The `P^(H)` of a solution is 11. lt is diluted by 1000 times. Then the `P^(H)` of resulting solution is

Options

A. 8
B. 14
C. 7
D. 7.04

Answer

To solve the problem, we need to determine the pH of a solution after it has been diluted by 1000 times. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the initial concentration of H⁺ ions
Given that the pH of the solution is 11, we can find the concentration of H⁺ ions using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
To find [H⁺], we rearrange the formula:
\[ [\text{H}^+] = 10^{-\text{pH}} \]
Substituting the given pH:
\[ [\text{H}^+] = 10^{-11} \, \text{M} \]

### Step 2: Calculate the concentration of OH⁻ ions
We know that the product of the concentrations of H⁺ and OH⁻ ions at 25°C is given by:
\[ [\text{H}^+][\text{OH}^-] = 10^{-14} \]
We can find the concentration of OH⁻ ions using the concentration of H⁺:
\[ [\text{OH}^-] = \frac{10^{-14}}{[\text{H}^+]} \]
Substituting the value of [H⁺]:
\[ [\text{OH}^-] = \frac{10^{-14}}{10^{-11}} = 10^{-3} \, \text{M} \]

### Step 3: Dilute the solution
The solution is diluted by 1000 times. Therefore, the new concentration of OH⁻ after dilution will be:
\[ [\text{OH}^-]_{\text{new}} = \frac{10^{-3}}{1000} = 10^{-6} \, \text{M} \]

### Step 4: Calculate the new concentration of H⁺ ions
Using the ionic product of water again:
\[ [\text{H}^+]_{\text{new}} = \frac{10^{-14}}{[\text{OH}^-]_{\text{new}}} \]
Substituting the new concentration of OH⁻:
\[ [\text{H}^+]_{\text{new}} = \frac{10^{-14}}{10^{-6}} = 10^{-8} \, \text{M} \]

### Step 5: Calculate the new pH
Finally, we can find the new pH using the concentration of H⁺ ions:
\[ \text{pH}_{\text{new}} = -\log[\text{H}^+]_{\text{new}} \]
Substituting the value of [H⁺]:
\[ \text{pH}_{\text{new}} = -\log(10^{-8}) = 8 \]

### Conclusion
The pH of the resulting solution after dilution is **8**.

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