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:

Options

A. 1
B. 3
C. 4
D. 2

Answer

To determine the total number of cations precipitated when concentrated HCl and H2S are added to the given aqueous solution containing ions such as Al^(3+), Zn^(2+), Ca^(2+), Fe^(3+), Ni^(2+), Ba^(2+), and Cu^(2+), we can follow these steps:

### Step 1: Identify the Groups of Cations
In qualitative analysis, cations are often classified into different groups based on their solubility and the reagents used. The groups relevant to this question are:

- **Group 1**: Precipitate with concentrated HCl (e.g., Pb^(2+), Ag^+, Hg2^(2+))
- **Group 2**: Precipitate with H2S (e.g., Cu^(2+), Cd^(2+), Pb^(2+))
- **Group 3**: Typically do not precipitate with HCl or H2S (e.g., Al^(3+), Fe^(3+))
- **Group 4**: Precipitate with ammonium hydroxide (e.g., Ni^(2+), Zn^(2+))
- **Group 5**: Remain in solution (e.g., Ca^(2+), Ba^(2+))

### Step 2: Analyze the Effect of Concentrated HCl
When concentrated HCl is added, it primarily precipitates cations from Group 1. However, none of the cations listed (Al^(3+), Zn^(2+), Ca^(2+), Fe^(3+), Ni^(2+), Ba^(2+), Cu^(2+)) belong to Group 1, so no precipitation occurs at this stage.

### Step 3: Analyze the Effect of H2S
Next, when H2S is added, it precipitates cations from Group 2. Among the given cations, only Cu^(2+) is known to precipitate as CuS when H2S is added. The other cations (Al^(3+), Zn^(2+), Ca^(2+), Fe^(3+), Ni^(2+), Ba^(2+)) do not precipitate with H2S.

### Step 4: Count the Total Number of Precipitated Cations
From the analysis:
- From concentrated HCl: 0 cations precipitated.
- From H2S: 1 cation (Cu^(2+)) precipitated.

Thus, the total number of cations precipitated during this reaction is **1**.

### Final Answer
The total number of cations precipitated during this reaction is **1**.

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