Which of the following will have nearly equal `H^(+)` concentration ?

Options

A. 100 ml 0.1 M HCl mixed with 50 ml water
B. 50 ml `0.1 M H_(2)SO_(4)` mixed with 50 ml water
C. 50 ml `0.1 M H_(2)SO_(4)` mixed with 100 ml water
D. 50 ml `0.1` M HCl mixed with 50 ml water

Explanation

To determine which of the given options has nearly equal \( H^+ \) concentration, we will calculate the \( H^+ \) concentration for each option step by step.

### Step 1: Calculate \( H^+ \) concentration for HCl (Option 1)
1. **Given**: Molarity of HCl = 0.1 M, Volume = 100 mL
2. **Convert volume to liters**:
\[ 100 \, \text{mL} = 100 \times 10^{-3} \, \text{L} = 0.1 \, \text{L} \]
3. **Calculate moles of HCl**:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 0.1 \times 0.1 = 0.01 \, \text{moles} \]
4. **Since HCl is a strong acid, it dissociates completely**:
\[ \text{Moles of } H^+ = \text{Moles of HCl} = 0.01 \, \text{moles} \]
5. **Total volume after dilution**:
\[ 100 \, \text{mL} + 50 \, \text{mL} = 150 \, \text{mL} = 0.15 \, \text{L} \]
6. **Calculate \( H^+ \) concentration**:
\[ [H^+] = \frac{\text{Moles of } H^+}{\text{Total Volume}} = \frac{0.01}{0.15} = \frac{1}{15} \, \text{M} \]

### Step 2: Calculate \( H^+ \) concentration for H2SO4 (Option 2)
1. **Given**: Molarity of H2SO4 = 0.1 M, Volume = 50 mL
2. **Convert volume to liters**:
\[ 50 \, \text{mL} = 50 \times 10^{-3} \, \text{L} = 0.05 \, \text{L} \]
3. **Calculate moles of H2SO4**:
\[ \text{Moles of H2SO4} = 0.1 \times 0.05 = 0.005 \, \text{moles} \]
4. **Since H2SO4 dissociates to give 2 moles of \( H^+ \)**:
\[ \text{Moles of } H^+ = 2 \times 0.005 = 0.01 \, \text{moles} \]
5. **Total volume after dilution**:
\[ 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]
6. **Calculate \( H^+ \) concentration**:
\[ [H^+] = \frac{0.01}{0.1} = 0.1 \, \text{M} \]

### Step 3: Calculate \( H^+ \) concentration for H2SO4 (Option 3)
1. **Given**: Molarity of H2SO4 = 0.1 M, Volume = 50 mL
2. **Convert volume to liters**:
\[ 50 \, \text{mL} = 0.05 \, \text{L} \]
3. **Calculate moles of H2SO4**:
\[ \text{Moles of H2SO4} = 0.1 \times 0.05 = 0.005 \, \text{moles} \]
4. **Moles of \( H^+ \)**:
\[ \text{Moles of } H^+ = 2 \times 0.005 = 0.01 \, \text{moles} \]
5. **Total volume after dilution**:
\[ 50 \, \text{mL} + 100 \, \text{mL} = 150 \, \text{mL} = 0.15 \, \text{L} \]
6. **Calculate \( H^+ \) concentration**:
\[ [H^+] = \frac{0.01}{0.15} = \frac{1}{15} \, \text{M} \]

### Step 4: Calculate \( H^+ \) concentration for HCl (Option 4)
1. **Given**: Molarity of HCl = 0.1 M, Volume = 50 mL
2. **Convert volume to liters**:
\[ 50 \, \text{mL} = 0.05 \, \text{L} \]
3. **Calculate moles of HCl**:
\[ \text{Moles of HCl} = 0.1 \times 0.05 = 0.005 \, \text{moles} \]
4. **Moles of \( H^+ \)**:
\[ \text{Moles of } H^+ = 0.005 \, \text{moles} \]
5. **Total volume after dilution**:
\[ 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]
6. **Calculate \( H^+ \) concentration**:
\[ [H^+] = \frac{0.005}{0.1} = 0.05 \, \text{M} \]

### Conclusion
From our calculations:
- Option 1: \( [H^+] = \frac{1}{15} \, \text{M} \)
- Option 2: \( [H^+] = 0.1 \, \text{M} \)
- Option 3: \( [H^+] = \frac{1}{15} \, \text{M} \)
- Option 4: \( [H^+] = 0.05 \, \text{M} \)

The options with nearly equal \( H^+ \) concentration are **Option 1 and Option 3**.