Caculate the pH of 0.001 N `H_(2)SO_(4)` solution.

Explanation

To calculate the pH of a 0.001 N \( H_2SO_4 \) solution, follow these steps:

### Step 1: Understand the Normality and Molarity Relationship
- \( H_2SO_4 \) is a dibasic acid, meaning it can donate two protons (H\(^+\)) per molecule.
- The normality (N) of a solution is related to the molarity (M) by the formula:
\[ \text{Normality} = \text{Molarity} \times \text{n-factor} \]
where the n-factor for \( H_2SO_4 \) is 2.

### Step 2: Convert Normality to Molarity
- Given that the normality of the \( H_2SO_4 \) solution is 0.001 N, we can find the molarity (M) as follows:
\[ \text{Molarity} = \frac{\text{Normality}}{\text{n-factor}} = \frac{0.001 \, \text{N}}{2} = 0.0005 \, \text{M} \]

### Step 3: Calculate the Concentration of \( H^+ \) Ions
- Since \( H_2SO_4 \) is a strong acid, it dissociates completely in solution:
\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]
- Therefore, the concentration of \( H^+ \) ions will be twice the molarity of the \( H_2SO_4 \):
\[ [H^+] = 2 \times 0.0005 \, \text{M} = 0.001 \, \text{M} \]

### Step 4: Calculate the pH
- The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
- Substituting the concentration of \( H^+ \):
\[ \text{pH} = -\log(0.001) = -\log(10^{-3}) = 3 \]

### Final Answer
- The pH of the 0.001 N \( H_2SO_4 \) solution is **3**.

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