A buffer solution contains `1` mole of `(NH_(4))_(2)SO_(4)` and `1` mole of `NH_(4)OH(K_(b)=10^(-5))`. The `pH` of solution will be:

Options

A. `5`
B. `9`
C. `5.3`
D. `8.7`

Explanation

To find the pH of the buffer solution containing 1 mole of \((NH_4)_2SO_4\) and 1 mole of \(NH_4OH\) (with \(K_b = 10^{-5}\)), we can follow these steps:

### Step 1: Identify the type of buffer
The solution contains a weak base (\(NH_4OH\)) and its salt (\((NH_4)_2SO_4\)), which means it is a basic buffer.

### Step 2: Use the buffer equation
For a basic buffer, the formula to calculate \(pOH\) is given by:
\[ pOH = pK_b + \log\left(\frac{[Salt]}{[Base]}\right) \]
Where:
- \([Salt]\) is the concentration of the salt (in this case, \((NH_4)_2SO_4\)).
- \([Base]\) is the concentration of the weak base (\(NH_4OH\)).

### Step 3: Calculate \(pK_b\)
To find \(pK_b\), we use the formula:
\[ pK_b = -\log(K_b) \]
Given \(K_b = 10^{-5}\):
\[ pK_b = -\log(10^{-5}) = 5 \]

### Step 4: Determine concentrations
Since we have 1 mole of each component in 1 liter of solution, the concentrations are:
- \([Salt] = 1 \, \text{M}\) (from \((NH_4)_2SO_4\))
- \([Base] = 1 \, \text{M}\) (from \(NH_4OH\))

### Step 5: Substitute values into the buffer equation
Now we can substitute the values into the buffer equation:
\[ pOH = 5 + \log\left(\frac{1}{1}\right) = 5 + \log(1) = 5 + 0 = 5 \]

### Step 6: Calculate pH from pOH
To find the pH, we use the relationship:
\[ pH + pOH = 14 \]
Thus,
\[ pH = 14 - pOH = 14 - 5 = 9 \]

### Final Answer
The pH of the buffer solution is **9**.

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