To prepare a buffer of pH `8.26`, amount of `(NH_(4))_(2)SO_(4)` to be added into 500mL of `0.01M NH_(4)OH` solution `[pK_(a)(NH_(4)^(+))=9.26]` is:

Options

A. `0.05mol e`
B. `0.025mol e`
C. `0.10mol e`
D. `0.005mol e`

Explanation

To prepare a buffer of pH 8.26 using 500 mL of 0.01 M NH₄OH, we need to determine the amount of (NH₄)₂SO₄ to be added. We will use the Henderson-Hasselbalch equation for a basic buffer.

### Step-by-step Solution:

1. **Identify the Given Values:**
- pH = 8.26
- Volume of NH₄OH solution = 500 mL = 0.5 L
- Concentration of NH₄OH = 0.01 M
- pKₐ (NH₄⁺) = 9.26

2. **Calculate pOH:**
\[ pOH = 14 - pH = 14 - 8.26 = 5.74 \]

3. **Use the Henderson-Hasselbalch Equation:**
For a basic buffer, the equation is:
\[ pOH = pK_b + \log \left( \frac{[A^-]}{[B]} \right) \]
Here, \( [A^-] \) is the concentration of the salt (NH₄)₂SO₄, and \( [B] \) is the concentration of the base (NH₄OH).

4. **Calculate pK_b:**
Since we have pKₐ for NH₄⁺, we can find pK_b using:
\[ pK_b = 14 - pK_a = 14 - 9.26 = 4.74 \]

5. **Substituting into the Equation:**
\[ 5.74 = 4.74 + \log \left( \frac{[A^-]}{[0.01]} \right) \]
Rearranging gives:
\[ \log \left( \frac{[A^-]}{[0.01]} \right) = 5.74 - 4.74 = 1 \]

6. **Exponentiate to Remove Logarithm:**
\[ \frac{[A^-]}{[0.01]} = 10^1 = 10 \]
Thus,
\[ [A^-] = 10 \times [0.01] = 0.1 \text{ M} \]

7. **Calculate the Amount of (NH₄)₂SO₄ Needed:**
Since we have 500 mL of solution:
\[ \text{Moles of } (NH₄)₂SO₄ = [A^-] \times \text{Volume} = 0.1 \times 0.5 = 0.05 \text{ moles} \]

8. **Convert Moles to Millimoles:**
\[ \text{Millimoles} = 0.05 \text{ moles} \times 1000 = 50 \text{ mmoles} \]

9. **Considering the Stoichiometry:**
Each mole of (NH₄)₂SO₄ provides 2 moles of NH₄⁺:
\[ \text{Millimoles of NH₄⁺} = 2 \times 50 = 100 \text{ mmoles} \]

10. **Final Calculation for (NH₄)₂SO₄:**
Since we need 50 mmoles of NH₄⁺ from (NH₄)₂SO₄, we need:
\[ \text{Millimoles of (NH₄)₂SO₄} = \frac{50}{2} = 25 \text{ mmoles} \]

11. **Convert to Moles:**
\[ \text{Moles of (NH₄)₂SO₄} = \frac{25}{1000} = 0.025 \text{ moles} \]

### Conclusion:
The amount of (NH₄)₂SO₄ to be added is **0.025 moles** or **25 mmoles**.