Compute the pH of a solution at `25^@C` which is twice as alkaline as pure water.

Answer

To compute the pH of a solution at 25°C that is twice as alkaline as pure water, we can follow these steps:

### Step 1: Understand the pH and pOH relationship
At 25°C, the pH of pure water is 7. This means that the concentration of hydroxide ions (OH⁻) in pure water is \( [OH⁻] = 10^{-7} \, \text{M} \).

### Step 2: Determine the hydroxide ion concentration for the alkaline solution
Since the solution is described as "twice as alkaline" as pure water, we can calculate the concentration of OH⁻ ions in this solution.

\[ [OH⁻] = 2 \times 10^{-7} \, \text{M} \]

### Step 3: Calculate the pOH of the solution
The pOH can be calculated using the formula:

\[ \text{pOH} = -\log[OH⁻] \]

Substituting the value of \( [OH⁻] \):

\[ \text{pOH} = -\log(2 \times 10^{-7}) \]

Using logarithmic properties, we can simplify this:

\[ \text{pOH} = -\log(2) - \log(10^{-7}) = -\log(2) + 7 \]

Using the approximate value \( \log(2) \approx 0.301 \):

\[ \text{pOH} \approx -0.301 + 7 = 6.699 \]

### Step 4: Calculate the pH of the solution
Using the relationship between pH and pOH:

\[ \text{pH} + \text{pOH} = 14 \]

We can find the pH:

\[ \text{pH} = 14 - \text{pOH} \approx 14 - 6.699 \approx 7.301 \]

### Final Answer
The pH of the solution at 25°C, which is twice as alkaline as pure water, is approximately **7.30**.

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