50 mL of 0.2M ammonia solution is treated with 25 mL of 0.2 M HCl. If `pK_(b)` of ammonia solution is 4.75 the pH of the mixture will be :

Options

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

Explanation

To solve the problem, we need to determine the pH of the mixture formed when 50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. Here’s a step-by-step solution:

### Step 1: Calculate the number of moles of ammonia (NH₃) and hydrochloric acid (HCl)

1. **For Ammonia (NH₃)**:
- Volume = 50 mL = 0.050 L
- Concentration = 0.2 M
- Moles of NH₃ = Concentration × Volume = 0.2 mol/L × 0.050 L = 0.01 moles

2. **For Hydrochloric Acid (HCl)**:
- Volume = 25 mL = 0.025 L
- Concentration = 0.2 M
- Moles of HCl = Concentration × Volume = 0.2 mol/L × 0.025 L = 0.005 moles

### Step 2: Determine the limiting reactant and the remaining moles after the reaction

- The reaction between NH₃ and HCl can be represented as:
\[ \text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4^+ + \text{Cl}^- \]

- From the calculations:
- Moles of NH₃ = 0.01 moles
- Moles of HCl = 0.005 moles

- Since HCl is the limiting reactant, it will completely react with NH₃.

- After the reaction:
- Remaining moles of NH₃ = 0.01 - 0.005 = 0.005 moles
- Moles of NH₄⁺ formed = 0.005 moles (equal to moles of HCl reacted)

### Step 3: Calculate the concentrations of NH₃ and NH₄⁺ in the final solution

- Total volume of the mixture = 50 mL + 25 mL = 75 mL = 0.075 L

1. **Concentration of NH₃**:
\[ \text{Concentration of NH}_3 = \frac{0.005 \text{ moles}}{0.075 \text{ L}} = \frac{0.005}{0.075} = 0.0667 \text{ M} \]

2. **Concentration of NH₄⁺**:
\[ \text{Concentration of NH}_4^+ = \frac{0.005 \text{ moles}}{0.075 \text{ L}} = \frac{0.005}{0.075} = 0.0667 \text{ M} \]

### Step 4: Use the Henderson-Hasselbalch equation to find the pH

- The Henderson-Hasselbalch equation for a basic buffer is:
\[ \text{pH} = \text{pK}_b + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]

- Given that \( pK_b = 4.75 \) and both concentrations are equal:
\[ \text{pH} = 14 - \left( pK_b + \log\left(\frac{[\text{NH}_4^+]}{[\text{NH}_3]}\right) \right) \]

- Since \([\text{NH}_4^+] = [\text{NH}_3]\):
\[ \text{pH} = 14 - (4.75 + \log(1)) = 14 - 4.75 = 9.25 \]

### Final Answer:
The pH of the mixture will be **9.25**.

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