Which of the following is insoluble in `NaHCO_(3)` ?

Options

A. Benzoic acid
B. Benzene sulphonic acid
C. o-Nitrophenol
D. Picric acid

Answer

To determine which compound is insoluble in sodium bicarbonate (NaHCO₃), we need to analyze the acidity of the given compounds. Sodium bicarbonate is a weak base and can react with acids to form salts and carbon dioxide. Therefore, the compound that is least acidic will be the one that does not react with NaHCO₃ and remains insoluble.

### Step-by-Step Solution:

1. **Identify the Compounds**: We have four compounds to analyze for their acidity:
- Benzoic acid
- Benzene sulfonic acid
- Ortho-nitrophenol
- Trichloroacetic acid

2. **Determine the pKa Values**: The acidity of a compound can be assessed by its pKa value. The lower the pKa, the stronger the acid:
- Benzoic acid: pKa ≈ 4.2
- Benzene sulfonic acid: pKa ≈ -2.8
- Ortho-nitrophenol: pKa ≈ 7.2
- Trichloroacetic acid: pKa ≈ 0.3

3. **Compare Acidity**:
- Benzene sulfonic acid is the strongest acid (pKa -2.8).
- Trichloroacetic acid is also quite acidic (pKa 0.3).
- Benzoic acid is moderately acidic (pKa 4.2).
- Ortho-nitrophenol is the least acidic (pKa 7.2).

4. **Identify the Least Acidic Compound**: Since NaHCO₃ reacts with acids, the compound that is least acidic will be insoluble in NaHCO₃. Here, ortho-nitrophenol has the highest pKa value (7.2), indicating it is the least acidic.

5. **Conclusion**: Therefore, ortho-nitrophenol is insoluble in NaHCO₃.

### Final Answer:
**Ortho-nitrophenol is insoluble in NaHCO₃.**