The `pH` of a `0.05 M` solution of `H_(2)SO_(4)` in water is nearly

Options

A. `0.05`
B. `1`
C. `-1`
D. `0`

Explanation

To find the pH of a 0.05 M solution of \( H_2SO_4 \), we can follow these steps:

### Step 1: Understand the nature of \( H_2SO_4 \)
Sulfuric acid (\( H_2SO_4 \)) is a strong acid. This means it completely dissociates in water.

### Step 2: Write the dissociation equation
The dissociation of \( H_2SO_4 \) in water can be represented as:
\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]
From this equation, we can see that one mole of \( H_2SO_4 \) produces two moles of hydrogen ions (\( H^+ \)).

### Step 3: Calculate the concentration of \( H^+ \) ions
Given that the concentration of \( H_2SO_4 \) is 0.05 M, the concentration of \( H^+ \) ions produced will be:
\[ [H^+] = 2 \times [H_2SO_4] = 2 \times 0.05 \, M = 0.10 \, M \]

### Step 4: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \( [H^+] \):
\[ \text{pH} = -\log(0.10) \]
Using logarithmic properties, we know:
\[ \log(0.10) = -1 \]
Thus,
\[ \text{pH} = -(-1) = 1 \]

### Conclusion
The pH of a 0.05 M solution of \( H_2SO_4 \) is nearly **1**.

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