100 mL of a solution with pH = 6 is diluted to 1000 mL by adding water. pH will

Options

A. decrease by 0.7 unit
B. increase by 1 unit
C. increase by 9 units
D. increase by 0.7 unit

Answer

To solve the problem of how the pH changes when a 100 mL solution with a pH of 6 is diluted to 1000 mL, we can follow these steps:

### Step 1: Understand the initial pH
The initial pH of the solution is given as 6. The pH is defined as:
\[ \text{pH} = -\log[H^+] \]
From this, we can find the concentration of hydrogen ions \([H^+]\):
\[ [H^+] = 10^{-\text{pH}} = 10^{-6} \, \text{mol/L} \]

### Step 2: Calculate the initial moles of \( H^+ \)
The volume of the initial solution is 100 mL, which is 0.1 L. Therefore, the initial moles of \( H^+ \) can be calculated as:
\[ \text{Moles of } H^+ = [H^+] \times \text{Volume} = 10^{-6} \, \text{mol/L} \times 0.1 \, \text{L} = 10^{-7} \, \text{mol} \]

### Step 3: Determine the new volume after dilution
When the solution is diluted to a total volume of 1000 mL (or 1 L), we need to find the new concentration of \( H^+ \) ions.

### Step 4: Calculate the new concentration of \( H^+ \)
The total volume after dilution is 1 L. The number of moles of \( H^+ \) remains the same (since dilution does not change the number of moles, only the concentration):
\[ [H^+]_{\text{new}} = \frac{\text{Moles of } H^+}{\text{New Volume}} = \frac{10^{-7} \, \text{mol}}{1 \, \text{L}} = 10^{-7} \, \text{mol/L} \]

### Step 5: Account for the contribution of \( H^+ \) from water
In pure water, the concentration of \( H^+ \) is \( 10^{-7} \, \text{mol/L} \). When we dilute the solution, the \( H^+ \) from the acid and the \( H^+ \) from water combine:
\[ [H^+]_{\text{total}} = [H^+]_{\text{acid}} + [H^+]_{\text{water}} = 10^{-7} + 10^{-7} = 2 \times 10^{-7} \, \text{mol/L} \]

### Step 6: Calculate the new pH
Now, we can calculate the new pH using the total concentration of \( H^+ \):
\[ \text{pH}_{\text{new}} = -\log[H^+]_{\text{total}} = -\log(2 \times 10^{-7}) \]
Using the logarithmic properties:
\[ \text{pH}_{\text{new}} = -\log(2) - \log(10^{-7}) = -\log(2) + 7 \]
Using \( \log(2) \approx 0.301 \):
\[ \text{pH}_{\text{new}} \approx 7 - 0.301 = 6.699 \]

### Conclusion
The new pH after dilution is approximately 6.7.