What is pH of `2xx10^(-8)` molar HCl solution ? Here log 2 = `0.301 and log 3 = 0.477`

Options

A. `5.4`
B. `7.7`
C. `6.92`
D. `9.5`

Explanation

To find the pH of a `2 x 10^(-8)` molar HCl solution, we need to consider both the H⁺ ions contributed by HCl and the H⁺ ions from the water itself, since the concentration of HCl is very low.

### Step-by-Step Solution:

1. **Identify the concentration of H⁺ ions from HCl**:
- The concentration of H⁺ ions from the HCl solution is `2 x 10^(-8)` M.

2. **Identify the concentration of H⁺ ions from water**:
- Pure water has a concentration of H⁺ ions of `1 x 10^(-7)` M at 25°C.

3. **Calculate the total concentration of H⁺ ions**:
- Since the HCl solution is very dilute, we need to add the H⁺ concentration from HCl and the H⁺ concentration from water:
\[ [H^+]_{total} = [H^+]_{HCl} + [H^+]_{water} = 2 \times 10^{-8} + 1 \times 10^{-7} = 1.2 \times 10^{-7} \text{ M} \]

4. **Calculate the pH**:
- The pH is calculated using the formula:
\[ pH = -\log[H^+]_{total} \]
- Substitute the total concentration of H⁺ ions:
\[ pH = -\log(1.2 \times 10^{-7}) \]

5. **Break down the logarithm**:
- We can express `1.2` as `4 x 3`:
\[ pH = -\log(1.2) + 7 \]
- Using the properties of logarithms:
\[ \log(1.2) = \log(4) + \log(3) = 2\log(2) + \log(3) \]
- Given that `log(2) = 0.301` and `log(3) = 0.477`, we can calculate:
\[ \log(4) = 2 \times 0.301 = 0.602 \]
\[ \log(1.2) = 0.602 + 0.477 = 1.079 \]

6. **Final calculation of pH**:
\[ pH = -1.079 + 7 = 6.921 \]
- Rounding gives us:
\[ pH \approx 6.92 \]

### Final Answer:
The pH of the `2 x 10^(-8)` molar HCl solution is approximately **6.92**.