To solve the question of which of the following solutions will have nearly equal \( H^+ \) concentrations, we will analyze each case step by step.
### Step 1: Analyze the first solution (100 ml of 1 M HCl mixed with 50 ml of water)
1. **Initial Concentration of \( H^+ \)**:
- HCl is a strong acid and dissociates completely. Therefore, the initial concentration of \( H^+ \) from HCl is 1 M.
- Volume of HCl = 100 ml = 0.1 L.
- Moles of \( H^+ \) from HCl = Concentration × Volume = \( 1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.1 \, \text{mol} \).
2. **Final Volume**:
- Total volume after dilution = 100 ml + 50 ml = 150 ml = 0.15 L.
3. **Final Concentration of \( H^+ \)**:
\[
[H^+] = \frac{\text{Moles of } H^+}{\text{Total Volume}} = \frac{0.1 \, \text{mol}}{0.15 \, \text{L}} = \frac{10}{150} = \frac{1}{15} \, \text{M}.
\]
### Step 2: Analyze the second solution (50 ml of 1 M \( H_2SO_4 \))
1. **Initial Concentration of \( H^+ \)**:
- \( H_2SO_4 \) is a diprotic acid and dissociates to give 2 moles of \( H^+ \) per mole of \( H_2SO_4 \).
- Volume of \( H_2SO_4 \) = 50 ml = 0.05 L.
- Moles of \( H^+ \) from \( H_2SO_4 \) = \( 2 \times (1 \, \text{mol/L} \times 0.05 \, \text{L}) = 0.1 \, \text{mol} \).
2. **Final Volume**:
- Total volume after dilution = 50 ml + 50 ml = 100 ml = 0.1 L.
3. **Final Concentration of \( H^+ \)**:
\[
[H^+] = \frac{0.1 \, \text{mol}}{0.1 \, \text{L}} = 1 \, \text{M}.
\]
### Step 3: Analyze the third solution (100 ml of 1 M \( H_2SO_4 \) mixed with 100 ml of water)
1. **Initial Concentration of \( H^+ \)**:
- Volume of \( H_2SO_4 \) = 100 ml = 0.1 L.
- Moles of \( H^+ \) = \( 2 \times (1 \, \text{mol/L} \times 0.1 \, \text{L}) = 0.2 \, \text{mol} \).
2. **Final Volume**:
- Total volume after dilution = 100 ml + 100 ml = 200 ml = 0.2 L.
3. **Final Concentration of \( H^+ \)**:
\[
[H^+] = \frac{0.2 \, \text{mol}}{0.2 \, \text{L}} = 1 \, \text{M}.
\]
### Step 4: Analyze the fourth solution (50 ml of 0.1 M \( HCl \) mixed with 50 ml of water)
1. **Initial Concentration of \( H^+ \)**:
- Volume of \( HCl \) = 50 ml = 0.05 L.
- Moles of \( H^+ \) = \( 0.1 \, \text{mol/L} \times 0.05 \, \text{L} = 0.005 \, \text{mol} \).
2. **Final Volume**:
- Total volume after dilution = 50 ml + 50 ml = 100 ml = 0.1 L.
3. **Final Concentration of \( H^+ \)**:
\[
[H^+] = \frac{0.005 \, \text{mol}}{0.1 \, \text{L}} = 0.05 \, \text{M}.
\]
### Conclusion
From the calculations:
- First solution: \( [H^+] = \frac{1}{15} \, \text{M} \)
- Second solution: \( [H^+] = 1 \, \text{M} \)
- Third solution: \( [H^+] = 1 \, \text{M} \)
- Fourth solution: \( [H^+] = 0.05 \, \text{M} \)
The solutions with nearly equal \( H^+ \) concentrations are the first and third solutions, both yielding \( \frac{1}{15} \, \text{M} \).
### Final Answer
The solutions that will have nearly equal \( H^+ \) concentrations are from the first and third cases.
