Calculate the pH of following solutions.
`N//100 H_(2)SO_(4)`

Answer

To calculate the pH of a \( \frac{N}{100} \) \( H_2SO_4 \) solution, we can follow these steps:

### Step 1: Understand the Normality
The given solution is \( \frac{N}{100} \) \( H_2SO_4 \), which is equivalent to a normality of \( 0.01 \) N (or \( 10^{-2} \) N).

### Step 2: Determine the n-factor for \( H_2SO_4 \)
The n-factor for sulfuric acid (\( H_2SO_4 \)) is 2 because it can donate two protons (\( H^+ \)) per molecule when it dissociates.

### Step 3: Calculate Molarity
To find the molarity (\( M \)) from normality (\( N \)), we use the formula:
\[ M = \frac{N}{n\text{-factor}} \]
Substituting the values:
\[ M = \frac{0.01}{2} = 0.005 \, M \, (or \, 5 \times 10^{-3} \, M) \]

### Step 4: Calculate the Concentration of \( H^+ \) Ions
Since \( H_2SO_4 \) dissociates completely, it produces two \( H^+ \) ions for every molecule of \( H_2SO_4 \):
\[ \text{Concentration of } H^+ = 2 \times [H_2SO_4] = 2 \times 0.005 = 0.01 \, M \, (or \, 10^{-2} \, M) \]

### Step 5: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the concentration of \( H^+ \):
\[ \text{pH} = -\log(10^{-2}) = 2 \]

### Final Answer
The pH of the \( \frac{N}{100} \) \( H_2SO_4 \) solution is **2**.

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