What is the pH of a `1.0 xx10^(-8) M` solution of NaOH?

Answer

To find the pH of a `1.0 x 10^(-8) M` solution of NaOH, we need to consider both the hydroxide ions contributed by NaOH and the hydroxide ions from the water itself. Here’s a step-by-step solution:

### Step 1: Understand the dissociation of NaOH
NaOH is a strong base and dissociates completely in water:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Thus, a `1.0 x 10^(-8) M` solution of NaOH will contribute `1.0 x 10^(-8) M` of OH⁻ ions.

### Step 2: Consider the contribution of OH⁻ from water
Pure water also contributes hydroxide ions:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]
At 25°C, the concentration of OH⁻ from water is:
\[ [\text{OH}^-] = 1.0 x 10^{-7} M \]

### Step 3: Calculate the total concentration of OH⁻
Now, we need to add the concentration of OH⁻ from NaOH and from water:
\[ [\text{OH}^-]_{\text{total}} = [\text{OH}^-]_{\text{NaOH}} + [\text{OH}^-]_{\text{water}} \]
\[ [\text{OH}^-]_{\text{total}} = 1.0 x 10^{-8} + 1.0 x 10^{-7} \]
To add these, we can express `1.0 x 10^{-7}` in terms of `10^{-8}`:
\[ [\text{OH}^-]_{\text{total}} = 1.0 x 10^{-8} + 10.0 x 10^{-8} = 11.0 x 10^{-8} \]
Thus,
\[ [\text{OH}^-]_{\text{total}} = 1.1 x 10^{-7} M \]

### Step 4: Calculate pOH
To find pOH, we use the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the total concentration of OH⁻:
\[ \text{pOH} = -\log(1.1 x 10^{-7}) \]

Using a calculator:
\[ \text{pOH} \approx 6.96 \]

### Step 5: Calculate pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
We can find pH:
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 6.96 \]
\[ \text{pH} \approx 7.04 \]

### Final Answer
The pH of a `1.0 x 10^(-8) M` solution of NaOH is approximately **7.04**.

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