`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 will follow these steps:

### Step 1: Calculate the millimoles of NH₄Cl and NaOH
- **For NH₄Cl**:
\[ \text{Millimoles of NH₄Cl} = \text{Volume (mL)} \times \text{Molarity (M)} = 50 \, \text{mL} \times 0.2 \, \text{M} = 10 \, \text{mmol} \]

- **For NaOH**:
\[ \text{Millimoles of NaOH} = \text{Volume (mL)} \times \text{Molarity (M)} = 75 \, \text{mL} \times 0.1 \, \text{M} = 7.5 \, \text{mmol} \]

### Step 2: Determine the limiting reagent
- Since both NH₄Cl and NaOH react in a 1:1 ratio, we see that NaOH (7.5 mmol) is the limiting reagent.

### Step 3: Calculate the remaining millimoles after the reaction
- **NH₄Cl remaining**:
\[ \text{Remaining NH₄Cl} = 10 \, \text{mmol} - 7.5 \, \text{mmol} = 2.5 \, \text{mmol} \]

- **NaOH remaining**:
\[ \text{Remaining NaOH} = 7.5 \, \text{mmol} - 7.5 \, \text{mmol} = 0 \, \text{mmol} \]

### Step 4: Identify the components in the solution
- After the reaction, we have 2.5 mmol of NH₄Cl remaining, which will dissociate to form NH₄⁺ ions. The NaCl formed does not affect the pH significantly as it is a neutral salt.

### Step 5: Use the Henderson-Hasselbalch equation
- The solution behaves as a basic buffer because we have a weak base (NH₄OH) and its salt (NH₄Cl).
- The pKₐ of NH₄⁺ can be calculated from the given pK₍b₎ of NH₃:
\[ pK_a + pK_b = 14 \implies pK_a = 14 - 4.74 = 9.26 \]

### Step 6: Calculate the pOH using the Henderson-Hasselbalch equation
- The formula for pOH in a basic buffer is:
\[ pOH = pK_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
- Here, we consider NH₄Cl as the salt and NH₄OH as the base. The concentrations are:
- Concentration of salt (NH₄Cl) = \(\frac{2.5 \, \text{mmol}}{(50 + 75) \, \text{mL}} = \frac{2.5}{125} = 0.02 \, \text{M}\)
- Concentration of base (NH₄OH) = \(\frac{7.5 \, \text{mmol}}{(50 + 75) \, \text{mL}} = \frac{7.5}{125} = 0.06 \, \text{M}\)

- Now substituting into the Henderson-Hasselbalch equation:
\[ pOH = 4.74 + \log\left(\frac{2.5}{7.5}\right) = 4.74 + \log\left(\frac{1}{3}\right) = 4.74 - 0.47 = 4.27 \]

### Step 7: Calculate the pH
- Finally, we can find the pH:
\[ pH = 14 - pOH = 14 - 4.27 = 9.73 \]

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

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