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The addition of NaOH to Cr^(3+) solution...

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

A

`Cr(OH)_(3)`

B

`CrO_(3)`

C

`Cr_(2)O_(3)(H_(2)O)_(n)`

D

`Cr_(2)O_(3)`

Text Solution

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The correct Answer is:
To solve the question regarding the addition of NaOH to a Cr³⁺ solution and the precipitate formed, we can follow these steps: ### Step 1: Understand the Reaction When sodium hydroxide (NaOH) is added to a solution containing chromium ions (Cr³⁺), the NaOH dissociates into sodium ions (Na⁺) and hydroxide ions (OH⁻). ### Step 2: Formation of Precipitate The hydroxide ions (OH⁻) will react with the chromium ions (Cr³⁺) in the solution to form chromium(III) hydroxide, which is represented by the formula Cr(OH)₃. This compound is insoluble in water and will precipitate out of the solution. **Chemical Equation:** \[ \text{Cr}^{3+} (aq) + 3 \text{OH}^- (aq) \rightarrow \text{Cr(OH)}_3 (s) \] ### Step 3: Identify the Precipitate The precipitate formed is chromium(III) hydroxide (Cr(OH)₃), which is typically a gelatinous, greenish solid. ### Step 4: Consider Excess NaOH If excess NaOH is added, the precipitate of Cr(OH)₃ can dissolve to form a soluble complex ion, specifically Cr(OH)₄⁻, which is green in color. However, the question specifically asks for the precipitate formed upon the addition of NaOH, not the behavior in excess NaOH. ### Final Answer The precipitate formed when NaOH is added to a Cr³⁺ solution is **Cr(OH)₃**. ---
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