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

Answer

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

### Step 1: Understand the Normality
The given normality of the solution is \( \frac{N}{100} \). This means that if we consider \( N = 1 \), the normality is \( \frac{1}{100} \) N, which is equivalent to \( 0.01 \) N.

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

### Step 3: Convert Normality to Molarity
Molarity (M) can be calculated from normality (N) using the formula:
\[ \text{Molarity} = \frac{\text{Normality}}{\text{n-factor}} \]
Substituting the values:
\[ \text{Molarity} = \frac{0.01 \, \text{N}}{2} = 0.005 \, \text{M} \]

### Step 4: Calculate the Concentration of \( H^+ \) Ions
Since \( H_2SO_4 \) is a strong acid, it completely dissociates in solution:
\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]
From the molarity calculated, the concentration of \( H^+ \) ions will be:
\[ \text{Concentration of } H^+ = 2 \times 0.005 \, \text{M} = 0.01 \, \text{M} \]

### Step 5: Calculate the pH
The pH can be calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the concentration of \( H^+ \):
\[ \text{pH} = -\log(0.01) = -\log(10^{-2}) = 2 \]

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

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