pH of `10^(-8)` N NaOH is

Options

A. `8.0`
B. `6.0`
C. `6.98`
D. `7.02`

Explanation

To find the pH of a `10^(-8)` N NaOH solution, we can follow these steps:

### Step 1: Convert Normality to Molarity
Since NaOH is a strong base and dissociates completely in solution, the molarity (M) of NaOH is the same as its normality (N). Thus, we have:
\[ \text{Molarity of NaOH} = 10^{-8} \, \text{M} \]

### Step 2: Calculate the Concentration of OH⁻ Ions
In a solution of NaOH, the concentration of hydroxide ions (OH⁻) will also be:
\[ [\text{OH}^-] = 10^{-8} \, \text{M} \]

### Step 3: Consider the Contribution of Water
Water also dissociates into H⁺ and OH⁻ ions, contributing to the overall concentration of OH⁻ in the solution. The dissociation of water at 25°C gives:
\[ K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \]
At neutral pH, the concentration of OH⁻ from water is:
\[ [\text{OH}^-]_{H_2O} = 10^{-7} \, \text{M} \]

### Step 4: Calculate Total OH⁻ Concentration
The total concentration of OH⁻ ions in the solution is the sum of the OH⁻ from NaOH and from water:
\[ [\text{OH}^-]_{total} = [\text{OH}^-]_{NaOH} + [\text{OH}^-]_{H_2O} \]
\[ [\text{OH}^-]_{total} = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \, \text{M} \]

### Step 5: Calculate the Concentration of H⁺ Ions
Using the ion product of water:
\[ K_w = [\text{H}^+][\text{OH}^-] \]
We can rearrange this to find the concentration of H⁺ ions:
\[ [\text{H}^+] = \frac{K_w}{[\text{OH}^-]_{total}} \]
Substituting the values:
\[ [\text{H}^+] = \frac{10^{-14}}{1.1 \times 10^{-7}} \approx 9.1 \times 10^{-8} \, \text{M} \]

### Step 6: Calculate the pH
Finally, we can calculate the pH using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the value we found:
\[ \text{pH} = -\log(9.1 \times 10^{-8}) \]
Calculating this gives:
\[ \text{pH} \approx 7.04 \]

### Final Result
Thus, the pH of `10^(-8)` N NaOH solution is approximately **7.04**.

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