The addition of `NaOH` to `Cr^(3+)` solution produces the precipitate of

Options

A. `Cr(OH)_(3)`
B. `CrO_(3)`
C. `Cr_(2)O_(3)(H_(2)O)_(n)`
D. `Cr_(2)O_(3)`

Answer

To solve the problem of what precipitate is formed when NaOH is added to a solution of Cr³⁺, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have a solution containing Cr³⁺ ions and we are adding NaOH to it. NaOH dissociates into Na⁺ and OH⁻ ions in solution.

2. **Reaction Between Cr³⁺ and OH⁻**:
- The Cr³⁺ ions will react with the OH⁻ ions from the NaOH. The reaction can be represented as:
\[ \text{Cr}^{3+} (aq) + 3 \text{OH}^- (aq) \rightarrow \text{Cr(OH)}_3 (s) \]
- Here, Cr(OH)₃ is the precipitate formed, which is chromium(III) hydroxide.

3. **Formation of Precipitate**:
- The product, Cr(OH)₃, is insoluble in water and will precipitate out of the solution. This is the first precipitate we are interested in.

4. **Dissolution in Excess NaOH**:
- If we add an excess of NaOH to the precipitate Cr(OH)₃, it can dissolve to form a complex ion:
\[ \text{Cr(OH)}_3 (s) + \text{OH}^- (aq) \rightarrow \text{Cr(OH)}_4^- (aq) \]
- This complex ion, Cr(OH)₄⁻, is soluble in excess hydroxide.

5. **Conclusion**:
- The precipitate formed initially when NaOH is added to Cr³⁺ solution is Cr(OH)₃.

### Final Answer:
The addition of NaOH to Cr³⁺ solution produces the precipitate of **Cr(OH)₃**.