pH of when 50mL of 0.10 M ammonia solution is treated with 50 mL of 0.05 M HCI solution :-
`(pK_(b) "of ammonia"=4.74)`

Options

A. 8. 26
B. 9. 26
C. 4 .74
D. None

Explanation

To find the pH of the solution when 50 mL of 0.10 M ammonia solution is mixed with 50 mL of 0.05 M HCl solution, we can follow these steps:

### Step 1: Calculate moles of ammonia and HCl

1. **Calculate moles of ammonia (NH₃)**:
\[ \text{Moles of NH}_3 = \text{Concentration} \times \text{Volume} = 0.10 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{mol} \]

2. **Calculate moles of HCl**:
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.05 \, \text{M} \times 0.050 \, \text{L} = 0.0025 \, \text{mol} \]

### Step 2: Determine the reaction between NH₃ and HCl

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

### Step 3: Calculate the remaining moles after the reaction

1. **Moles of NH₃ after reaction**:
\[ \text{Remaining moles of NH}_3 = 0.005 \, \text{mol} - 0.0025 \, \text{mol} = 0.0025 \, \text{mol} \]

2. **Moles of NH₄⁺ produced**:
\[ \text{Moles of NH}_4^+ = 0.0025 \, \text{mol} \quad (\text{equal to moles of HCl reacted}) \]

### Step 4: Calculate the total volume of the solution

The total volume after mixing is:
\[ \text{Total volume} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]

### Step 5: Calculate the concentrations of NH₃ and NH₄⁺

1. **Concentration of NH₃**:
\[ [\text{NH}_3] = \frac{0.0025 \, \text{mol}}{0.1 \, \text{L}} = 0.025 \, \text{M} \]

2. **Concentration of NH₄⁺**:
\[ [\text{NH}_4^+] = \frac{0.0025 \, \text{mol}}{0.1 \, \text{L}} = 0.025 \, \text{M} \]

### Step 6: Use the Henderson-Hasselbalch equation to find pH

The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]

First, we need to find \( \text{pK}_a \) from \( \text{pK}_b \):
\[ \text{pK}_a + \text{pK}_b = 14 \implies \text{pK}_a = 14 - 4.74 = 9.26 \]

Now, substituting the values into the Henderson-Hasselbalch equation:
\[ \text{pH} = 9.26 + \log\left(\frac{0.025}{0.025}\right) = 9.26 + \log(1) = 9.26 + 0 = 9.26 \]

### Final Answer
The pH of the solution is **9.26**.

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