The pH of the solution containing 10 mL of 0.1 N NaOH and 10 mL of 0.05 N `H_(2)SO_(4)` would be

Options

A. 1
B. `0`
C. 7
D. `gt7`

Explanation

To find the pH of the solution containing 10 mL of 0.1 N NaOH and 10 mL of 0.05 N H₂SO₄, we can follow these steps:

### Step 1: Calculate the milliequivalents of NaOH
- **Formula**: Milliequivalents (mEq) = Normality (N) × Volume (mL)
- For NaOH:
\[ \text{Milliequivalents of NaOH} = 0.1 \, \text{N} \times 10 \, \text{mL} = 1 \, \text{mEq} \]

### Step 2: Calculate the milliequivalents of H₂SO₄
- For H₂SO₄:
\[ \text{Milliequivalents of H₂SO₄} = 0.05 \, \text{N} \times 10 \, \text{mL} = 0.5 \, \text{mEq} \]

### Step 3: Determine the neutralization reaction
- The reaction between NaOH and H₂SO₄ can be represented as:
\[ \text{NaOH} + \text{H₂SO₄} \rightarrow \text{Na₂SO₄} + \text{H₂O} \]
- Each mole of H₂SO₄ can neutralize two moles of NaOH, but since we are dealing with milliequivalents, we consider the equivalents directly.

### Step 4: Compare the milliequivalents
- We have:
- Milliequivalents of NaOH = 1 mEq
- Milliequivalents of H₂SO₄ = 0.5 mEq
- Since 1 mEq (NaOH) > 0.5 mEq (H₂SO₄), NaOH is in excess.

### Step 5: Calculate the remaining milliequivalents of NaOH after neutralization
- Remaining milliequivalents of NaOH after neutralization:
\[ \text{Remaining NaOH} = 1 \, \text{mEq} - 0.5 \, \text{mEq} = 0.5 \, \text{mEq} \]

### Step 6: Determine the pH of the solution
- Since there is excess NaOH, the solution will be basic. The pH of a solution with excess NaOH can be calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
- To find the concentration of OH⁻ ions, we need to convert milliequivalents to molarity:
- Total volume of the solution = 10 mL (NaOH) + 10 mL (H₂SO₄) = 20 mL = 0.020 L
- Concentration of OH⁻ ions:
\[ [\text{OH}^-] = \frac{0.5 \, \text{mEq}}{0.020 \, \text{L}} = 25 \, \text{mEq/L} = 0.025 \, \text{N} \]
- Now, calculate pOH:
\[ \text{pOH} = -\log(0.025) \approx 1.60 \]
- Finally, calculate pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 1.60 = 12.40 \]

### Conclusion
The pH of the solution containing 10 mL of 0.1 N NaOH and 10 mL of 0.05 N H₂SO₄ is approximately **12.40**, which is greater than 7.