The `P^(H )`of a solution is 6. Its `[H_(3)O^(+)]` is decreased by 1000 times. Its `P^(H)` will be

Options

A. 9
B. 6.96
C. 7.04
D. 8

Answer

To solve the problem step by step, we will follow the concepts of pH and hydronium ion concentration.

### Step 1: Understand the initial pH and calculate the initial concentration of \([H_3O^+]\).

The pH of the solution is given as 6. The relationship between pH and hydronium ion concentration \([H_3O^+]\) is given by the formula:

\[ \text{pH} = -\log[H_3O^+] \]

To find the concentration of \([H_3O^+]\):

\[ [H_3O^+] = 10^{-\text{pH}} = 10^{-6} \, \text{M} \]

### Step 2: Decrease the concentration of \([H_3O^+]\) by 1000 times.

If the concentration of \([H_3O^+]\) is decreased by 1000 times, the new concentration will be:

\[ [H_3O^+]_{\text{new}} = \frac{10^{-6}}{1000} = 10^{-6} \times 10^{-3} = 10^{-9} \, \text{M} \]

### Step 3: Consider the contribution of \([H_3O^+]\) from water.

In pure water, the concentration of \([H_3O^+]\) is:

\[ [H_3O^+]_{\text{water}} = 10^{-7} \, \text{M} \]

### Step 4: Calculate the total concentration of \([H_3O^+]\).

The total concentration of \([H_3O^+]\) in the solution will be the sum of the new concentration from the acid and the concentration from water:

\[ [H_3O^+]_{\text{total}} = [H_3O^+]_{\text{new}} + [H_3O^+]_{\text{water}} = 10^{-9} + 10^{-7} \]

To add these concentrations, we can express \(10^{-9}\) in terms of \(10^{-7}\):

\[ [H_3O^+]_{\text{total}} = 10^{-9} + 10^{-7} = 10^{-9} + 100 \times 10^{-9} = 101 \times 10^{-9} = 1.01 \times 10^{-7} \, \text{M} \]

### Step 5: Calculate the new pH.

Now we can calculate the new pH using the total concentration of \([H_3O^+]\):

\[ \text{pH}_{\text{new}} = -\log[H_3O^+]_{\text{total}} = -\log(1.01 \times 10^{-7}) \]

Using logarithmic properties:

\[ \text{pH}_{\text{new}} \approx 7 - \log(1.01) \approx 7 - 0.00432 \approx 6.96 \]

### Final Answer:

The new pH of the solution is approximately **6.96**.

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