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. 0.05 mole
B. 0.025 mole
C. 0.10 mole
D. 0.0005 mole

Explanation

To prepare a buffer solution of pH 8.26 using ammonium sulfate and a 0.01 M solution of ammonium hydroxide (NH₄OH), we will follow these steps:

### Step-by-Step Solution:

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

2. **Convert pH to pOH**:
- Use the relationship:
\[ pOH = 14 - pH \]
- Calculate:
\[ pOH = 14 - 8.26 = 5.74 \]

3. **Calculate pKₑ**:
- Use the relationship:
\[ pK_b = 14 - pK_a \]
- Calculate:
\[ pK_b = 14 - 9.26 = 4.74 \]

4. **Use the Buffer Equation**:
- For a basic buffer, the equation is:
\[ pOH = pK_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
- Rearranging gives:
\[ \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) = pOH - pK_b \]
- Substitute the values:
\[ \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) = 5.74 - 4.74 = 1 \]

5. **Convert the Logarithmic Equation**:
- Convert from logarithmic form:
\[ \frac{[\text{Salt}]}{[\text{Base}]} = 10^1 = 10 \]
- This means:
\[ [\text{Salt}] = 10 \times [\text{Base}] \]

6. **Calculate the Concentration of Base**:
- The concentration of NH₄OH (the base) is:
\[ [\text{Base}] = 0.01 \, \text{M} \]
- Therefore:
\[ [\text{Salt}] = 10 \times 0.01 = 0.1 \, \text{M} \]

7. **Calculate the Amount of Salt Required**:
- The volume of the solution is 500 mL (0.5 L).
- The moles of salt needed:
\[ \text{Moles of Salt} = [\text{Salt}] \times \text{Volume} = 0.1 \, \text{M} \times 0.5 \, \text{L} = 0.05 \, \text{moles} \]

8. **Convert Moles of Salt to Millimoles**:
- Convert moles to millimoles:
\[ 0.05 \, \text{moles} = 50 \, \text{mmol} \]

9. **Determine the Amount of Ammonium Sulfate**:
- Each molecule of ammonium sulfate \((NH₄)_2SO₄\) provides 2 ammonium ions (NH₄⁺).
- Therefore, the millimoles of ammonium ions needed:
\[ \text{Millimoles of } NH₄^+ = 50 \, \text{mmol} \text{ (from salt)} \]
- Since each mole of ammonium sulfate provides 2 moles of NH₄⁺:
\[ \text{Millimoles of } (NH₄)_2SO₄ = \frac{50}{2} = 25 \, \text{mmol} \]

10. **Convert Millimoles to Moles**:
- Convert to moles:
\[ \text{Moles of } (NH₄)_2SO₄ = \frac{25}{1000} = 0.025 \, \text{moles} \]

### Final Answer:
The amount of ammonium sulfate \((NH₄)_2SO₄\) to be added is **0.025 moles**.