100ml of 0.75 N `NH_(4)OH` is mixed with 100ml of 0.25 N HCl. The `K_(b)` of `NH_(4) OH = 2 xx 10^(-5)` . Hence pH of solution is .

Answer

To solve the problem step-by-step, we will follow the outlined procedure to find the pH of the solution formed when 100 mL of 0.75 N NH₄OH is mixed with 100 mL of 0.25 N HCl.

### Step 1: Calculate the moles of NH₄OH and HCl

1. **Calculate moles of NH₄OH:**
\[ \text{Normality (N)} = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} \]
\[ \text{Moles of NH₄OH} = 0.75 \, \text{N} \times 0.1 \, \text{L} = 0.075 \, \text{moles} \]

2. **Calculate moles of HCl:**
\[ \text{Moles of HCl} = 0.25 \, \text{N} \times 0.1 \, \text{L} = 0.025 \, \text{moles} \]

### Step 2: Determine the limiting reactant

Since NH₄OH is a weak base and HCl is a strong acid, we need to determine how much of each reactant will remain after the reaction.

- **Moles of NH₄OH available:** 0.075 moles
- **Moles of HCl available:** 0.025 moles

### Step 3: Calculate the moles after neutralization

The reaction between NH₄OH and HCl can be represented as:
\[ \text{NH₄OH} + \text{HCl} \rightarrow \text{NH₄Cl} + \text{H₂O} \]

- **Moles of NH₄OH after reaction:**
\[ \text{Remaining NH₄OH} = 0.075 - 0.025 = 0.050 \, \text{moles} \]

- **Moles of HCl after reaction:**
\[ \text{Remaining HCl} = 0.025 - 0.025 = 0 \, \text{moles} \]

### Step 4: Calculate the concentration of OH⁻ ions

Since we have remaining NH₄OH, we can calculate the concentration of OH⁻ ions using the base dissociation constant \( K_b \).

1. **Calculate the concentration of NH₄OH in the total volume:**
\[ \text{Total volume} = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L} \]
\[ \text{Concentration of NH₄OH} = \frac{0.050 \, \text{moles}}{0.2 \, \text{L}} = 0.25 \, \text{N} \]

2. **Use the formula for \( K_b \) to find the concentration of OH⁻:**
\[ K_b = \frac{[NH₄^+][OH^-]}{[NH₄OH]} \]
Let \( x \) be the concentration of OH⁻ produced.
\[ K_b = \frac{x^2}{0.25 - x} \approx \frac{x^2}{0.25} \quad (\text{since } x \text{ is small}) \]
\[ 2 \times 10^{-5} = \frac{x^2}{0.25} \]
\[ x^2 = 2 \times 10^{-5} \times 0.25 = 5 \times 10^{-6} \]
\[ x = \sqrt{5 \times 10^{-6}} \approx 0.00224 \, \text{N} \quad (\text{this is } [OH^-]) \]

### Step 5: Calculate the concentration of H⁺ ions

Since HCl was completely neutralized, we need to find the concentration of H⁺ ions in the solution.

1. **Using the relationship \( K_w = [H^+][OH^-] \):**
\[ K_w = 1 \times 10^{-14} \]
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1 \times 10^{-14}}{0.00224} \approx 4.46 \times 10^{-12} \, \text{N} \]

### Step 6: Calculate the pH

Finally, we can find the pH:
\[ \text{pH} = -\log[H^+] = -\log(4.46 \times 10^{-12}) \approx 11.35 \]

### Final Answer
The pH of the solution is approximately **11.35**.