`100ml` of `0.2 M H_(2)SO_(4)` is added to `100 ml` of `0.2 M NaOH`. The resulting solution will be

Options

A. Acidic
B. Basic
C. Neutral
D. Slightly basic

Explanation

To solve the problem of determining the resulting solution when `100 ml` of `0.2 M H₂SO₄` is mixed with `100 ml` of `0.2 M NaOH`, we will follow these steps:

### Step 1: Calculate the milliequivalents of H₂SO₄
- H₂SO₄ is a diprotic acid, meaning it can donate 2 protons (H⁺ ions).
- The formula to calculate milliequivalents (meq) is:
\[ \text{meq} = \text{Molarity (M)} \times \text{Volume (L)} \times n \]
where \( n \) is the number of protons the acid can donate.

- For H₂SO₄:
\[ \text{meq of H₂SO₄} = 0.2 \, \text{M} \times 0.1 \, \text{L} \times 2 = 0.04 \, \text{equivalents} = 40 \, \text{meq} \]

### Step 2: Calculate the milliequivalents of NaOH
- NaOH is a strong base that can donate 1 hydroxide ion (OH⁻).
- Using the same formula:
\[ \text{meq of NaOH} = 0.2 \, \text{M} \times 0.1 \, \text{L} \times 1 = 0.02 \, \text{equivalents} = 20 \, \text{meq} \]

### Step 3: Determine the excess H⁺ ions
- The total milliequivalents of H⁺ from H₂SO₄ is 40 meq, and the total milliequivalents of OH⁻ from NaOH is 20 meq.
- The remaining H⁺ ions after neutralization can be calculated as:
\[ \text{Excess H⁺} = \text{meq of H₂SO₄} - \text{meq of NaOH} = 40 \, \text{meq} - 20 \, \text{meq} = 20 \, \text{meq} \]

### Step 4: Calculate the total volume of the solution
- The total volume after mixing is:
\[ \text{Total Volume} = 100 \, \text{ml} + 100 \, \text{ml} = 200 \, \text{ml} = 0.2 \, \text{L} \]

### Step 5: Calculate the concentration of H⁺ ions in the resulting solution
- The concentration of H⁺ ions is given by:
\[ \text{Concentration of H⁺} = \frac{\text{Excess H⁺}}{\text{Total Volume}} = \frac{20 \, \text{meq}}{200 \, \text{ml}} = \frac{20}{200} = 0.1 \, \text{M} \]

### Step 6: Calculate the pH of the solution
- The pH can be calculated using the formula:
\[ \text{pH} = -\log[\text{H⁺}] \]
where \([\text{H⁺}] = 0.1 \, \text{M}\):
\[ \text{pH} = -\log(0.1) = 1 \]

### Conclusion
Since the pH of the resulting solution is 1, which is less than 7, the solution is acidic.

### Final Answer
The resulting solution will be **Acidic**.

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