Calculate the pH of `10^(-2)N H_(2)SO_(4)`.

Explanation

To calculate the pH of `10^(-2)N H2SO4`, we can follow these steps:

### Step 1: Understand the Ionization of H2SO4
H2SO4 (sulfuric acid) is a strong acid that completely ionizes in solution. It dissociates as follows:
\[ \text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-} \]

This means that for every mole of H2SO4, we get 2 moles of H⁺ ions.

### Step 2: Determine the Concentration of H⁺ Ions
Given that the normality (N) of the solution is `10^(-2)N`, we can determine the concentration of H⁺ ions produced from the ionization of H2SO4:
- Since sulfuric acid provides 2 H⁺ ions for each molecule, the concentration of H⁺ ions will be:
\[ [\text{H}^+] = 2 \times 10^{-2} \, \text{N} = 2 \times 10^{-2} \, \text{M} \]

### 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:
\[ \text{pH} = -\log(2 \times 10^{-2}) \]

### Step 4: Simplify the Logarithm
Using the properties of logarithms:
\[ \text{pH} = -\log(2) - \log(10^{-2}) \]
\[ \text{pH} = -\log(2) + 2 \]

Using the approximate value of \(\log(2) \approx 0.301\):
\[ \text{pH} = -0.301 + 2 \]
\[ \text{pH} \approx 1.699 \]

### Step 5: Final Result
Thus, rounding to two decimal places, the pH of the solution is:
\[ \text{pH} \approx 1.70 \]

### Summary
The pH of `10^(-2)N H2SO4` is approximately **1.70**.

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