The pH of `10^(-8) M` NaOH will be

Options

A. -4
B. -10
C. 4
D. 7

Explanation

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

### Step 1: Understand the dissociation of NaOH
NaOH dissociates completely in water to give:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
This means that a `10^(-8) M` NaOH solution will produce `10^(-8) M` of OH⁻ ions.

### Step 2: Consider the autoionization of water
Water also contributes OH⁻ ions through its autoionization:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]
At 25°C, the concentration of H⁺ and OH⁻ from water is `10^(-7) M`.

### Step 3: Calculate the total concentration of OH⁻ ions
The total concentration of OH⁻ ions in the solution is the sum of the OH⁻ from NaOH and the OH⁻ from water:
\[ [\text{OH}^-]_{\text{total}} = [\text{OH}^-]_{\text{NaOH}} + [\text{OH}^-]_{\text{water}} \]
\[ [\text{OH}^-]_{\text{total}} = 10^{-8} + 10^{-7} \]

### Step 4: Simplify the total concentration
To add these concentrations, factor out `10^(-7)`:
\[ [\text{OH}^-]_{\text{total}} = 10^{-7}(1 + 0.1) \]
\[ [\text{OH}^-]_{\text{total}} = 10^{-7}(1.1) = 1.1 \times 10^{-7} \, M \]

### Step 5: Calculate pOH
Using the formula for pOH:
\[ \text{pOH} = -\log[\text{OH}^-] \]
\[ \text{pOH} = -\log(1.1 \times 10^{-7}) \]
Using properties of logarithms:
\[ \text{pOH} = -\log(1.1) - \log(10^{-7}) \]
\[ \text{pOH} \approx -0.041 - (-7) \]
\[ \text{pOH} \approx 7 - 0.041 \approx 6.959 \]

### Step 6: Calculate pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 6.959 \]
\[ \text{pH} \approx 7.041 \]

### Conclusion
The pH of a `10^(-8) M` NaOH solution is approximately 7.041, which can be rounded to 7.

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