Assuming complete ionisation, the pH of 0.1 M HCl is 1. The molarity of `H_(2)SO_(4)` with the same pH is

Options

A. 0.2
B. 0.1
C. `2.0`
D. 0.05

Explanation

To find the molarity of \( H_2SO_4 \) that results in the same pH as a 0.1 M \( HCl \) solution, we can follow these steps:

### Step 1: Understand the pH of 0.1 M HCl
Given that the pH of 0.1 M \( HCl \) is 1, we can confirm that:
\[ \text{pH} = -\log[H^+] \]
From the pH, we can find the concentration of hydrogen ions \( [H^+] \):
\[ 1 = -\log[H^+] \]
Taking the antilog:
\[ [H^+] = 10^{-1} = 0.1 \, M \]

### Step 2: Determine the contribution of \( H_2SO_4 \) to \( [H^+] \)
Sulfuric acid \( H_2SO_4 \) is a strong acid that ionizes completely in two steps:
1. \( H_2SO_4 \rightarrow H^+ + HSO_4^- \)
2. \( HSO_4^- \rightarrow H^+ + SO_4^{2-} \)

From the first ionization, 1 mole of \( H_2SO_4 \) produces 1 mole of \( H^+ \). From the second ionization, 1 mole of \( HSO_4^- \) can produce another mole of \( H^+ \). Therefore, 1 mole of \( H_2SO_4 \) can produce up to 2 moles of \( H^+ \).

### Step 3: Set up the equation for \( H_2SO_4 \)
Let the molarity of \( H_2SO_4 \) be \( x \). The total concentration of \( H^+ \) ions from \( H_2SO_4 \) will be:
\[ [H^+] = x + x = 2x \]
We want this to equal the \( [H^+] \) from the 0.1 M \( HCl \):
\[ 2x = 0.1 \]

### Step 4: Solve for \( x \)
To find \( x \):
\[ x = \frac{0.1}{2} = 0.05 \, M \]

### Conclusion
The molarity of \( H_2SO_4 \) that results in the same pH of 1 is:
\[ \boxed{0.05 \, M} \]