The pH of solution at `25^@C` which has twice as many hydroxide ion as in pure water at `25^@C` , will be

Options

A. 14
B. 9
C. 6.699
D. 7.301

Answer

To find the pH of a solution at 25°C that has twice as many hydroxide ions as in pure water, we can follow these steps:

### Step-by-Step Solution:

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

2. **Calculate the concentration of hydroxide ions in the given solution:**
- According to the problem, the solution has twice as many hydroxide ions as in pure water. Therefore, the concentration of hydroxide ions in the solution is:
\[ [OH^-] = 2 \times (1 \times 10^{-7}) = 2 \times 10^{-7} \, \text{mol/L} \]

3. **Calculate the pOH of the solution:**
- The formula for pOH is:
\[ pOH = -\log[OH^-] \]
- Substituting the value of \([OH^-]\):
\[ pOH = -\log(2 \times 10^{-7}) = -\log(2) - \log(10^{-7}) \]
- This simplifies to:
\[ pOH = -\log(2) + 7 \]
- Using \(\log(2) \approx 0.301\):
\[ pOH \approx -0.301 + 7 = 6.699 \]

4. **Calculate the pH of the solution:**
- We know that:
\[ pH + pOH = 14 \]
- Therefore, we can find pH:
\[ pH = 14 - pOH = 14 - 6.699 = 7.301 \]

### Final Answer:
The pH of the solution is approximately **7.301**.

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