The `pH` of `10^(-8)M` solution of `HCl` in water is

Options

A. 8
B. 6
C. Between 6 and 7
D. Between 7 and 8

Answer

To find the pH of a \(10^{-8} M\) 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:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Given a \(10^{-8} M\) solution of HCl, the concentration of \(\text{H}^+\) ions contributed by HCl is \(10^{-8} M\).

### Step 2: Consider the autoionization of water
Pure water also contributes \(\text{H}^+\) ions due to 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 \(\text{H}^+\)
To find the total concentration of \(\text{H}^+\) ions in the solution, we need to add the contributions from both HCl and water:
\[ [\text{H}^+]_{\text{total}} = [\text{H}^+]_{\text{HCl}} + [\text{H}^+]_{\text{water}} = 10^{-8} M + 10^{-7} M \]
This can be simplified:
\[ [\text{H}^+]_{\text{total}} = 10^{-7} M + 10^{-8} M = 10^{-7} M (1 + 0.1) = 1.1 \times 10^{-7} M \]

### Step 4: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the total concentration of \(\text{H}^+\):
\[ \text{pH} = -\log(1.1 \times 10^{-7}) \]
Using logarithmic properties:
\[ \text{pH} = -\log(1.1) - \log(10^{-7}) = -\log(1.1) + 7 \]
Calculating \(-\log(1.1)\) (approximately 0.0414):
\[ \text{pH} \approx 7 - 0.0414 \approx 6.9586 \]

### Step 5: Final pH value
Rounding the value, we get:
\[ \text{pH} \approx 6.95 \]

### Conclusion
Thus, the pH of a \(10^{-8} M\) solution of HCl in water is approximately **6.95**.