Upon mixing equal volumes of aqueous solutions of 0.1 M HCl and 0.2 M `H_(2)SO_(4)`, the concentration of `H^(+)` in the resulting solution is-

Options

A. `0.30mol//L`
B. `0.25mol//L`
C. `0.15mol//L`
D. `0.10mol//L`

Explanation

To find the concentration of \( H^+ \) ions in the resulting solution after mixing equal volumes of 0.1 M HCl and 0.2 M \( H_2SO_4 \), we can follow these steps:

### Step 1: Determine the moles of \( H^+ \) from HCl
- The concentration of HCl is 0.1 M.
- When mixed, let's assume we take a volume \( V \) liters of HCl.
- The moles of \( H^+ \) from HCl can be calculated as:
\[ \text{Moles of } H^+ \text{ from HCl} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times V \, \text{L} = 0.1V \]

### Step 2: Determine the moles of \( H^+ \) from \( H_2SO_4 \)
- The concentration of \( H_2SO_4 \) is 0.2 M.
- \( H_2SO_4 \) dissociates to produce 2 moles of \( H^+ \) ions per mole of \( H_2SO_4 \).
- Therefore, the moles of \( H^+ \) from \( H_2SO_4 \) can be calculated as:
\[ \text{Moles of } H^+ \text{ from } H_2SO_4 = 2 \times \text{Concentration} \times \text{Volume} = 2 \times 0.2 \, \text{mol/L} \times V \, \text{L} = 0.4V \]

### Step 3: Calculate the total moles of \( H^+ \)
- Now, we can sum the moles of \( H^+ \) from both acids:
\[ \text{Total moles of } H^+ = 0.1V + 0.4V = 0.5V \]

### Step 4: Determine the total volume of the resulting solution
- Since we mixed equal volumes of both solutions, the total volume after mixing is:
\[ \text{Total Volume} = V + V = 2V \]

### Step 5: Calculate the concentration of \( H^+ \) in the resulting solution
- The concentration of \( H^+ \) ions in the resulting solution can be calculated using the formula:
\[ \text{Concentration of } H^+ = \frac{\text{Total moles of } H^+}{\text{Total Volume}} = \frac{0.5V}{2V} \]
- Simplifying this gives:
\[ \text{Concentration of } H^+ = \frac{0.5}{2} = 0.25 \, \text{M} \]

### Final Answer:
The concentration of \( H^+ \) in the resulting solution is **0.25 M**.

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