The pH of 0.005 molar aqueous solution of sulphuric acid is approximately

Options

A. `0.005`
B. `2`
C. `1`
D. ` 0.01`

Answer

To find the pH of a 0.005 molar aqueous solution of sulfuric acid (H₂SO₄), we can follow these steps:

### Step 1: Understand the dissociation of sulfuric acid
Sulfuric acid is a strong acid and dissociates completely in water. The dissociation reaction can be represented as:
\[ \text{H}_2\text{SO}_4 \rightarrow 2 \text{H}^+ + \text{SO}_4^{2-} \]

### Step 2: Determine the concentration of hydrogen ions
From the dissociation equation, we see that one mole of sulfuric acid produces two moles of hydrogen ions (H⁺). Therefore, if we have a 0.005 M solution of H₂SO₄, the concentration of H⁺ ions will be:
\[ \text{Concentration of H}^+ = 2 \times 0.005 \, \text{M} = 0.01 \, \text{M} \]

### Step 3: Calculate the pH
The pH of a solution is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the concentration of H⁺ ions into the formula:
\[ \text{pH} = -\log(0.01) \]

### Step 4: Solve for pH
Calculating the logarithm:
\[ \text{pH} = -\log(0.01) = -(-2) = 2 \]

### Conclusion
Thus, the pH of a 0.005 molar aqueous solution of sulfuric acid is approximately **2**.

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