An aqueous solution of `HCl` is `10^(-9) M HCl`. The `pH` of the solution should be

Options

A. `9`
B. Between `6` and `7`
C. `7`
D. Unpredictable

Explanation

To find the pH of a `10^(-9) M` HCl solution, we need to consider both the contribution of HCl and the autoionization of water. Here’s a step-by-step solution:

### Step 1: Understand the Contribution of HCl
HCl is a strong acid and dissociates completely in water. Therefore, a `10^(-9) M` solution of HCl will provide `10^(-9) M` of H⁺ ions.

### Step 2: Consider the Autoionization of Water
Water also contributes H⁺ ions due to its autoionization:
\[ H_2O \rightleftharpoons H^+ + OH^- \]
At 25°C, the concentration of H⁺ and OH⁻ from water is `10^(-7) M`.

### Step 3: Calculate the Total H⁺ Concentration
The total concentration of H⁺ ions in the solution will be the sum of H⁺ from HCl and H⁺ from water:
\[ [H^+]_{total} = [H^+]_{HCl} + [H^+]_{H_2O} \]
Substituting the values:
\[ [H^+]_{total} = 10^{-9} + 10^{-7} \]
Since `10^(-7)` is much larger than `10^(-9)`, we can approximate:
\[ [H^+]_{total} \approx 10^{-7} \text{ (from water)} \]

### Step 4: Add the Contributions
Now, we need to add the contributions:
\[ [H^+]_{total} = 10^{-9} + 10^{-7} \approx 10^{-7} + 10^{-7} = 1.01 \times 10^{-7} \text{ M} \]

### Step 5: Calculate the pH
Now we can calculate the pH using the formula:
\[ pH = -\log[H^+] \]
Substituting the total H⁺ concentration:
\[ pH = -\log(1.01 \times 10^{-7}) \]
Using the logarithmic property:
\[ pH = -\log(1.01) - \log(10^{-7}) \approx -0.00432 + 7 \approx 6.99568 \approx 6.9 \]

### Final Answer
Thus, the pH of the solution is approximately **6.9**.

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