In an acid base titration 50ml of 2N-NaOH solution is completely neutralised by 20ml of `H_(2)SO_(4)` . What is the molarity of sulphuric acid solution?

Options

A. 2M
B. 0.2M
C. 2.5M
D. 0.25M

Answer

To find the molarity of the sulfuric acid solution in the given acid-base titration problem, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Volume of NaOH solution (V1) = 50 mL
- Normality of NaOH solution (N1) = 2 N
- Volume of H2SO4 solution (V2) = 20 mL
- Normality of H2SO4 solution (N2) = ? (this is what we need to find)

2. **Use the Neutralization Equation:**
The neutralization reaction can be represented by the equation:
\[ N_1 \times V_1 = N_2 \times V_2 \]
Where:
- \(N_1\) = Normality of NaOH
- \(V_1\) = Volume of NaOH
- \(N_2\) = Normality of H2SO4
- \(V_2\) = Volume of H2SO4

3. **Substitute the Known Values:**
\[ 2 \, \text{N} \times 50 \, \text{mL} = N_2 \times 20 \, \text{mL} \]

4. **Calculate Normality of H2SO4 (N2):**
\[ 100 = N_2 \times 20 \]
\[ N_2 = \frac{100}{20} = 5 \, \text{N} \]

5. **Determine the n-factor for H2SO4:**
- H2SO4 dissociates into 2 H⁺ ions and 1 SO4²⁻ ion.
- Therefore, the n-factor for H2SO4 = 2.

6. **Relate Normality to Molarity:**
The relationship between normality (N) and molarity (M) is given by:
\[ N = M \times n \]
Where n is the n-factor.

7. **Calculate Molarity (M):**
\[ 5 = M \times 2 \]
\[ M = \frac{5}{2} = 2.5 \, \text{M} \]

### Final Answer:
The molarity of the sulfuric acid solution is **2.5 M**.

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