One mole of ammonia was completely absorbed in one litre solution each of `(1) 1 M HCI , (2) 1 M CH_(3)COOH " and "(3) 1 M H_(2)SO_(4)` at 298 K. The decreasing order for the pH of the resulting solution is
(Given , `K_(b) (NH_(3)) = 4.74 )`

Options

A. `2 gt 3 gt 1`
B. `1 gt 2 gt 3`
C. `2 gt 1 gt 3`
D. `3 gt 2 gt1`

Explanation

To determine the decreasing order of pH for the resulting solutions after 1 mole of ammonia is absorbed in 1 liter of each of the given acids (1 M HCl, 1 M CH₃COOH, and 1 M H₂SO₄), we can follow these steps:

### Step 1: Identify the products formed
When ammonia (NH₃) reacts with the acids, it forms the corresponding ammonium salts:
- With HCl: NH₄Cl (Ammonium Chloride)
- With CH₃COOH: NH₄CH₃COO (Ammonium Acetate)
- With H₂SO₄: (NH₄)₂SO₄ (Ammonium Sulfate)

### Step 2: Understand the nature of the acids
- HCl is a strong acid and will completely dissociate in solution.
- CH₃COOH is a weak acid and will partially dissociate.
- H₂SO₄ is a strong acid and will also completely dissociate, but it is diprotic, meaning it can release two protons (H⁺) per molecule.

### Step 3: Calculate the pH for each solution
1. **For HCl (1 M):**
- Since HCl is a strong acid, it will completely dissociate:
\[ [H^+] = 1 \, \text{M} \]
- pH = -log[H⁺] = -log(1) = 0.

2. **For CH₃COOH (1 M):**
- The dissociation of acetic acid can be represented as:
\[ CH₃COOH \rightleftharpoons H^+ + CH₃COO^- \]
- The equilibrium expression is:
\[ K_a = \frac{[H^+][CH₃COO^-]}{[CH₃COOH]} \]
- Given that Kb for NH₃ is 4.74, we can find Ka for acetic acid using:
\[ K_w = K_a \cdot K_b \Rightarrow K_a = \frac{K_w}{K_b} \]
- Assuming \(K_w = 1.0 \times 10^{-14}\):
\[ K_a = \frac{1.0 \times 10^{-14}}{4.74} \approx 2.11 \times 10^{-15} \]
- Since acetic acid is weak, we can use the approximation for weak acids to find [H⁺]:
\[ [H^+] \approx \sqrt{K_a \cdot C} \approx \sqrt{2.11 \times 10^{-15} \cdot 1} \approx 1.45 \times 10^{-8} \, \text{M} \]
- pH = -log(1.45 × 10⁻⁸) ≈ 7.84.

3. **For H₂SO₄ (1 M):**
- H₂SO₄ is a strong acid and will dissociate completely:
\[ H₂SO₄ \rightarrow 2H^+ + SO₄^{2-} \]
- Thus, the concentration of H⁺ ions will be:
\[ [H^+] = 2 \, \text{M} \]
- pH = -log(2) ≈ -0.301.

### Step 4: Compare the pH values
- pH of HCl solution = 0
- pH of CH₃COOH solution ≈ 7.84
- pH of H₂SO₄ solution ≈ -0.301

### Step 5: Determine the order of decreasing pH
From the calculated pH values, we can conclude:
1. pH of CH₃COOH (≈ 7.84) is the highest.
2. pH of HCl (0) is next.
3. pH of H₂SO₄ (≈ -0.301) is the lowest.

### Final Order of Decreasing pH:
1. CH₃COOH (Ammonium Acetate)
2. HCl (Ammonium Chloride)
3. H₂SO₄ (Ammonium Sulfate)

### Summary:
The decreasing order for the pH of the resulting solutions is:
**CH₃COOH > HCl > H₂SO₄**