Which of the following mixtures of ions in solution can be separated by using NaOH solution?

Options

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

Answer

To determine which mixtures of ions in solution can be separated using NaOH solution, we need to analyze the behavior of different cations when they react with NaOH. Here’s a step-by-step solution:

### Step 1: Identify the Cations
We need to identify the cations present in the mixtures. Common cations that can be tested with NaOH include Fe³⁺, Pb²⁺, Al³⁺, and Cu²⁺.

### Step 2: React each Cation with NaOH
We will examine how each cation reacts with NaOH:

1. **Fe³⁺ (Iron(III) ion)**:
- When Fe³⁺ reacts with NaOH, it forms a reddish-brown precipitate of Fe(OH)₃.
- Reaction: Fe³⁺ + 3NaOH → Fe(OH)₃ (s) + 3Na⁺

2. **Pb²⁺ (Lead(II) ion)**:
- Pb²⁺ reacts with NaOH to form a white precipitate of Pb(OH)₂.
- Reaction: Pb²⁺ + 2NaOH → Pb(OH)₂ (s) + 2Na⁺

3. **Al³⁺ (Aluminum ion)**:
- Al³⁺ reacts with NaOH to form a white precipitate of Al(OH)₃.
- Reaction: Al³⁺ + 3NaOH → Al(OH)₃ (s) + 3Na⁺

4. **Cu²⁺ (Copper(II) ion)**:
- Cu²⁺ reacts with NaOH to form a light blue precipitate of Cu(OH)₂.
- Reaction: Cu²⁺ + 2NaOH → Cu(OH)₂ (s) + 2Na⁺

### Step 3: Analyze the Precipitates
The formation of different colored precipitates allows for the separation of these ions:

- **Fe³⁺**: Reddish-brown precipitate (Fe(OH)₃)
- **Pb²⁺**: White precipitate (Pb(OH)₂)
- **Al³⁺**: White precipitate (Al(OH)₃)
- **Cu²⁺**: Light blue precipitate (Cu(OH)₂)

### Step 4: Conclusion
Based on the above reactions, we can conclude that the mixtures containing Fe³⁺, Pb²⁺, Al³⁺, and Cu²⁺ can be separated by using NaOH solution due to the formation of distinct precipitates.

### Final Answer
The mixtures of ions that can be separated by using NaOH solution are those containing Fe³⁺, Pb²⁺, Al³⁺, and Cu²⁺.

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