50 mL of `H_(2)O` is added to `50 mL` of `1 xx 10^(-3)M` barium hydroxide solution. What is the pH of the resulting solution?

Options

A. 3
B. 3.3
C. 11
D. 11.7

Explanation

To find the pH of the resulting solution when 50 mL of water is added to 50 mL of 1 x 10^(-3) M barium hydroxide solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Chemical Formula**:
The formula for barium hydroxide is \( \text{Ba(OH)}_2 \). Each mole of barium hydroxide produces 2 moles of hydroxide ions (\( \text{OH}^- \)).

2. **Calculate Moles of Barium Hydroxide**:
\[ \text{Moles of Ba(OH)}_2 = \text{Molarity} \times \text{Volume} \]
Given:
- Molarity = \( 1 \times 10^{-3} \, \text{M} \)
- Volume = \( 50 \, \text{mL} = 0.050 \, \text{L} \)

\[ \text{Moles of Ba(OH)}_2 = 1 \times 10^{-3} \, \text{mol/L} \times 0.050 \, \text{L} = 5 \times 10^{-5} \, \text{mol} \]

3. **Calculate Moles of Hydroxide Ions**:
Since each mole of barium hydroxide produces 2 moles of hydroxide ions:
\[ \text{Moles of OH}^- = 2 \times \text{Moles of Ba(OH)}_2 = 2 \times 5 \times 10^{-5} = 1 \times 10^{-4} \, \text{mol} \]

4. **Calculate the Final Volume**:
The final volume after mixing 50 mL of barium hydroxide solution with 50 mL of water is:
\[ \text{Final Volume} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.100 \, \text{L} \]

5. **Calculate the Final Concentration of Hydroxide Ions**:
\[ \text{Concentration of OH}^- = \frac{\text{Moles of OH}^-}{\text{Final Volume}} = \frac{1 \times 10^{-4} \, \text{mol}}{0.100 \, \text{L}} = 1 \times 10^{-3} \, \text{M} \]

6. **Calculate pOH**:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log(1 \times 10^{-3}) = 3 \]

7. **Calculate pH**:
Using the relationship \( \text{pH} + \text{pOH} = 14 \):
\[ \text{pH} = 14 - \text{pOH} = 14 - 3 = 11 \]

### Final Answer:
The pH of the resulting solution is **11**.

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