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

Options

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

Explanation

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

### Step 1: Calculate pOH from pH
Given that the pH of the solution is 9.25, we can calculate the pOH using the relationship:
\[ pH + pOH = 14 \]
Thus,
\[ pOH = 14 - pH = 14 - 9.25 = 4.75 \]

### Step 2: Understand the dissociation of the components
The dissociation of \( NH_4OH \) and \( NH_4Cl \) can be represented as:
- \( NH_4OH \rightleftharpoons NH_4^+ + OH^- \)
- \( NH_4Cl \rightleftharpoons NH_4^+ + Cl^- \)

Both solutions contribute \( NH_4^+ \) ions, which creates a common ion effect.

### Step 3: Apply the formula for pOH
The formula relating pOH, \( pK_b \), and the concentrations of the salt and base is:
\[ pOH = pK_b + \log\left(\frac{[Salt]}{[Base]}\right) \]
In this case, since both the salt \( NH_4Cl \) and the base \( NH_4OH \) have the same concentration of \( 0.1 N \), we can substitute:
\[ pOH = pK_b + \log\left(\frac{0.1}{0.1}\right) \]
Since \( \log(1) = 0 \), this simplifies to:
\[ pOH = pK_b \]

### Step 4: Substitute the value of pOH
From Step 1, we found that \( pOH = 4.75 \). Therefore, we can conclude:
\[ pK_b = 4.75 \]

### Final Answer
Thus, the \( pK_b \) of \( NH_4OH \) is \( 4.75 \).

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