When 100 ml of 10 M solution of `H_(2)SO_(4)` and 100 ml of 1 0 M solution of `NaOH` are mixed the resulting solution will be

Options

A. Acidic
B. Basic
C. Neutral
D. Can not predicted

Explanation

To solve the problem of mixing 100 ml of 10 M `H₂SO₄` and 100 ml of 10 M `NaOH`, we need to follow these steps:

### Step 1: Calculate the number of moles of `H₂SO₄`
- The concentration of `H₂SO₄` is 10 M (molar).
- Volume = 100 ml = 0.1 L (since 1 L = 1000 ml).
- Number of moles of `H₂SO₄` = Concentration × Volume = 10 mol/L × 0.1 L = 1 mole.

### Step 2: Calculate the number of moles of `NaOH`
- The concentration of `NaOH` is also 10 M.
- Volume = 100 ml = 0.1 L.
- Number of moles of `NaOH` = Concentration × Volume = 10 mol/L × 0.1 L = 1 mole.

### Step 3: Determine the reaction between `H₂SO₄` and `NaOH`
- The balanced chemical equation for the reaction is:
\[ H₂SO₄ + 2 NaOH \rightarrow Na₂SO₄ + 2 H₂O \]
- From the equation, we see that 1 mole of `H₂SO₄` reacts with 2 moles of `NaOH`.

### Step 4: Determine the limiting reactant
- We have 1 mole of `H₂SO₄` and 1 mole of `NaOH`.
- According to the stoichiometry of the reaction, 1 mole of `H₂SO₄` requires 2 moles of `NaOH` to completely react.
- Since we only have 1 mole of `NaOH`, it is the limiting reactant.

### Step 5: Calculate the remaining moles after the reaction
- `NaOH` will completely react with half a mole of `H₂SO₄` (since 2 moles of `NaOH` are required for 1 mole of `H₂SO₄`).
- Therefore, after the reaction:
- Moles of `H₂SO₄` remaining = 1 mole - 0.5 mole = 0.5 mole.
- Moles of `NaOH` remaining = 0 moles (it is completely consumed).

### Step 6: Determine the nature of the resulting solution
- Since there is excess `H₂SO₄` (0.5 mole remaining) and no `NaOH` left, the resulting solution will be acidic.

### Conclusion
The resulting solution after mixing 100 ml of 10 M `H₂SO₄` and 100 ml of 10 M `NaOH` will be acidic.

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