If an aqueous solution at `25^(@)C` has twice as many `OH^(-)` as pure water its `pOH` will be

Options

A. `6.7`
B. `7.3`
C. 7
D. `6.98`

Answer

To solve the question, we need to find the pOH of an aqueous solution at 25°C that has twice the concentration of OH⁻ ions compared to pure water.

### Step-by-Step Solution:

1. **Understand the concentration of OH⁻ in pure water**:
- At 25°C, the concentration of OH⁻ ions in pure water is \( [OH^-] = 10^{-7} \, \text{mol/L} \).

2. **Calculate the concentration of OH⁻ in the given solution**:
- The problem states that the solution has twice as many OH⁻ ions as pure water. Therefore:
\[ [OH^-] = 2 \times 10^{-7} \, \text{mol/L} \]

3. **Use the formula for pOH**:
- The pOH is calculated using the formula:
\[ pOH = -\log[OH^-] \]

4. **Substitute the concentration into the pOH formula**:
- Now, substituting the concentration of OH⁻ into the pOH formula:
\[ pOH = -\log(2 \times 10^{-7}) \]

5. **Break down the logarithm**:
- Using the properties of logarithms, we can separate the terms:
\[ pOH = -\log(2) - \log(10^{-7}) \]
- Since \(-\log(10^{-7}) = 7\), we have:
\[ pOH = -\log(2) + 7 \]

6. **Calculate \(-\log(2)\)**:
- The value of \(-\log(2)\) is approximately \(0.301\). Therefore:
\[ pOH = 7 - 0.301 = 6.699 \]

7. **Round to appropriate significant figures**:
- Rounding \(6.699\) gives us \(6.70\).

8. **Final Answer**:
- Thus, the pOH of the solution is approximately \(6.70\).

### Conclusion:
The pOH of the aqueous solution with twice the concentration of OH⁻ ions compared to pure water is **6.70**.