`10^(-6)M HCl` is diluted to `100` times. Its `pH` is:

Options

A. `6.0`
B. `8.0`
C. `6.95`
D. `9.5`

Explanation

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

### Step 1: Determine the initial concentration of HCl
The initial concentration of HCl is given as \(10^{-6} \, M\).

### Step 2: Calculate the concentration after dilution
Since the solution is diluted 100 times, the new concentration of HCl can be calculated as follows:
\[ \text{Concentration after dilution} = \frac{10^{-6} \, M}{100} = 10^{-8} \, M \]

### Step 3: Account for the contribution of H\(^+\) ions from water
In pure water at 25°C, the concentration of H\(^+\) ions is \(10^{-7} \, M\). Since we are dealing with a solution, we need to consider the contribution from both HCl and water.

### Step 4: Calculate the total concentration of H\(^+\) ions
The total concentration of H\(^+\) ions in the solution is the sum of the H\(^+\) ions from HCl and from water:
\[ \text{Total } [H^+] = [H^+] \text{ from HCl} + [H^+] \text{ from water} = 10^{-8} \, M + 10^{-7} \, M \]
\[ \text{Total } [H^+] = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \, M \]

### Step 5: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the total concentration of H\(^+\):
\[ \text{pH} = -\log(1.1 \times 10^{-7}) \]

Using logarithmic properties:
\[ \text{pH} = -\log(1.1) - \log(10^{-7}) = -\log(1.1) + 7 \]

Since \(\log(1.1) \approx 0.041\):
\[ \text{pH} \approx 7 - 0.041 = 6.959 \]

### Step 6: Final result
Rounding to two decimal places, the pH of the solution is approximately:
\[ \text{pH} \approx 6.96 \]

### Summary
The pH of the \(10^{-6} \, M\) HCl solution diluted 100 times is approximately **6.96**.

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