The pH of a `10 ^(−10)` molar HCl solution is approximately

Answer

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

### Step 1: Understand the dissociation of HCl
HCl is a strong acid and dissociates completely in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
For a \(10^{-10}\) molar HCl solution, the concentration of \(\text{H}^+\) ions from HCl is:
\[ [\text{H}^+] = 10^{-10} \, \text{M} \]

### Step 2: Consider the contribution of water
Pure water also contributes \(\text{H}^+\) ions due to self-ionization:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]
At 25°C, the concentration of \(\text{H}^+\) ions from water is:
\[ [\text{H}^+] = 10^{-7} \, \text{M} \]

### Step 3: Calculate the total \(\text{H}^+\) concentration
The total concentration of \(\text{H}^+\) ions in the solution is the sum of the contributions from HCl and water:
\[ [\text{H}^+]_{\text{total}} = [\text{H}^+]_{\text{HCl}} + [\text{H}^+]_{\text{water}} = 10^{-10} + 10^{-7} \]
Since \(10^{-10}\) is much smaller than \(10^{-7}\), 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:
\[ \text{pH} = -\log(10^{-7}) = 7 \]
However, since we have a small contribution from HCl, we need to refine our calculation. The total concentration can be approximated as:
\[ [\text{H}^+]_{\text{total}} = 10^{-7} + 10^{-10} \approx 10^{-7}(1 + 10^{-3}) \approx 10^{-7} \times 1.001 \]
Thus, the pH becomes:
\[ \text{pH} \approx -\log(1.001 \times 10^{-7}) \approx 6.999 \]

### Conclusion
The pH of a \(10^{-10}\) molar HCl solution is approximately:
\[ \text{pH} \approx 6.99 \]