Asolution with pH 2.699 is diluted two times, then calculate pH of the resulting solution. [Given antilog of 0.3010 = 2

Answer

To solve the problem of calculating the pH of a solution after it has been diluted, follow these steps:

### Step 1: Understand the relationship between pH and [H⁺]
The pH of a solution is defined as:
\[ \text{pH} = -\log[H^+] \]
where \([H^+]\) is the concentration of hydrogen ions in moles per liter.

### Step 2: Calculate the initial concentration of [H⁺]
Given the pH of the original solution is 2.699, we can find the concentration of hydrogen ions using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-2.699} \]

### Step 3: Simplify the exponent
To calculate \(10^{-2.699}\), we can break it down:
\[ 10^{-2.699} = 10^{-2} \times 10^{-0.699} \]
We know that \(10^{-2} = 0.01\).

Now, we need to calculate \(10^{-0.699}\). Using the given information that \(10^{0.3010} = 2\), we can find:
\[ 10^{-0.699} = \frac{1}{10^{0.699}} = \frac{1}{10^{1 - 0.3010}} = \frac{1}{2} \approx 0.5 \]
Thus, we can estimate:
\[ 10^{-0.699} \approx 0.2 \]
So,
\[ [H^+] \approx 0.01 \times 0.2 = 0.002 \text{ M} \]

### Step 4: Calculate the concentration after dilution
When the solution is diluted two times, the concentration of hydrogen ions will be halved:
\[ [H^+]_{\text{diluted}} = \frac{0.002}{2} = 0.001 \text{ M} \]

### Step 5: Calculate the pH of the diluted solution
Now, we can find the pH of the diluted solution:
\[ \text{pH}_{\text{diluted}} = -\log[H^+]_{\text{diluted}} = -\log(0.001) \]
Since \(0.001 = 10^{-3}\), we have:
\[ \text{pH}_{\text{diluted}} = 3 \]

### Final Answer
The pH of the resulting solution after dilution is **3**.

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