Which one of the following solutions of sulphuric acid will exactly neutralise 25 ml of 0.2 M sodium hydroxide solution ?

Options

A. 12.5ml of 0.1M solution
B. 25ml of 0.1 M solution
C. 25ml of 0.2 M solution
D. 50 ml of 0.2 M solution

Answer

To determine which solution of sulfuric acid will exactly neutralize 25 ml of 0.2 M sodium hydroxide (NaOH) solution, we can follow these steps:

### Step 1: Calculate the milli equivalents of NaOH
We know that:
- Molarity (M) of NaOH = 0.2 M
- Volume (V) of NaOH = 25 ml
- N factor for NaOH = 1 (since it releases one hydroxide ion, OH⁻)

Using the formula for milli equivalents:
\[ \text{Milli equivalents} = \text{Molarity} \times \text{N factor} \times \text{Volume (ml)} \]
Substituting the values:
\[ \text{Milli equivalents of NaOH} = 0.2 \times 1 \times 25 = 5 \text{ milli equivalents} \]

### Step 2: Determine the required milli equivalents of sulfuric acid (H₂SO₄)
Sulfuric acid (H₂SO₄) has a basicity of 2, meaning it can release 2 H⁺ ions. Therefore, the N factor for H₂SO₄ is 2.

To neutralize 5 milli equivalents of NaOH, we need an equal amount of H⁺ ions, which means we need 5 milli equivalents of H₂SO₄.

### Step 3: Calculate the volume of sulfuric acid required to provide 5 milli equivalents
Using the formula again for H₂SO₄:
\[ \text{Milli equivalents} = \text{Molarity} \times \text{N factor} \times \text{Volume (ml)} \]
Let’s denote the molarity of the sulfuric acid solution as \( C \) and the volume we need to find as \( V \).

Setting up the equation for H₂SO₄:
\[ 5 = C \times 2 \times V \]
Rearranging gives:
\[ V = \frac{5}{C \times 2} \]

### Step 4: Evaluate the options
Now, we need to evaluate the given options for sulfuric acid solutions to find which one gives us exactly 5 milli equivalents when calculated using the above formula.

1. If the molarity of H₂SO₄ is 0.1 M:
\[ V = \frac{5}{0.1 \times 2} = \frac{5}{0.2} = 25 \text{ ml} \]

2. If the molarity of H₂SO₄ is 0.2 M:
\[ V = \frac{5}{0.2 \times 2} = \frac{5}{0.4} = 12.5 \text{ ml} \]

3. If the molarity of H₂SO₄ is 0.5 M:
\[ V = \frac{5}{0.5 \times 2} = \frac{5}{1} = 5 \text{ ml} \]

4. If the molarity of H₂SO₄ is 1 M:
\[ V = \frac{5}{1 \times 2} = \frac{5}{2} = 2.5 \text{ ml} \]

### Conclusion
From the calculations, we find that the solution of sulfuric acid with a molarity of 0.2 M requires 12.5 ml to exactly neutralize 25 ml of 0.2 M NaOH solution.