`pH` of solution made by mixing `50mL` of `0.2M NH_(4)CI` and `75mL` of `0.1M NaOH` is `[pK_(b) of NH_(3)(aq) = 4.74. log 3 = 0.47]`

Options

A. `7.02`
B. `13.0`
C. `7.02`
D. `9.73`

Explanation

To find the pH of the solution made by mixing 50 mL of 0.2 M NH₄Cl and 75 mL of 0.1 M NaOH, we can follow these steps:

### Step 1: Calculate the moles of NH₄Cl and NaOH
1. **Moles of NH₄Cl**:
\[ \text{Moles of NH₄Cl} = \text{Volume (L)} \times \text{Concentration (M)} = 0.050 \, \text{L} \times 0.2 \, \text{M} = 0.01 \, \text{mol} \]

2. **Moles of NaOH**:
\[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.075 \, \text{L} \times 0.1 \, \text{M} = 0.0075 \, \text{mol} \]

### Step 2: Determine the reaction between NH₄Cl and NaOH
The reaction can be represented as:
\[ \text{NH₄Cl} + \text{NaOH} \rightarrow \text{NH₄OH} + \text{NaCl} \]
In this reaction, NH₄Cl (a weak acid) reacts with NaOH (a strong base).

### Step 3: Calculate the remaining moles after the reaction
- **Moles of NH₄Cl remaining**:
\[ \text{Remaining NH₄Cl} = 0.01 \, \text{mol} - 0.0075 \, \text{mol} = 0.0025 \, \text{mol} \]

- **Moles of NaOH remaining**:
\[ \text{Remaining NaOH} = 0.0075 \, \text{mol} - 0.0075 \, \text{mol} = 0 \, \text{mol} \]

### Step 4: Calculate the concentration of NH₄⁺ and OH⁻
The total volume of the solution after mixing is:
\[ \text{Total Volume} = 50 \, \text{mL} + 75 \, \text{mL} = 125 \, \text{mL} = 0.125 \, \text{L} \]

- **Concentration of NH₄⁺**:
\[ [\text{NH₄}^+] = \frac{0.0025 \, \text{mol}}{0.125 \, \text{L}} = 0.02 \, \text{M} \]

### Step 5: Use the Henderson-Hasselbalch equation
Since NH₄⁺ is a weak acid, we can use the Henderson-Hasselbalch equation to find the pOH:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
Where:
- \(\text{pK}_b = 4.74\)
- \([\text{Salt}] = [\text{NH₄}^+] = 0.02 \, \text{M}\)
- \([\text{Base}] = 0 \, \text{M}\) (since all NaOH has reacted)

However, since we have a weak base (NH₄OH), we can also use the relationship:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{NH₄}^+]}{[\text{OH}^-]}\right) \]
Here, we need to find the concentration of OH⁻ produced from the dissociation of NH₄OH.

### Step 6: Calculate pOH and then pH
Using the values:
\[ \text{pOH} = 4.74 + \log\left(\frac{0.0025}{0.0075}\right) \]
Calculating the log:
\[ \log\left(\frac{0.0025}{0.0075}\right) = \log\left(\frac{1}{3}\right) \approx -0.47 \]
Thus:
\[ \text{pOH} = 4.74 - 0.47 = 4.27 \]

Finally, we find pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 4.27 = 9.73 \]

### Final Answer
The pH of the solution is **9.73**.

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