The pH of a `0.005 M H_(2)SO_(4)` solution is

Options

A. `3.3`
B. `5.0`
C. `2.0`
D. `4.0`

Explanation

To find the pH of a `0.005 M H₂SO₄` solution, follow these steps:

### Step 1: Understand the Ionization of Sulfuric Acid
Sulfuric acid (H₂SO₄) is a strong acid and it completely ionizes in solution. The ionization can be represented as:
\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]
This means that for every 1 mole of H₂SO₄, 2 moles of hydrogen ions (H⁺) are produced.

### Step 2: Calculate the Concentration of Hydrogen Ions
Given the concentration of H₂SO₄ is `0.005 M`, we can calculate the concentration of hydrogen ions produced:
\[ \text{Concentration of } H^+ = 2 \times \text{Concentration of } H_2SO_4 \]
\[ \text{Concentration of } H^+ = 2 \times 0.005 \, M = 0.01 \, M \]

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

### Step 4: Calculate the pH
Now, calculate the logarithm:
\[ \text{pH} = -\log(10^{-2}) \]
Since \(-\log(10^{-2}) = 2\):
\[ \text{pH} = 2 \]

### Final Answer
The pH of a `0.005 M H₂SO₄` solution is **2**.

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