The pH of a `10^(-9)`M solution of HCl in water is :

Options

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

Explanation

To find the pH of a \(10^{-9}\) M solution of HCl in water, 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 dissociation of HCl
HCl is a strong acid and dissociates completely in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
This means that a \(10^{-9}\) M solution of HCl will contribute \(10^{-9}\) M of \(\text{H}^+\) ions.

### Step 2: Consider the autoionization of water
Water also contributes \(\text{H}^+\) ions through its autoionization:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]
At 25°C, the concentration of \(\text{H}^+\) ions from water is \(10^{-7}\) M.

### Step 3: Calculate the total concentration of \(\text{H}^+\)
The total concentration of \(\text{H}^+\) ions in the solution is the sum of the contributions from HCl and water:
\[ [\text{H}^+]_{\text{total}} = [\text{H}^+]_{\text{HCl}} + [\text{H}^+]_{\text{water}} = 10^{-9} + 10^{-7} \]
To add these, we can factor out \(10^{-7}\):
\[ [\text{H}^+]_{\text{total}} = 10^{-7} (1 + 0.01) = 1.01 \times 10^{-7} \text{ M} \]

### Step 4: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the total \(\text{H}^+\) concentration:
\[ \text{pH} = -\log(1.01 \times 10^{-7}) \]
Using the logarithmic property \(\log(ab) = \log a + \log b\):
\[ \text{pH} = -\log(1.01) - \log(10^{-7}) = -\log(1.01) + 7 \]
Since \(\log(1.01)\) is slightly greater than 0 (approximately 0.0043), we can approximate:
\[ \text{pH} \approx 7 - 0.0043 \approx 6.9957 \]
Thus, the pH is approximately:
\[ \text{pH} \approx 6.99 \]

### Conclusion
The pH of a \(10^{-9}\) M solution of HCl in water is approximately 6.99.

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