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The problem asks to calculate the pH of a sodium hydroxide solution (NaOH) given the following details:

- **Mass of NaOH dissolved**: 699 mg
- **Volume of solution**: 180 mL
- **Temperature**: 25°C (though not directly needed for the calculation since pH for NaOH is not dependent on temperature).

### Step-by-Step Solution:

1. **Convert the mass of NaOH to moles**:
   The molar mass of NaOH is approximately 40.00 g/mol.
   \[
   \text{moles of NaOH} = \frac{\text{mass of NaOH (in grams)}}{\text{molar mass of NaOH}}
   \]
   First, convert 699 mg to grams:
   \[
   699 \, \text{mg} = 0.699 \, \text{g}
   \]
   Now calculate the moles of NaOH:
   \[
   \text{moles of NaOH} = \frac{0.699 \, \text{g}}{40.00 \, \text{g/mol}} = 0.017475 \, \text{mol}
   \]

2. **Calculate the concentration of NaOH**:
   The concentration is calculated by dividing the moles of NaOH by the volume of the solution (in liters).
   \[
   \text{Volume of solution} = 180 \, \text{mL} = 0.180 \, \text{L}
   \]
   Therefore, the concentration of NaOH is:
   \[
   \text{Concentration of NaOH} = \frac{0.017475 \, \text{mol}}{0.180 \, \text{L}} = 0.09708 \, \text{mol/L}
   \]

3. **Determine the concentration of OH⁻ ions**:
   NaOH is a strong base and dissociates completely in water, so the concentration of OH⁻ is the same as the concentration of NaOH:
   \[
   [\text{OH}^-] = 0.09708 \, \text{mol/L}
   \]

4. **Calculate the pOH**:
   The pOH is calculated using the formula:
   \[
   \text{pOH} = -\log[\text{OH}^-]
   \]
   \[
   \text{pOH} = -\log(0.09708) \approx 1.012
   \]

5. **Calculate the pH**:
   Since \( \text{pH} + \text{pOH} = 14 \) (at 25°C), we can calculate the pH as:
   \[
   \text{pH} = 14 - 1.012 \approx 12.99
   \]

### Final Answer:
The pH of the solution is approximately **12.99**, rounded to the correct number of significant figures based on the input data (3 significant figures).

---

Would you like more details or have any further questions? Here are five related questions to consider:

1. What happens to the pH if you increase the volume of the solution while keeping the mass of NaOH constant?
2. How would the pH change if a weaker base were used instead of NaOH?
3. Why does NaOH dissociate completely in water, and how does that affect its pH?
4. How does the temperature affect the dissociation of water and the pH of strong bases like NaOH?
5. How would you calculate the pH if the base were only partially dissociated?

**Tip:** Always check units when converting between milliliters and liters, and between milligrams and grams, to avoid calculation errors.

A chemist dissolves 699 mg of pure sodium hydroxide in enough water to make up 180 mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25°C.)

Learn how to calculate the pH of a sodium hydroxide (NaOH) solution where 699 mg of NaOH is dissolved in 180 mL of water. This solution involves concentration calculations, mole conversions, and logarithmic functions, suitable for college-level chemistry.

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