100 ml of a solution of HCl with pH value 3 is diluted with 500 ml of water. The new `P^(H)` of the solution is

Options

A. `3.7`
B. `5.3`
C. `4.32`
D. `5.6`

Explanation

To find the new pH of the diluted HCl solution, we can follow these steps:

### Step 1: Calculate the initial concentration of H⁺ ions
Given that the pH of the initial HCl solution is 3, we can find the concentration of H⁺ ions using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Rearranging gives:
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-3} \, \text{mol/L} \]

### Step 2: Calculate the number of moles of H⁺ in the initial solution
We know the volume of the initial solution is 100 mL (or 0.1 L). The number of moles of H⁺ can be calculated using:
\[ \text{Number of moles} = [\text{H}^+] \times \text{Volume} = 10^{-3} \, \text{mol/L} \times 0.1 \, \text{L} = 10^{-4} \, \text{mol} \]

### Step 3: Calculate the total volume after dilution
The initial volume is 100 mL, and we are adding 500 mL of water. Therefore, the total volume after dilution is:
\[ \text{Total Volume} = 100 \, \text{mL} + 500 \, \text{mL} = 600 \, \text{mL} = 0.6 \, \text{L} \]

### Step 4: Calculate the new concentration of H⁺ ions after dilution
The concentration of H⁺ ions after dilution can be calculated using the number of moles and the total volume:
\[ [\text{H}^+]_{\text{new}} = \frac{\text{Number of moles}}{\text{Total Volume}} = \frac{10^{-4} \, \text{mol}}{0.6 \, \text{L}} = \frac{10^{-4}}{0.6} \approx 1.67 \times 10^{-4} \, \text{mol/L} \]

### Step 5: Calculate the new pH
Now we can find the new pH using the new concentration of H⁺ ions:
\[ \text{pH}_{\text{new}} = -\log[\text{H}^+]_{\text{new}} = -\log(1.67 \times 10^{-4}) \]
Calculating this gives:
\[ \text{pH}_{\text{new}} \approx 3.78 \]

### Final Answer
The new pH of the solution after dilution is approximately **3.78**.

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