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The problem asks for the **pH of a solution** where **318 mg of pure potassium hydroxide (KOH)** is dissolved in **80.0 mL of water** at **25°C**. Let’s walk through the solution step-by-step.

---

### Step 1: Convert the mass of KOH to moles  
The **molar mass of KOH** is:
\[
M(KOH) = 39.1 + 16.0 + 1.0 = 56.1 \, \text{g/mol}
\]

Now, convert **318 mg** (or **0.318 g**) of KOH into moles:
\[
\text{moles of KOH} = \frac{0.318 \, \text{g}}{56.1 \, \text{g/mol}} = 5.67 \times 10^{-3} \, \text{mol}
\]

---

### Step 2: Find the concentration of OH⁻ ions  
Since **KOH dissociates completely** in water:
\[
\text{KOH} \rightarrow \text{K}^+ + \text{OH}^-
\]
The concentration of OH⁻ will be the same as the concentration of KOH.

Now, calculate the molarity (M) of the solution:
\[
\text{Molarity (OH⁻)} = \frac{\text{moles of KOH}}{\text{volume in liters}}
= \frac{5.67 \times 10^{-3} \, \text{mol}}{80.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}}}
= 0.0709 \, \text{M}
\]

---

### Step 3: Calculate the pOH  
Using the definition of **pOH**:
\[
\text{pOH} = -\log[\text{OH}^-]
\]
\[
\text{pOH} = -\log(0.0709) \approx 1.15
\]

---

### Step 4: Calculate the pH  
Since:
\[
\text{pH} + \text{pOH} = 14 \quad \text{(at 25°C)}
\]
\[
\text{pH} = 14 - 1.15 = 12.85
\]

---

### Final Answer
The **pH of the solution** is:
\[
\boxed{12.85}
\]

---

Would you like further details or have any questions?  

Here are 5 related questions to explore:  
1. What would happen to the pH if you used only 40 mL of water?  
2. How does temperature affect the pH of water-based solutions?  
3. Why does KOH fully dissociate in water?  
4. How do you calculate the pH of weak bases?  
5. What are the applications of potassium hydroxide solutions in chemistry?  

**Tip:** Remember, for strong bases like KOH, the concentration of OH⁻ equals the concentration of the base since they dissociate completely.

A chemist dissolves 318 mg of pure potassium hydroxide in enough water to make up 80.0 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 solution where 318 mg of potassium hydroxide (KOH) is dissolved in 80.0 mL of water at 25°C. This solution involves molarity calculations, pOH, and pH relations, covering essential stoichiometry and solution chemistry concepts. Ideal for Grades 10-12.

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