Which of the following mixtures of ions cann be separated by using an excess of an NaOH solution?

Options

A. `Pb^(2+) and Zn^(2+)`
B. `Al^(3+) and Zn^(2+)`
C. `Fe^(3+) annd Al^(3+)`
D. `Sn^(2+) and Pb^(2+)`

Answer

To determine which of the given mixtures of ions can be separated using an excess of NaOH solution, we will analyze each pair of ions based on their reactions with NaOH.

### Step-by-Step Solution:

1. **Identify the Ions in Each Option**:
- Option 1: Lead (II) ions (Pb²⁺) and Zinc (II) ions (Zn²⁺)
- Option 2: Aluminium (III) ions (Al³⁺) and Zinc (II) ions (Zn²⁺)
- Option 3: Iron (III) ions (Fe³⁺) and Aluminium (III) ions (Al³⁺)
- Option 4: Tin (II) ions (Sn²⁺) and Lead (II) ions (Pb²⁺)

2. **Reactions with NaOH**:
- **Option 1: Pb²⁺ and Zn²⁺**:
- Pb²⁺ + 2 NaOH → Pb(OH)₂ (white precipitate)
- Zn²⁺ + 2 NaOH → Zn(OH)₂ (white precipitate)
- Both lead hydroxide and zinc hydroxide are soluble in excess NaOH, thus they cannot be separated.

- **Option 2: Al³⁺ and Zn²⁺**:
- Al³⁺ + 3 NaOH → Al(OH)₃ (white precipitate)
- Zn²⁺ + 2 NaOH → Zn(OH)₂ (white precipitate)
- Both aluminum hydroxide and zinc hydroxide are soluble in excess NaOH, thus they cannot be separated.

- **Option 3: Fe³⁺ and Al³⁺**:
- Fe³⁺ + 3 NaOH → Fe(OH)₃ (brown precipitate)
- Al³⁺ + 3 NaOH → Al(OH)₃ (white precipitate)
- Aluminum hydroxide is soluble in excess NaOH, while iron (III) hydroxide is insoluble. Therefore, they can be separated.

- **Option 4: Sn²⁺ and Pb²⁺**:
- Sn²⁺ + 2 NaOH → Sn(OH)₂ (white precipitate)
- Pb²⁺ + 2 NaOH → Pb(OH)₂ (white precipitate)
- Both tin (II) hydroxide and lead (II) hydroxide are soluble in excess NaOH, thus they cannot be separated.

3. **Conclusion**:
- The only mixture that can be separated using excess NaOH is **Option 3: Iron (III) ions (Fe³⁺) and Aluminium (III) ions (Al³⁺)**.

### Final Answer:
The mixture of ions that can be separated by using an excess of NaOH solution is **Iron (III) ions (Fe³⁺) and Aluminium (III) ions (Al³⁺)**.