`10 ml` of `M/10 NH_(4)OH` is mixed with`4ml` of `M/10H_(2)SO_(4)` solution. The `pH` of the resulting solution is `(pK_(b)NH_(4)OH=4.76), (log2=0.3)`.

Options

A. 5.36
B. 8.64
C. 9.26
D. 4.74

Explanation

To solve the problem, we need to find the pH of a solution formed by mixing 10 ml of M/10 NH₄OH with 4 ml of M/10 H₂SO₄. We will follow these steps:

### Step 1: Calculate the moles of NH₄OH and H₂SO₄

1. **Moles of NH₄OH**:
\[ \text{Moles of NH₄OH} = \text{Volume (L)} \times \text{Molarity (M)} = 0.010 \, \text{L} \times \frac{1}{10} \, \text{M} = 0.001 \, \text{mol} \]

2. **Moles of H₂SO₄**:
\[ \text{Moles of H₂SO₄} = \text{Volume (L)} \times \text{Molarity (M)} = 0.004 \, \text{L} \times \frac{1}{10} \, \text{M} = 0.0004 \, \text{mol} \]

### Step 2: Determine the neutralization reaction

The reaction between NH₄OH and H₂SO₄ can be represented as:
\[ 2 \, \text{NH₄OH} + \text{H₂SO₄} \rightarrow \text{(NH₄)₂SO₄} + 2 \, \text{H₂O} \]

From the stoichiometry of the reaction, 2 moles of NH₄OH react with 1 mole of H₂SO₄.

### Step 3: Calculate the limiting reactant

- **Moles of NH₄OH required for 0.0004 moles of H₂SO₄**:
\[ \text{Moles of NH₄OH required} = 2 \times 0.0004 = 0.0008 \, \text{mol} \]

Since we have 0.001 mol of NH₄OH available, NH₄OH is in excess, and H₂SO₄ is the limiting reactant.

### Step 4: Calculate the remaining moles after the reaction

- **Remaining moles of NH₄OH**:
\[ \text{Remaining NH₄OH} = 0.001 - 0.0008 = 0.0002 \, \text{mol} \]

- **Moles of (NH₄)₂SO₄ formed**:
\[ \text{Moles of (NH₄)₂SO₄} = 0.0004 \, \text{mol} \]

### Step 5: Calculate the total volume of the solution

Total volume = 10 ml + 4 ml = 14 ml = 0.014 L

### Step 6: Calculate the concentrations of NH₄⁺ and NH₄OH

1. **Concentration of NH₄⁺** (from (NH₄)₂SO₄):
\[ [\text{NH₄}^+] = \frac{0.0004 \, \text{mol}}{0.014 \, \text{L}} = 0.02857 \, \text{M} \]

2. **Concentration of NH₄OH**:
\[ [\text{NH₄OH}] = \frac{0.0002 \, \text{mol}}{0.014 \, \text{L}} = 0.01429 \, \text{M} \]

### Step 7: Use the Henderson-Hasselbalch equation to find pH

The Henderson-Hasselbalch equation for a weak base and its conjugate acid is:
\[ \text{pH} = \text{pK}_b + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]

Given:
- \( \text{pK}_b = 4.76 \)
- \( [\text{Base}] = [\text{NH₄OH}] = 0.01429 \, \text{M} \)
- \( [\text{Acid}] = [\text{NH₄}^+] = 0.02857 \, \text{M} \)

Substituting these values:
\[ \text{pH} = 4.76 + \log\left(\frac{0.01429}{0.02857}\right) \]

Calculating the log term:
\[ \log\left(\frac{0.01429}{0.02857}\right) = \log(0.5) = -0.301 \]

Thus:
\[ \text{pH} = 4.76 - 0.301 = 4.459 \]

### Step 8: Calculate pOH and final pH

Since \( \text{pH} + \text{pOH} = 14 \):
\[ \text{pOH} = 14 - 4.459 = 9.541 \]

### Final Answer

The pH of the resulting solution is approximately **9.54**.

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