250 mL of 0.2 M `NaOH` and 100 mL of `0.25 M Ba (OH)_(2)` are mixed. Calculate the molarity of hydroxyl ion in the mixture.

Answer

To solve the problem of finding the molarity of hydroxyl ions (OH⁻) in a mixture of sodium hydroxide (NaOH) and barium hydroxide (Ba(OH)₂), we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the moles of NaOH:**
- The formula to calculate moles is:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (in L)} \]
- For NaOH:
\[ \text{Moles of NaOH} = 0.2 \, \text{M} \times 0.250 \, \text{L} = 0.050 \, \text{moles} \]

2. **Calculate the moles of Ba(OH)₂:**
- For Ba(OH)₂, we use the same formula:
\[ \text{Moles of Ba(OH)₂} = 0.25 \, \text{M} \times 0.100 \, \text{L} = 0.025 \, \text{moles} \]

3. **Determine the moles of hydroxyl ions from NaOH:**
- Each mole of NaOH produces one mole of OH⁻:
\[ \text{Moles of OH⁻ from NaOH} = 0.050 \, \text{moles} \]

4. **Determine the moles of hydroxyl ions from Ba(OH)₂:**
- Each mole of Ba(OH)₂ produces two moles of OH⁻:
\[ \text{Moles of OH⁻ from Ba(OH)₂} = 0.025 \, \text{moles} \times 2 = 0.050 \, \text{moles} \]

5. **Calculate the total moles of hydroxyl ions (OH⁻):**
- Add the moles of OH⁻ from both sources:
\[ \text{Total moles of OH⁻} = 0.050 \, \text{moles (from NaOH)} + 0.050 \, \text{moles (from Ba(OH)₂)} = 0.100 \, \text{moles} \]

6. **Calculate the total volume of the mixture:**
- The total volume is the sum of the volumes of both solutions:
\[ \text{Total Volume} = 250 \, \text{mL} + 100 \, \text{mL} = 350 \, \text{mL} = 0.350 \, \text{L} \]

7. **Calculate the molarity of hydroxyl ions (OH⁻) in the mixture:**
- Use the formula for molarity:
\[ \text{Molarity (M)} = \frac{\text{Total moles}}{\text{Total volume (in L)}} \]
- Therefore:
\[ \text{Molarity of OH⁻} = \frac{0.100 \, \text{moles}}{0.350 \, \text{L}} \approx 0.2857 \, \text{M} \]

### Final Answer:
The molarity of hydroxyl ions (OH⁻) in the mixture is approximately **0.286 M**.