Solution of `0.1 N NH_(4)OH` and `0.1 N NH_(4)Cl` has `pH 9.25`, then find out `K_(b)` of `NH_(4)OH`.

Options

A. `9.25`
B. `4.75`
C. `3.75`
D. `8.25`

Explanation

To find the \( K_b \) of \( NH_4OH \) given that a solution of \( 0.1 N \) \( NH_4OH \) and \( 0.1 N \) \( NH_4Cl \) has a \( pH \) of \( 9.25 \), we can follow these steps:

### Step 1: Understand the Buffer System
The solution consists of a weak base \( NH_4OH \) and its conjugate acid \( NH_4Cl \). This creates a basic buffer system.

### Step 2: Calculate \( pOH \)
Using the relationship between \( pH \) and \( pOH \):
\[ pOH = 14 - pH \]
Substituting the given \( pH \):
\[ pOH = 14 - 9.25 = 4.75 \]

### Step 3: Use the Henderson-Hasselbalch Equation
For a basic buffer, the Henderson-Hasselbalch equation is:
\[ pH = pK_b + \log \left( \frac{[Salt]}{[Base]} \right) \]
Here, \( [Salt] \) is the concentration of \( NH_4Cl \) and \( [Base] \) is the concentration of \( NH_4OH \). Given both concentrations are \( 0.1 N \), we can substitute:
\[ pH = pK_b + \log \left( \frac{0.1}{0.1} \right) \]
Since \( \log(1) = 0 \):
\[ pH = pK_b \]

### Step 4: Relate \( pK_b \) to \( pOH \)
Since we have calculated \( pOH \) as \( 4.75 \):
\[ pK_b = pOH = 4.75 \]

### Step 5: Calculate \( K_b \)
To find \( K_b \), we use the relationship:
\[ K_b = 10^{-pK_b} \]
Substituting \( pK_b = 4.75 \):
\[ K_b = 10^{-4.75} \]

### Step 6: Final Calculation
Calculating \( K_b \):
\[ K_b \approx 1.78 \times 10^{-5} \]

### Conclusion
The value of \( K_b \) for \( NH_4OH \) is approximately \( 1.78 \times 10^{-5} \).

---