Which of the following compounds are soluble in `NaHCO_(3)`?

Options

A. Picric acid
B.
C.
D.

Explanation

To determine which compounds are soluble in sodium bicarbonate (NaHCO₃), we need to identify which of the given compounds can react with NaHCO₃ to produce carbon dioxide (CO₂) gas. Generally, this occurs with acidic compounds, particularly strong acids.

### Step-by-Step Solution:

1. **Understand the Reaction**: Sodium bicarbonate reacts with acids to produce carbon dioxide gas. Therefore, we need to identify which of the given compounds are acidic enough to react with NaHCO₃.

2. **Identify the Compounds**: The compounds in question are:
- A: Picric acid
- B: Sulfonic acid
- C: Ortho-nitrophenol
- D: Ortho-nitrobenzoic acid

3. **Analyze Each Compound**:
- **Picric Acid (A)**: This is a strong acid due to the presence of three nitro groups, which increase its acidity. It will react with NaHCO₃, producing CO₂. **(Soluble)**

- **Sulfonic Acid (B)**: Sulfonic acids are also strong acids. They can react with NaHCO₃, producing CO₂. **(Soluble)**

- **Ortho-Nitrophenol (C)**: Although it has a hydroxyl (-OH) group, it is less acidic due to intramolecular hydrogen bonding, which stabilizes the molecule and decreases its acidity. Therefore, it will not react with NaHCO₃. **(Not Soluble)**

- **Ortho-Nitrobenzoic Acid (D)**: This compound contains a carboxylic acid group (-COOH), which is acidic. The presence of the nitro group increases its acidity, allowing it to react with NaHCO₃ and produce CO₂. **(Soluble)**

4. **Conclusion**: The compounds that are soluble in NaHCO₃ are:
- Picric acid (A)
- Sulfonic acid (B)
- Ortho-nitrobenzoic acid (D)

### Final Answer:
The compounds that are soluble in NaHCO₃ are A (Picric acid), B (Sulfonic acid), and D (Ortho-nitrobenzoic acid).

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