How may millilitres of a `9 N H_(2)SO_(4)` solution will be required to neutralize completely `20mL` of a `3.6 N NaOH` solution ?

Options

A. `18.0mL`
B. `8. 0 Ml`
C. `16.0mL`
D. `80.0 mL`

Explanation

To solve the problem of how many milliliters of a `9 N H₂SO₄` solution are required to neutralize `20 mL` of a `3.6 N NaOH` solution, we can use the formula based on the concept of normality:

### Step-by-Step Solution:

1. **Identify the given values:**
- Normality of H₂SO₄ (N₁) = 9 N
- Volume of NaOH (V₂) = 20 mL
- Normality of NaOH (N₂) = 3.6 N
- Volume of H₂SO₄ (V₁) = ? (this is what we need to find)

2. **Use the neutralization formula:**
The formula to use is:
\[ N₁V₁ = N₂V₂ \]
Here, \(N₁\) and \(V₁\) correspond to the acid (H₂SO₄) and \(N₂\) and \(V₂\) correspond to the base (NaOH).

3. **Substitute the known values into the formula:**
\[ 9 \, V₁ = 3.6 \times 20 \]

4. **Calculate the right side of the equation:**
\[ 3.6 \times 20 = 72 \]
So now we have:
\[ 9 \, V₁ = 72 \]

5. **Solve for \(V₁\):**
To find \(V₁\), divide both sides by 9:
\[ V₁ = \frac{72}{9} = 8 \, \text{mL} \]

### Conclusion:
The volume of `9 N H₂SO₄` solution required to neutralize `20 mL` of `3.6 N NaOH` solution is **8 mL**.

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