100 ml of 0.3 M HCl solution is mixed with 200 ml of 0.3 M `H_(2)SO_(4)` solution. What is the molariyt of `H^(+)` in resultant solution ?

Options

A. `0.9`
B. `0.6`
C. `0.4`
D. `0.5`

Answer

To find the molarity of \( H^+ \) in the resultant solution after mixing 100 ml of 0.3 M HCl with 200 ml of 0.3 M \( H_2SO_4 \), we can follow these steps:

### Step 1: Calculate the moles of \( H^+ \) from HCl
1. **Determine the moles of \( HCl \)**:
\[ \text{Moles of } HCl = \text{Molarity} \times \text{Volume} = 0.3 \, \text{mol/L} \times 0.1 \, \text{L} = 0.03 \, \text{mol} \]
2. **Since HCl dissociates completely into \( H^+ \) and \( Cl^- \)**, the moles of \( H^+ \) from HCl will also be 0.03 mol.

### Step 2: Calculate the moles of \( H^+ \) from \( H_2SO_4 \)
1. **Determine the moles of \( H_2SO_4 \)**:
\[ \text{Moles of } H_2SO_4 = \text{Molarity} \times \text{Volume} = 0.3 \, \text{mol/L} \times 0.2 \, \text{L} = 0.06 \, \text{mol} \]
2. **Since \( H_2SO_4 \) dissociates into 2 \( H^+ \) ions and 1 \( SO_4^{2-} \)**, the moles of \( H^+ \) from \( H_2SO_4 \) will be:
\[ \text{Moles of } H^+ = 2 \times \text{Moles of } H_2SO_4 = 2 \times 0.06 \, \text{mol} = 0.12 \, \text{mol} \]

### Step 3: Calculate the total moles of \( H^+ \)
- **Total moles of \( H^+ \)** from both solutions:
\[ \text{Total moles of } H^+ = \text{Moles from } HCl + \text{Moles from } H_2SO_4 = 0.03 \, \text{mol} + 0.12 \, \text{mol} = 0.15 \, \text{mol} \]

### Step 4: Calculate the total volume of the resultant solution
- **Total volume**:
\[ \text{Total Volume} = 100 \, \text{ml} + 200 \, \text{ml} = 300 \, \text{ml} = 0.3 \, \text{L} \]

### Step 5: Calculate the molarity of \( H^+ \) in the resultant solution
1. **Molarity of \( H^+ \)**:
\[ \text{Molarity} = \frac{\text{Total moles of } H^+}{\text{Total volume in L}} = \frac{0.15 \, \text{mol}}{0.3 \, \text{L}} = 0.5 \, \text{M} \]

### Final Answer
The molarity of \( H^+ \) in the resultant solution is **0.5 M**.

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