pH of a `1.0xx10^(-8)` M solution of HCl is

Options

A. 7.02
B. 6.958
C. 7.4
D. 6.8

Answer

To find the pH of a `1.0 x 10^(-8)` M solution of HCl, we need to consider both the contribution of HCl and the autoionization of water. Here’s a step-by-step solution:

### Step 1: Identify the concentration of H⁺ from HCl
The concentration of H⁺ ions from HCl is given as:
\[ [H^+]_{HCl} = 1.0 \times 10^{-8} \, \text{M} \]

### Step 2: Identify the concentration of H⁺ from water
In pure water, the concentration of H⁺ ions is:
\[ [H^+]_{H_2O} = 1.0 \times 10^{-7} \, \text{M} \]
This is due to the autoionization of water.

### Step 3: Calculate the total concentration of H⁺ ions
Since both contributions are significant in this very dilute solution, we need to add them together:
\[ [H^+]_{total} = [H^+]_{HCl} + [H^+]_{H_2O} \]
\[ [H^+]_{total} = 1.0 \times 10^{-8} + 1.0 \times 10^{-7} \]

### Step 4: Simplify the total concentration
To simplify the addition, we can express \( 1.0 \times 10^{-8} \) as \( 0.1 \times 10^{-7} \):
\[ [H^+]_{total} = 0.1 \times 10^{-7} + 1.0 \times 10^{-7} = 1.1 \times 10^{-7} \, \text{M} \]

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

### Step 6: Apply logarithmic properties
Using the properties of logarithms:
\[ \text{pH} = -\log(1.1) - \log(10^{-7}) \]
\[ \text{pH} = -\log(1.1) + 7 \]

### Step 7: Calculate the value of \(-\log(1.1)\)
Using a calculator, we find:
\[ -\log(1.1) \approx -0.0414 \]
Thus:
\[ \text{pH} \approx 7 - 0.0414 = 6.9586 \]

### Final Result
Rounding to three significant figures, we get:
\[ \text{pH} \approx 6.96 \]