`NaOH` can be used to seprate `Al(OH)_(3) `and `Zn(OH)_(2)` . (T/F)

Explanation

To determine whether the statement "NaOH can be used to separate Al(OH)₃ and Zn(OH)₂" is true or false, we need to analyze the solubility of both aluminum hydroxide and zinc hydroxide in sodium hydroxide.

### Step-by-Step Solution:

1. **Understanding the Compounds**:
- Aluminum hydroxide (Al(OH)₃) and zinc hydroxide (Zn(OH)₂) are both amphoteric hydroxides, meaning they can react with both acids and bases.

2. **Reaction with NaOH**:
- When aluminum hydroxide is treated with sodium hydroxide (NaOH), it forms soluble sodium aluminate (NaAlO₂) or sodium tetrahydroxoaluminate (NaAl(OH)₄).
- The reaction can be represented as:
\[ \text{Al(OH)}_3 + \text{NaOH} \rightarrow \text{NaAlO}_2 + \text{H}_2\text{O} \]
or
\[ \text{Al(OH)}_3 + 4\text{NaOH} \rightarrow \text{NaAl(OH)}_4 + 3\text{H}_2\text{O} \]

3. **Zinc Hydroxide Reaction**:
- Similarly, when zinc hydroxide reacts with sodium hydroxide, it forms soluble sodium zincate (Na₂ZnO₂) or sodium tetrahydroxozincate (Na₂Zn(OH)₄).
- The reaction can be represented as:
\[ \text{Zn(OH)}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + 2\text{H}_2\text{O} \]
or
\[ \text{Zn(OH)}_2 + 4\text{NaOH} \rightarrow \text{Na}_2\text{Zn(OH)}_4 + 2\text{H}_2\text{O} \]

4. **Conclusion**:
- Since both Al(OH)₃ and Zn(OH)₂ dissolve in NaOH to form soluble complexes, NaOH cannot be used to separate them. Therefore, the statement is **False**.

### Final Answer:
**False** - NaOH cannot be used to separate Al(OH)₃ and Zn(OH)₂.