The pH of a ` 10 ^(-8) ` molar solution of HCl in water is

Options

A. 8
B. `-8`
C. between7 and 8
D. between 6 and 7 .

Explanation

To find the pH of a \( 10^{-8} \) molar solution of HCl in water, we need to consider the contributions of hydronium ions from both the HCl and the water itself. Here’s a step-by-step solution:

### Step 1: Understand the contribution of HCl
HCl is a strong acid and completely dissociates in water. Therefore, a \( 10^{-8} \) molar solution of HCl will contribute \( 10^{-8} \) moles of \( H^+ \) ions per liter.

### Step 2: Consider the contribution of water
Pure water has a pH of 7, which corresponds to a hydronium ion concentration of \( 10^{-7} \) moles per liter. This means that even in pure water, there are \( 10^{-7} \) moles of \( H^+ \) ions present.

### Step 3: Calculate the total hydronium ion concentration
To find the total hydronium ion concentration in the solution, we need to add the contributions from HCl and water:
\[ [H^+]_{\text{total}} = [H^+]_{\text{HCl}} + [H^+]_{\text{water}} = 10^{-8} + 10^{-7} \]
To add these, we can express \( 10^{-7} \) in terms of \( 10^{-8} \):
\[ 10^{-7} = 10 \times 10^{-8} \]
Thus,
\[ [H^+]_{\text{total}} = 10^{-8} + 10 \times 10^{-8} = 11 \times 10^{-8} \]

### Step 4: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the total hydronium ion concentration:
\[ \text{pH} = -\log(11 \times 10^{-8}) \]
Using the logarithmic property:
\[ \text{pH} = -(\log 11 + \log(10^{-8})) = -\log 11 + 8 \]
We know that \( \log 11 \approx 1.04 \), so:
\[ \text{pH} = -1.04 + 8 = 6.96 \]

### Final Answer
The pH of a \( 10^{-8} \) molar solution of HCl in water is approximately **6.96**.