The pH of `10^(-8)` M acid solution lies between .........and .............. .

Answer

To find the pH of a `10^(-8)` M acid solution, we need to consider the contribution of hydrogen ions from both the acid and the water itself. Here’s a step-by-step solution:

### Step 1: Understanding pH and H⁺ Concentration
The pH of a solution is defined as:
\[ \text{pH} = -\log[H^+] \]
where \([H^+]\) is the concentration of hydrogen ions in moles per liter.

### Step 2: Initial H⁺ Concentration from the Acid
For a `10^(-8)` M acid solution, the initial concentration of H⁺ ions from the acid is:
\[ [H^+]_{\text{acid}} = 10^{-8} \, \text{M} \]

### Step 3: H⁺ Concentration from Water
Pure water also contributes H⁺ ions. The concentration of H⁺ ions in pure water at 25°C is:
\[ [H^+]_{\text{water}} = 10^{-7} \, \text{M} \]

### Step 4: Total H⁺ Concentration
Since the acid concentration is very low, we must consider the contribution from both the acid and the water:
\[ [H^+]_{\text{total}} = [H^+]_{\text{acid}} + [H^+]_{\text{water}} \]
\[ [H^+]_{\text{total}} = 10^{-8} + 10^{-7} \]
\[ [H^+]_{\text{total}} = 1.1 \times 10^{-7} \, \text{M} \]

### Step 5: Calculating pH
Now we can calculate the pH using the total H⁺ concentration:
\[ \text{pH} = -\log(1.1 \times 10^{-7}) \]

Using logarithmic properties:
\[ \text{pH} \approx 7 - \log(1.1) \]

Calculating \(-\log(1.1)\):
- \(-\log(1.1) \approx -0.041\) (approximately)

Thus:
\[ \text{pH} \approx 7 - 0.041 \]
\[ \text{pH} \approx 6.96 \]

### Step 6: Conclusion
The pH of the `10^(-8)` M acid solution lies between:
\[ 6 \text{ and } 7 \]