The pH of a solution which is twice as acidic as pure water is (Given : log 2 = 0.3)

Options

A. `7.0`
B. `3.5`
C. `6.7`
D. `14.0`

Answer

To find the pH of a solution that is twice as acidic as pure water, we can follow these steps:

### Step 1: Understand the pH of Pure Water
The pH of pure water is 7. This is because the concentration of hydrogen ions \([H^+]\) in pure water is \(10^{-7} \, \text{M}\).

### Step 2: Determine the Hydrogen Ion Concentration for the New Solution
Since the solution is twice as acidic as pure water, the concentration of hydrogen ions in this solution will be:
\[ [H^+] = 2 \times [H^+]_{\text{pure water}} = 2 \times 10^{-7} \, \text{M} = 2 \times 10^{-7} \, \text{M} \]

### Step 3: Calculate the pH of the New Solution
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ \text{pH} = -\log(2 \times 10^{-7}) \]

### Step 4: Use Logarithmic Properties
We can separate the logarithm:
\[ \text{pH} = -\log(2) - \log(10^{-7}) \]
Using the property of logarithms, we know:
\[ \log(10^{-7}) = -7 \]
Thus,
\[ \text{pH} = -\log(2) + 7 \]

### Step 5: Substitute the Value of \(\log(2)\)
Given that \(\log(2) = 0.3\):
\[ \text{pH} = -0.3 + 7 \]
Calculating this gives:
\[ \text{pH} = 7 - 0.3 = 6.7 \]

### Final Answer
The pH of the solution which is twice as acidic as pure water is **6.7**.

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