When `100 ml` of `1 M NaOH` solution is mixed with `10 ml` of `10 M H_(2)SO_(4)`, the resulting mixture will be

Options

A. Acidic
B. Alkaline
C. Neutral
D. Strongly alkaline

Explanation

To solve the problem of mixing 100 ml of 1 M NaOH solution with 10 ml of 10 M H₂SO₄, we will follow these steps:

### Step 1: Calculate the Normality of NaOH and H₂SO₄
- **NaOH**:
- Molarity = 1 M
- x-factor (basicity) = 1 (since NaOH provides 1 OH⁻)
- Normality (N) = Molarity × x-factor = 1 M × 1 = 1 N

- **H₂SO₄**:
- Molarity = 10 M
- x-factor (basicity) = 2 (since H₂SO₄ provides 2 H⁺)
- Normality (N) = Molarity × x-factor = 10 M × 2 = 20 N

### Step 2: Calculate the Milli-equivalents of NaOH and H₂SO₄
- **NaOH**:
- Volume = 100 ml
- Milli-equivalents = Normality × Volume (in ml) = 1 N × 100 ml = 100 milli-equivalents

- **H₂SO₄**:
- Volume = 10 ml
- Milli-equivalents = Normality × Volume (in ml) = 20 N × 10 ml = 200 milli-equivalents

### Step 3: Determine the Limiting Reagent
- **NaOH** has 100 milli-equivalents.
- **H₂SO₄** has 200 milli-equivalents.
- Since NaOH has fewer milli-equivalents, it is the limiting reagent.

### Step 4: Determine the Resulting Mixture
- The reaction between NaOH and H₂SO₄ is a neutralization reaction:
\[ H₂SO₄ + 2 NaOH \rightarrow Na₂SO₄ + 2 H₂O \]
- From the equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, 200 milli-equivalents of H₂SO₄ would require 100 milli-equivalents of NaOH to fully react.
- Since we have exactly 100 milli-equivalents of NaOH, it will completely react with 100 milli-equivalents of H₂SO₄, leaving 100 milli-equivalents of H₂SO₄ unreacted.

### Conclusion
The resulting mixture will be acidic due to the excess H₂SO₄ remaining after the reaction.

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