Which of the following statement(s) is/are correct ?

Options

A. Yellow precipitated of silver arsenite is soluble in both nitric acid and ammonia
B. Potassium cyanide when added in very small quantity to copper sulphate solution, produces first yellow precipitate which quickly converts in to white precipitate.
C. Black precipitate of `Bil_(3)` turns orange on heating with water.
D. White precipitate of `Bi(OH)_(3)` turns yellowish brown, when boiled.

Explanation

To determine which of the statements are correct, we will analyze each statement one by one based on the provided information.

### Step-by-Step Solution:

1. **Statement 1: Yellow precipitate of silver arsenate is soluble in both nitric acid and ammonia.**
- Silver arsenate (Ag3AsO3) is indeed a yellow precipitate.
- When dissolved in ammonia (NH3), it forms a complex with silver ions (Ag+), making it soluble. The reaction can be represented as:
\[ \text{Ag}_3\text{AsO}_3 + \text{NH}_3 \rightarrow \text{Ag(NH}_3)_3^{+} + \text{AsO}_3^{3-} \]
- Similarly, when treated with nitric acid (HNO3), it also dissolves, forming silver nitrate (AgNO3) and arsenic trioxide (As2O3).
\[ \text{Ag}_3\text{AsO}_3 + 3\text{HNO}_3 \rightarrow 3\text{AgNO}_3 + \text{As}_2\text{O}_3 + 3\text{H}_2\text{O} \]
- Therefore, **Statement 1 is correct.**

2. **Statement 2: Potassium cyanide when added in small quantity to copper sulfate solution first produces a yellow precipitate which quickly converts to a white precipitate.**
- When potassium cyanide (KCN) is added to copper sulfate (CuSO4), it initially forms cuprous cyanide (CuCN), which is a yellow precipitate:
\[ \text{Cu}^{2+} + 2\text{CN}^{-} \rightarrow \text{CuCN} \downarrow \text{ (yellow precipitate)} \]
- This yellow precipitate is unstable and can convert to a white precipitate (CuCN) upon shaking or heating, releasing cyanogen gas (C2N2).
- Therefore, **Statement 2 is correct.**

3. **Statement 3: Black precipitate of bismuth iodide turns orange on heating with water.**
- Bismuth iodide (BiI3) is a black precipitate. When heated with water, it forms bismuth oxyiodide (BiOI) and hydrogen iodide (HI):
\[ \text{BiI}_3 + \text{H}_2\text{O} \rightarrow \text{BiOI} + 2\text{HI} \]
- The resulting product is indeed orange in color.
- Therefore, **Statement 3 is correct.**

4. **Statement 4: White precipitate of bismuth hydroxide turns yellowish brown when boiled.**
- Bismuth hydroxide (Bi(OH)3) is a white precipitate. Upon heating, it does not turn yellowish brown but rather forms bismuth oxide (Bi2O3), which can appear yellowish or brownish depending on the conditions.
- Therefore, **Statement 4 is incorrect.**

### Conclusion:
The correct statements are:
- Statement 1: Correct
- Statement 2: Correct
- Statement 3: Correct
- Statement 4: Incorrect

### Final Answer:
The correct statements are 1, 2, and 3.

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