).

Note: If an ion is insoluble on the basis of solubility, then it forms a solid(precipitate) with an ion from the other reactant present. While if all the ions in a reaction are soluble, i.e. they will remain in aqueous ion form, then no precipitation will occur.