`10^(-5)M HCI` solution at `25^(@)C` is diluted 1000 times. The pH of the diluted solution will

Options

A. be equal to 8
B. lie between 7 and 8
C. lie between 6 and 7
D. remain unchanged

Answer

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

### Step 1: Determine the initial concentration of HCl
The initial concentration of HCl is given as \( 10^{-5} \, M \).

### Step 2: Calculate the pH of the initial solution
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
For the initial concentration:
\[ \text{pH} = -\log(10^{-5}) = 5 \]

### Step 3: Dilute the solution
The solution is diluted 1000 times. When a solution is diluted, the concentration of the solute decreases. The new concentration \( M_2 \) can be calculated using the dilution formula:
\[ M_1V_1 = M_2V_2 \]
Here, \( V_2 = 1000V_1 \) (since it is diluted 1000 times). Thus:
\[ M_2 = \frac{M_1V_1}{V_2} = \frac{10^{-5} \cdot V_1}{1000 \cdot V_1} = 10^{-8} \, M \]

### Step 4: Consider the contribution of \( H^+ \) ions from water
Since \( HCl \) is a strong acid, it fully dissociates in solution. However, at very low concentrations, we must also consider the \( H^+ \) ions contributed by the autoionization of water, which is \( 10^{-7} \, M \).

### Step 5: Calculate the total concentration of \( H^+ \) ions
The total concentration of \( H^+ \) ions in the solution after dilution will be:
\[ [H^+]_{\text{total}} = [H^+]_{\text{from HCl}} + [H^+]_{\text{from water}} = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \, M \]

### Step 6: Calculate the new pH
Now, we can calculate the new pH using the total concentration of \( H^+ \):
\[ \text{pH} = -\log(1.1 \times 10^{-7}) \]
Using the logarithmic property:
\[ \text{pH} = -(-7 + \log(1.1)) \approx 7 - 0.04 = 6.96 \]

### Final Answer
Thus, the pH of the diluted solution is approximately **6.96**, which lies between 6 and 7.