A 100 mL 0.1m solution of ammonium nitrate is diluted by adding 100 ml of water. The pH of the resulting solution will be (`pK_a` of acetic acid is nearly equal of `NH_4OH` )

Options

A. 4.9
B. `5.0`
C. `7.0`
D. `10.0`

Answer

To find the pH of the resulting solution after diluting a 100 mL 0.1 m solution of ammonium nitrate by adding 100 mL of water, we can follow these steps:

### Step 1: Understand the Nature of Ammonium Nitrate
Ammonium nitrate (NH4NO3) is a salt formed from a weak base (NH4OH) and a strong acid (HNO3). When dissolved in water, it dissociates into NH4+ and NO3- ions. The NH4+ ion can react with water to produce NH4OH and H+ ions, affecting the pH of the solution.

### Step 2: Calculate the New Concentration After Dilution
When you dilute a solution, the concentration of the solute decreases. The initial concentration of ammonium nitrate is 0.1 m in 100 mL. When you add 100 mL of water, the total volume becomes 200 mL.

\[ \text{New Concentration} = \frac{\text{Initial Concentration} \times \text{Initial Volume}}{\text{Final Volume}} = \frac{0.1 \, \text{m} \times 100 \, \text{mL}}{200 \, \text{mL}} = 0.05 \, \text{m} \]

### Step 3: Determine the pKa and pKb
Since the problem states that the pKa of acetic acid is nearly equal to the pKb of ammonium hydroxide, we can denote:

\[ pK_a = pK_b \]

Let’s denote this common value as \( pK \).

### Step 4: Use the Formula for pH of a Salt Solution
For a salt formed from a weak base and a strong acid, the pH can be calculated using the following formula:

\[ pH = 7 + \frac{1}{2} (pK_a - pK_b) \]

Since \( pK_a = pK_b \), we can simplify this to:

\[ pH = 7 + \frac{1}{2} (pK - pK) = 7 \]

### Step 5: Conclusion
Thus, the pH of the diluted solution of ammonium nitrate is:

\[ \text{pH} = 7 \]