To an aqueous solution containing ions such as Al$^{3+}$, Zn$^{2+}$, Ca$^{2+}$, Fe$^{3+}$, Ni$^{2+}$, Ba$^{2+}$ and Cu$^{2+}$ was added conc. HCl, followed by H$_2$S. The total number of cations precipitated during this reaction is/are :
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Remember the common ion effect in qualitative analysis. Adding a strong acid (HCl) to H$_2$S solution decreases the sulfide ion concentration, allowing for the selective precipitation of metal sulfides with very low solubility products (Group II cations).
This question describes the conditions for the precipitation of Group II cations in the qualitative analysis scheme.
The group reagent for Group II cations is H$_2$S gas in the presence of a dilute acid (like HCl).
The added acid (HCl) increases the H$^+$ ion concentration in the solution. According to Le Chatelier's principle, this suppresses the dissociation of the weak electrolyte H$_2$S:
H$_2$S $\rightleftharpoons$ 2H$^+$ + S$^{2-}$
This results in a very low concentration of sulfide ions (S$^{2-}$).
This low concentration of S$^{2-}$ is sufficient to exceed the solubility product (K$_{sp}$) of only the Group II metal sulfides, which have very low K$_{sp}$ values.
The cations given are: Al$^{3+}$ (Group III), Zn$^{2+}$ (Group IV), Ca$^{2+}$ (Group V), Fe$^{3+}$ (Group III), Ni$^{2+}$ (Group IV), Ba$^{2+}$ (Group V), and Cu$^{2+}$ (Group II).
From the list, only Cu$^{2+}$ is a Group II cation. Therefore, only copper(II) sulfide (CuS) will precipitate under these conditions.
The total number of cations precipitated is 1.
