15 ml of NaOH solution gets complete neutralised with 10 ml of HCI solution. What volume of the same HCI soluction will be required to neutralise 30 ml of same NaOH solution ?

Options

A. 5 ml
B. 10 ml
C. 15 ml
D. 20 ml

Explanation

To solve the problem step by step, we will use the concept of neutralization reactions and the relationship between the volumes and concentrations of the acid and base involved.

### Step-by-Step Solution:

1. **Understanding the Neutralization Reaction**:
- The neutralization reaction between NaOH (a base) and HCl (an acid) can be represented as:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- From the reaction, we can see that 1 mole of NaOH reacts with 1 mole of HCl.

2. **Using Given Data**:
- We know that 15 ml of NaOH solution is completely neutralized by 10 ml of HCl solution.
- This means that the ratio of NaOH to HCl is:
\[ \frac{\text{Volume of NaOH}}{\text{Volume of HCl}} = \frac{15 \text{ ml}}{10 \text{ ml}} = \frac{3}{2} \]

3. **Finding the Volume of HCl for 30 ml of NaOH**:
- We need to find out how much HCl is required to neutralize 30 ml of NaOH.
- Since the ratio of NaOH to HCl remains constant, we can set up a proportion:
\[ \frac{15 \text{ ml NaOH}}{10 \text{ ml HCl}} = \frac{30 \text{ ml NaOH}}{V \text{ ml HCl}} \]
- Here, \( V \) is the volume of HCl we need to find.

4. **Cross-Multiplying to Solve for V**:
- Cross-multiplying gives us:
\[ 15 \times V = 30 \times 10 \]
- Simplifying this:
\[ 15V = 300 \]
- Now, divide both sides by 15:
\[ V = \frac{300}{15} = 20 \text{ ml} \]

5. **Conclusion**:
- Therefore, the volume of HCl required to neutralize 30 ml of NaOH is **20 ml**.

### Final Answer:
The volume of HCl solution required to neutralize 30 ml of NaOH solution is **20 ml**.