The correct order of `K_(a)` is :
(i) `CH_(3)-CH_(2)-COOH`
(ii) `OHC-CH_(2)-COOH`
(iii) `Ph-CH_(2)-COOH`

Options

A. `iiigtiigti`
B. `igtiigtiii`
C. `iigtiiigti`
D. `igtiiigtii`

Answer

To determine the correct order of acidity (K_a) for the given carboxylic acids, we need to analyze the stability of their conjugate bases after deprotonation. The more stable the conjugate base, the stronger the acid.

### Step-by-Step Solution:

1. **Identify the Compounds**:
- (i) `CH₃-CH₂-COOH` (Propanoic acid)
- (ii) `OHC-CH₂-COOH` (2-Hydroxypropanoic acid or Lactic acid)
- (iii) `Ph-CH₂-COOH` (Benzyl acid)

2. **Deprotonation**:
- When each acid donates a proton (H⁺), we need to analyze the resulting conjugate bases:
- (i) `CH₃-CH₂-COO⁻` (Propanoate)
- (ii) `OHC-CH₂-COO⁻` (2-Hydroxypropanoate)
- (iii) `Ph-CH₂-COO⁻` (Benzylate)

3. **Stability of Conjugate Bases**:
- **Compound (i)**: The conjugate base `CH₃-CH₂-COO⁻` has two alkyl groups (electron-donating), which destabilize the negative charge.
- **Compound (ii)**: The conjugate base `OHC-CH₂-COO⁻` has an aldehyde group (−I effect), which is electron-withdrawing, stabilizing the negative charge.
- **Compound (iii)**: The conjugate base `Ph-CH₂-COO⁻` has a benzyl group, which can also stabilize the negative charge through resonance, but less effectively than the aldehyde in compound (ii).

4. **Comparing Stability**:
- The stability order of the conjugate bases is:
- (ii) > (iii) > (i)
- This means that the acidity order (K_a) will be the reverse of the stability order:
- (i) < (iii) < (ii)

5. **Final Order of Acidity (K_a)**:
- Therefore, the correct order of K_a is:
- `K_a (ii) > K_a (iii) > K_a (i)`

### Conclusion:
The correct order of K_a is:
- (ii) `OHC-CH₂-COOH` > (iii) `Ph-CH₂-COOH` > (i) `CH₃-CH₂-COOH`.