At `25^(@)C`, pH of a `10^(-8)M` aqueous KOH solution will be

Options

A. `6.0`
B. 7.01
C. 8.02
D. 9.02

Explanation

To determine the pH of a \(10^{-8} M\) aqueous KOH solution at \(25^\circ C\), we can follow these steps:

### Step 1: Understand the Ionization of KOH
KOH is a strong base and will completely dissociate in water:
\[ \text{KOH} \rightarrow \text{K}^+ + \text{OH}^- \]
Since the concentration of KOH is \(10^{-8} M\), the concentration of \(\text{OH}^-\) produced from KOH will also be \(10^{-8} M\).

### Step 2: Consider the Contribution of Water
Pure water also contributes \(\text{OH}^-\) ions. At \(25^\circ C\), the concentration of \(\text{OH}^-\) from water is:
\[ [\text{OH}^-] = 10^{-7} M \]
Thus, the total concentration of \(\text{OH}^-\) ions in the solution will be:
\[ [\text{OH}^-]_{\text{total}} = [\text{OH}^-]_{\text{KOH}} + [\text{OH}^-]_{\text{water}} = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} M \]

### Step 3: Calculate the pOH
Now, we can calculate the pOH using the total concentration of \(\text{OH}^-\):
\[ \text{pOH} = -\log(1.1 \times 10^{-7}) \]
Using logarithmic properties, we can approximate:
\[ \text{pOH} \approx 7 - \log(1.1) \approx 7 - 0.041 = 6.959 \]

### Step 4: Calculate the pH
Finally, we can find the pH using the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
Thus,
\[ \text{pH} = 14 - \text{pOH} = 14 - 6.959 \approx 7.041 \]

### Conclusion
Therefore, the pH of a \(10^{-8} M\) aqueous KOH solution at \(25^\circ C\) is approximately **7.04**.

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