A buffer of pH 9.26 is made by dissolving x moles of ammonium sulphate and 0.1 mole of ammonia into 100 mL solution. If `pK_b` of ammonia is 4.74, calculate value of x.

Answer

To solve the problem, we need to find the value of \( x \) (the number of moles of ammonium sulfate) that contributes to a buffer solution with a pH of 9.26. We know that the solution contains 0.1 moles of ammonia and is made up to a total volume of 100 mL.

### Step-by-Step Solution:

1. **Calculate pOH from pH**:
\[ \text{pOH} = 14 - \text{pH} = 14 - 9.26 = 4.74 \]

2. **Identify the relationship between pOH and \( pK_b \)**:
Given that \( pK_b \) of ammonia is 4.74, we can see that:
\[ \text{pOH} = pK_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]

3. **Set up the equation using the Henderson-Hasselbalch equation**:
Here, the salt is ammonium sulfate which provides ammonium ions (\( NH_4^+ \)), and the base is ammonia (\( NH_3 \)). The concentration of the salt is given by:
\[ [\text{Salt}] = \frac{2x}{0.1} \quad \text{(since 1 mole of ammonium sulfate gives 2 moles of } NH_4^+ \text{)} \]
The concentration of the base (ammonia) is:
\[ [\text{Base}] = \frac{0.1}{0.1} = 1 \]

4. **Substituting values into the equation**:
\[ 4.74 = 4.74 + \log\left(\frac{2x}{1}\right) \]

5. **Simplifying the equation**:
\[ 0 = \log(2x) \]
This implies:
\[ 2x = 1 \quad \text{(since } \log(1) = 0\text{)} \]

6. **Solving for \( x \)**:
\[ x = \frac{1}{2} = 0.5 \]

### Final Answer:
The value of \( x \) is **0.5 moles**.