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 from the HCl and the autoionization of water. Here’s a step-by-step solution:

### Step 1: Understand the contributions to H⁺ concentration
HCl is a strong acid and will dissociate completely in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
In a `10^(-9)` M solution of HCl, the concentration of H⁺ ions contributed by HCl is `10^(-9)` M.

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

### Step 3: Calculate the total H⁺ concentration
Now, we need to add the contributions from both the HCl and the water:
\[ [\text{H}^+]_{\text{total}} = [\text{H}^+]_{\text{HCl}} + [\text{H}^+]_{\text{water}} = 10^{-9} + 10^{-7} \]
Since `10^(-7)` is much larger than `10^(-9)`, we can approximate:
\[ [\text{H}^+]_{\text{total}} \approx 10^{-7} \text{ M} \]

### Step 4: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the total concentration of H⁺ ions:
\[ \text{pH} = -\log(1.01 \times 10^{-7}) \]
Calculating this:
\[ \text{pH} \approx -\log(10^{-7}) - \log(1.01) \approx 7 - 0.004 \approx 6.996 \]
Thus, we can round this to:
\[ \text{pH} \approx 6.99 \]

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

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