To determine which of the given options will have nearly equal pH, we need to calculate the concentration of hydrogen ions (H⁺) in each solution. The pH of a solution is inversely related to the concentration of H⁺ ions, so solutions with the same H⁺ concentration will have nearly equal pH.
### Step-by-Step Solution:
1. **Option A: 100 ml of 0.1 M HCl mixed with 50 ml of water**
- Calculate the total volume:
\[
V_{\text{total}} = 100 \, \text{ml} + 50 \, \text{ml} = 150 \, \text{ml}
\]
- Calculate the moles of H⁺ from HCl:
\[
\text{Moles of H⁺} = M \times V = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol}
\]
- Calculate the concentration of H⁺:
\[
[H⁺] = \frac{\text{Moles of H⁺}}{V_{\text{total}}} = \frac{0.01 \, \text{mol}}{0.150 \, \text{L}} = \frac{0.01}{0.150} = \frac{1}{15} \, \text{mol/L}
\]
2. **Option B: 50 ml of 0.1 M H₂SO₄**
- Calculate the normality of H₂SO₄:
\[
\text{Normality} = 0.1 \, \text{mol/L} \times 2 = 0.2 \, \text{N}
\]
- Calculate the total volume (assuming it is mixed with 50 ml of water):
\[
V_{\text{total}} = 50 \, \text{ml} + 50 \, \text{ml} = 100 \, \text{ml}
\]
- Calculate the moles of H⁺:
\[
\text{Moles of H⁺} = 0.2 \, \text{N} \times 0.050 \, \text{L} = 0.01 \, \text{mol}
\]
- Calculate the concentration of H⁺:
\[
[H⁺] = \frac{0.01 \, \text{mol}}{0.100 \, \text{L}} = \frac{0.01}{0.100} = \frac{1}{10} \, \text{mol/L}
\]
3. **Option C: 50 ml of 0.1 M H₂SO₄**
- Similar to Option B, the normality is:
\[
\text{Normality} = 0.2 \, \text{N}
\]
- Calculate the total volume (assuming it is mixed with 100 ml of water):
\[
V_{\text{total}} = 50 \, \text{ml} + 100 \, \text{ml} = 150 \, \text{ml}
\]
- Calculate the moles of H⁺:
\[
\text{Moles of H⁺} = 0.2 \, \text{N} \times 0.050 \, \text{L} = 0.01 \, \text{mol}
\]
- Calculate the concentration of H⁺:
\[
[H⁺] = \frac{0.01 \, \text{mol}}{0.150 \, \text{L}} = \frac{0.01}{0.150} = \frac{1}{15} \, \text{mol/L}
\]
4. **Option D: 50 ml of 0.1 M acetic acid mixed with 50 ml of water**
- Calculate the total volume:
\[
V_{\text{total}} = 50 \, \text{ml} + 50 \, \text{ml} = 100 \, \text{ml}
\]
- Calculate the moles of acetic acid:
\[
\text{Moles of acetic acid} = 0.1 \, \text{mol/L} \times 0.050 \, \text{L} = 0.005 \, \text{mol}
\]
- Acetic acid is a weak acid, and we need to consider its dissociation:
\[
\text{Let } x \text{ be the amount that dissociates. Then: } [H⁺] \approx \sqrt{K_a \cdot C}
\]
where \( K_a \) for acetic acid is approximately \( 1.8 \times 10^{-5} \).
- The concentration of acetic acid after dilution is:
\[
[\text{Acetic Acid}] = \frac{0.005 \, \text{mol}}{0.100 \, \text{L}} = 0.05 \, \text{mol/L}
\]
- Using the approximation for weak acids, we can find that the concentration of H⁺ will be significantly less than that of the strong acids.
### Conclusion:
From the calculations, we find:
- Option A: [H⁺] = 1/15 mol/L
- Option B: [H⁺] = 1/10 mol/L
- Option C: [H⁺] = 1/15 mol/L
- Option D: [H⁺] is significantly lower than the others.
Thus, **Options A and C will have nearly equal pH**.
