Which is decreasing order of acidity in,
`HCOOH(I), CH_(3)COOH(II), CH_(3)CH_(2)COOH(III) and C_(6)H_(5)COOH(IV)`?

Answer

To determine the decreasing order of acidity for the compounds HCOOH (formic acid), CH₃COOH (acetic acid), CH₃CH₂COOH (propanoic acid), and C₆H₅COOH (benzoic acid), we need to analyze the stability of their conjugate bases. The more stable the conjugate base, the stronger the acid.

### Step-by-Step Solution:

1. **Identify the Compounds and Their Conjugate Bases:**
- HCOOH → Conjugate base: HCOO⁻ (formate ion)
- CH₃COOH → Conjugate base: CH₃COO⁻ (acetate ion)
- CH₃CH₂COOH → Conjugate base: CH₃CH₂COO⁻ (propionate ion)
- C₆H₅COOH → Conjugate base: C₆H₅COO⁻ (benzoate ion)

2. **Analyze the Electron-Withdrawing and Electron-Donating Effects:**
- **HCOOH:** No electron-donating groups. The conjugate base (HCOO⁻) is stable due to resonance stabilization.
- **CH₃COOH:** The methyl group (CH₃) is an electron-donating group, which slightly destabilizes the conjugate base (CH₃COO⁻) compared to HCOO⁻.
- **CH₃CH₂COOH:** The ethyl group (CH₃CH₂) is also an electron-donating group, further decreasing acidity compared to acetic acid.
- **C₆H₅COOH:** The phenyl group (C₆H₅) acts as an electron-withdrawing group due to resonance, stabilizing the conjugate base (C₆H₅COO⁻) and increasing acidity.

3. **Order of Acidity Based on Stability of Conjugate Bases:**
- The most stable conjugate base corresponds to the strongest acid.
- **C₆H₅COOH (benzoic acid)** > **HCOOH (formic acid)** > **CH₃COOH (acetic acid)** > **CH₃CH₂COOH (propanoic acid)**.

4. **Final Decreasing Order of Acidity:**
- C₆H₅COOH > HCOOH > CH₃COOH > CH₃CH₂COOH

### Summary of the Decreasing Order of Acidity:
C₆H₅COOH (IV) > HCOOH (I) > CH₃COOH (II) > CH₃CH₂COOH (III)