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 solve the question regarding 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. **Understand the concentration of hydroxide ions in pure water**:
At 25°C, the ion product of water (Kw) is given by:
\[ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \]
In pure water, the concentrations of hydrogen ions \([H^+]\) and hydroxide ions \([OH^-]\) are equal:
\[ [H^+] = [OH^-] = 1.0 \times 10^{-7} \, \text{M} \]

2. **Determine the concentration of hydroxide ions in the new solution**:
According to the problem, the concentration of hydroxide ions in the new solution is twice that of pure water:
\[ [OH^-] = 2 \times [OH^-]_{\text{pure water}} = 2 \times 1.0 \times 10^{-7} = 2.0 \times 10^{-7} \, \text{M} \]

3. **Calculate the concentration of hydrogen ions**:
Using the ion product of water, we can find the concentration of hydrogen ions in the new solution:
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-7}} \]
Performing the calculation:
\[ [H^+] = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-7}} = 5.0 \times 10^{-8} \, \text{M} \]

4. **Calculate the pH**:
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ \text{pH} = -\log(5.0 \times 10^{-8}) \]
Using a calculator or logarithm tables, we find:
\[ \text{pH} \approx 7.301 \]

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