A solution contains `10mL` of `0.1N NaOH` and `10mL` of `0.05Na_(2)SO_(4), pH` of this solution is

Options

A. `7`
B. Less than `7`
C. Greater than `7`
D. Zero

Explanation

To find the pH of a solution containing 10 mL of 0.1 N NaOH and 10 mL of 0.05 N Na₂SO₄, we can follow these steps:

### Step 1: Understand the Components
- **NaOH** is a strong base, which will affect the pH of the solution.
- **Na₂SO₄** is a salt derived from a strong acid (H₂SO₄) and a strong base (NaOH). Salts from strong acids and bases are neutral and do not affect the pH significantly.

### Step 2: Calculate the Total Volume of the Solution
- The total volume of the solution after mixing the two components is:
\[ V_{\text{total}} = 10 \, \text{mL} + 10 \, \text{mL} = 20 \, \text{mL} \]

### Step 3: Calculate the Molarity of NaOH in the New Volume
Using the dilution formula \( M_1V_1 = M_2V_2 \):
- \( M_1 = 0.1 \, \text{N} \) (initial concentration of NaOH)
- \( V_1 = 10 \, \text{mL} \) (initial volume of NaOH)
- \( V_2 = 20 \, \text{mL} \) (final volume after mixing)

We need to find \( M_2 \):
\[ M_2 = \frac{M_1 \cdot V_1}{V_2} = \frac{0.1 \, \text{N} \cdot 10 \, \text{mL}}{20 \, \text{mL}} = 0.05 \, \text{N} \]

### Step 4: Determine the Concentration of OH⁻ Ions
Since NaOH dissociates completely in solution:
\[ [\text{OH}^-] = 0.05 \, \text{N} \]

### Step 5: Calculate the pOH
Using the formula for pOH:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log(0.05) \]
Calculating this gives:
\[ \text{pOH} \approx 1.30 \]

### Step 6: Calculate the pH
Using the relationship between pH and pOH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 1.30 = 12.70 \]

### Conclusion
The pH of the solution is **12.70**, indicating that it is basic.

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