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How many of the following pairs of ions ...

How many of the following pairs of ions can be separated by `H_(2)S` in dilute `HCl`?
`Bi^(3+)` and `Sn^(4+),Al^(3+)` and `Hg^(2+),Cd^(2+)` and `Zn^(2+),Fe^(3+)` and `Cu^(2+),As^(3+)` and `Sb^(3+)`

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To determine how many pairs of ions can be separated by `H₂S` in dilute `HCl`, we will analyze each pair of ions provided in the question. ### Step-by-Step Solution: 1. **Pair 1: Bi³⁺ and Sn⁴⁺** - When Bi³⁺ and Sn⁴⁺ are treated with `H₂S` in dilute `HCl`, they form Bi₂S₃ (black precipitate) and SnS₂ (yellow precipitate). - Since both Bi₂S₃ and SnS₂ precipitate out, they cannot be separated by `H₂S`. - **Conclusion**: This pair cannot be separated. 2. **Pair 2: Al³⁺ and Hg²⁺** - When Al³⁺ and Hg²⁺ react with `H₂S` in dilute `HCl`, HgS (black precipitate) forms while Al³⁺ remains in solution. - Since Al³⁺ stays in solution and only Hg²⁺ precipitates out, this pair can be separated. - **Conclusion**: This pair can be separated. 3. **Pair 3: Cd²⁺ and Zn²⁺** - When Cd²⁺ and Zn²⁺ are treated with `H₂S` in dilute `HCl`, CdS (yellow precipitate) forms, while Zn²⁺ remains in solution. - Since Zn²⁺ stays in solution and Cd²⁺ precipitates out, this pair can be separated. - **Conclusion**: This pair can be separated. 4. **Pair 4: Fe³⁺ and Cu²⁺** - When Fe³⁺ and Cu²⁺ react with `H₂S` in dilute `HCl`, Cu₂S (black precipitate) forms, while Fe³⁺ remains in solution. - Since Fe³⁺ stays in solution and Cu²⁺ precipitates out, this pair can be separated. - **Conclusion**: This pair can be separated. 5. **Pair 5: As³⁺ and Sb³⁺** - When As³⁺ and Sb³⁺ are treated with `H₂S` in dilute `HCl`, As₂S₃ (yellow precipitate) and Sb₂S₃ (orange precipitate) form. - Since both As₂S₃ and Sb₂S₃ precipitate out, they cannot be separated by `H₂S`. - **Conclusion**: This pair cannot be separated. ### Summary of Results: - **Pairs that can be separated**: Al³⁺ and Hg²⁺, Cd²⁺ and Zn²⁺, Fe³⁺ and Cu²⁺. - **Pairs that cannot be separated**: Bi³⁺ and Sn⁴⁺, As³⁺ and Sb³⁺. ### Final Answer: **3 pairs can be separated by `H₂S` in dilute `HCl`.**

To determine how many pairs of ions can be separated by `H₂S` in dilute `HCl`, we will analyze each pair of ions provided in the question. ### Step-by-Step Solution: 1. **Pair 1: Bi³⁺ and Sn⁴⁺** - When Bi³⁺ and Sn⁴⁺ are treated with `H₂S` in dilute `HCl`, they form Bi₂S₃ (black precipitate) and SnS₂ (yellow precipitate). - Since both Bi₂S₃ and SnS₂ precipitate out, they cannot be separated by `H₂S`. - **Conclusion**: This pair cannot be separated. ...
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