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 (H₂SO₄) solution, we can use the relationship between normality and molarity, along with the concept of neutralization in titration. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
- 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) = ?

### Step 2: Use the Neutralization Equation
In a neutralization reaction, the number of equivalents of the acid equals the number of equivalents of the base. This can be expressed as:
\[ N₁ \times V₁ = N₂ \times V₂ \]

### Step 3: Substitute the Known Values
Substituting the known values into the equation:
\[ 2 \, \text{N} \times 50 \, \text{mL} = N₂ \times 20 \, \text{mL} \]

### Step 4: Calculate N₂ (Normality of H₂SO₄)
Now, we can solve for N₂:
\[ 100 = N₂ \times 20 \]
\[ N₂ = \frac{100}{20} = 5 \, \text{N} \]

### Step 5: Calculate the Molarity of H₂SO₄
To find the molarity (M) of H₂SO₄, we use the relationship between normality and molarity:
\[ \text{Molarity} = \frac{\text{Normality}}{n_f} \]
Where \( n_f \) is the number of hydrogen ions contributed by H₂SO₄. Since H₂SO₄ can donate 2 protons (H⁺), \( n_f = 2 \).

Substituting the values:
\[ \text{Molarity} = \frac{5 \, \text{N}}{2} = 2.5 \, \text{M} \]

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

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