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:

First we note the classical qualitative analysis scheme for metal cations. The cations are divided into analytical groups according to the reagent that first causes their precipitation. We recall the two steps involved here:

Step 1 - Addition of concentrated HCl. The reagent supplies a very high concentration of $$\text{Cl}^-$$ ions. According to the scheme, only the “Group I” cations—Ag}^+,\,$$\text{Pb}^{2+}$$,\,$$\text{Hg}_2^{2+—form$$ practically insoluble chlorides and precipitate at this stage:

Ag $$^+ + \text{Cl}^- \longrightarrow \text{AgCl}\downarrow$$

Pb $$^{2+} + 2\text{Cl}^- \longrightarrow \text{PbCl}_2\downarrow$$

None of the ions present in the given mixture—Al}^{3+},\,$$\text{Zn}^{2+}$$,\,$$\text{Ca}^{2+}$$,\,$$\text{Fe}^{3+}$$,\,$$\text{Ni}^{2+}$$,\,$$\text{Ba}^{2+}$$,\,$$\text{Cu}^{2+—belong$$ to this chloride group. Hence after adding conc. HCl no precipitate forms so far; every ion remains in solution.

Step 2 - Passing H$$_2$$S through the strongly acidic solution. We first state the relevant equilibrium that controls the availability of sulfide ions:

H $$_2\text{S}\rightleftharpoons\text{H}^+ + \text{HS}^- \quad\text{and}\quad \text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-}$$

Because concentrated HCl supplies a large $$[\text{H}^+]$$, the above equilibria lie far to the left, so the actual concentration of $$\text{S}^{2-}$$ ions in the solution is extremely low. Therefore, only those metal sulfides whose solubility product $$K_{sp}$$ is exceedingly small (Group II cations) can satisfy

$$[$$ M $$^{n+}]\,[$$ S $$^{2-}] \gt K_{sp}$$

and precipitate in this acidic medium. The recognised Group II cations are Cu}^{2+},\,$$\text{Cd}^{2+}$$,\,$$\text{Bi}^{3+}$$,\,$$\text{Pb}^{2+}$$,\,$$\text{Hg}^{2+}$$,\,$$\text{As}^{3+/5+}$$,\,$$\text{Sb}^{3+/5+}$$,\,$$\text{Sn}^{2+/4+$$. Among our list, only

Cu $$^{2+} + \text{S}^{2-} \longrightarrow \text{CuS}\downarrow$$

belongs to this category; CuS has such a tiny $$K_{sp}$$ that it precipitates even with the minute sulfide ion concentration present.

The other given ions behave differently:

• $$\text{Fe}^{3+}$$ and $$\text{Al}^{3+}$$ form hydroxide precipitates in basic medium (Group III), not sulfides in acid.
• $$\text{Ni}^{2+}$$ and $$\text{Zn}^{2+}$$ form sulfides only when the medium is basic (Group IV); in acidic solution their sulfides remain far too soluble.
• $$\text{Ca}^{2+}$$ and $$\text{Ba}^{2+}$$ precipitate later as carbonates (Group V); their sulfides are readily soluble in acid.

Hence, out of the entire set, only one cation—Cu$$^{2+}$$—precipitates as CuS at this stage.

Hence, the correct answer is Option A (1).