In a basic buffer, 0.0025 mole of `NH_(4)Cl` and 0.15 mole of `NH_(4) OH` are present. The pH of the solution will be `(pK_(a)=4.74)`

Options

A. 11.04
B. 10.24
C. 6.62
D. 5.48

Explanation

To find the pH of the basic buffer solution containing 0.0025 moles of NH₄Cl and 0.15 moles of NH₄OH, we can use the Henderson-Hasselbalch equation, which is given by:

\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]

In this case, NH₄OH is the base and NH₄Cl is the acid.

### Step 1: Identify the components
- **Base (B)**: NH₄OH (ammonium hydroxide)
- **Acid (A)**: NH₄Cl (ammonium chloride)

### Step 2: Calculate the concentrations
Since we have the moles of each component, we can directly use them in the Henderson-Hasselbalch equation without converting to concentrations, assuming the volume of the solution is constant.

- Moles of Base (NH₄OH) = 0.15 moles
- Moles of Acid (NH₄Cl) = 0.0025 moles

### Step 3: Use the Henderson-Hasselbalch equation
We know that \( pK_a = 4.74 \) (given). To find \( pK_b \), we can use the relationship:

\[ pK_a + pK_b = 14 \]

Thus,

\[ pK_b = 14 - pK_a = 14 - 4.74 = 9.26 \]

### Step 4: Substitute values into the equation
Now we can substitute the values into the Henderson-Hasselbalch equation:

\[ \text{pH} = pK_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]

Substituting the values we have:

\[ \text{pH} = 4.74 + \log\left(\frac{0.15}{0.0025}\right) \]

### Step 5: Calculate the ratio
Calculate the ratio:

\[ \frac{0.15}{0.0025} = 60 \]

### Step 6: Calculate the logarithm
Now calculate the logarithm:

\[ \log(60) \approx 1.78 \]

### Step 7: Final calculation of pH
Now substitute this back into the equation:

\[ \text{pH} = 4.74 + 1.78 = 6.52 \]

Thus, the pH of the solution is approximately **6.52**.

### Summary of Steps:
1. Identify the components (Base and Acid).
2. Calculate the ratio of moles of Base to Acid.
3. Use the Henderson-Hasselbalch equation.
4. Calculate the logarithm of the ratio.
5. Add to \( pK_a \) to find the pH.