What is the nature of the resulting solution prepared from 100ml to `0.2M` `H_2SO_4` added to 100ml of `0.2 M` `NaOH`.

Options

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

Answer

To determine the nature of the resulting solution when 100 mL of 0.2 M H₂SO₄ is added to 100 mL of 0.2 M NaOH, we can follow these steps:

### Step 1: Calculate the milliequivalents of H₂SO₄
- **Formula**: Milliequivalents (mEq) = n-factor × Volume (mL) × Molarity (M)
- For H₂SO₄, the n-factor is 2 (since it can donate 2 protons).
- Volume = 100 mL, Molarity = 0.2 M

\[ \text{mEq of H₂SO₄} = 2 \times 100 \, \text{mL} \times 0.2 \, \text{M} = 40 \, \text{mEq} \]

### Step 2: Convert milliequivalents to equivalents
- To convert mEq to equivalents, divide by 1000.

\[ \text{Equivalents of H₂SO₄} = \frac{40 \, \text{mEq}}{1000} = 0.04 \, \text{Eq} \]

### Step 3: Calculate the milliequivalents of NaOH
- For NaOH, the n-factor is 1 (since it can donate 1 proton).
- Volume = 100 mL, Molarity = 0.2 M

\[ \text{mEq of NaOH} = 1 \times 100 \, \text{mL} \times 0.2 \, \text{M} = 20 \, \text{mEq} \]

### Step 4: Convert milliequivalents to equivalents for NaOH
\[ \text{Equivalents of NaOH} = \frac{20 \, \text{mEq}}{1000} = 0.02 \, \text{Eq} \]

### Step 5: Determine the remaining H⁺ concentration
- The total H⁺ from H₂SO₄ is 0.04 Eq, and the total OH⁻ from NaOH is 0.02 Eq.
- The remaining H⁺ after neutralization:

\[ \text{Remaining H⁺} = 0.04 \, \text{Eq} - 0.02 \, \text{Eq} = 0.02 \, \text{Eq} \]

### Step 6: Calculate the total volume of the solution
- Total volume = 100 mL + 100 mL = 200 mL

### Step 7: Calculate the concentration of H⁺ ions
\[ \text{Concentration of H⁺} = \frac{0.02 \, \text{Eq}}{0.2 \, \text{L}} = 0.1 \, \text{M} \]

### Step 8: Calculate the pH of the solution
- pH is calculated using the formula:

\[ \text{pH} = -\log[\text{H⁺}] \]

\[ \text{pH} = -\log(0.1) = 1 \]

### Step 9: Determine the nature of the solution
- Since the pH is 1, which is less than 7, the solution is acidic.

### Final Answer:
The resulting solution is **acidic**.

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