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We are tasked with calculating the pH of a solution containing potassium hydroxide (KOH). Here's a breakdown of the steps:

### Step 1: Determine the number of moles of KOH.
Given that the mass of KOH is \(528 \, \text{mg}\), we need to convert this to grams:
\[
528 \, \text{mg} = 0.528 \, \text{g}
\]

The molar mass of KOH is:
\[
M_{\text{KOH}} = 39.10 \, \text{g/mol (K)} + 15.999 \, \text{g/mol (O)} + 1.008 \, \text{g/mol (H)} = 56.107 \, \text{g/mol}
\]

So, the number of moles of KOH is:
\[
\text{moles of KOH} = \frac{0.528 \, \text{g}}{56.107 \, \text{g/mol}} = 0.00941 \, \text{mol}
\]

### Step 2: Calculate the molarity of the solution.
The volume of the solution is \(120 \, \text{mL} = 0.120 \, \text{L}\).

The molarity (concentration in mol/L) is:
\[
\text{Molarity} = \frac{0.00941 \, \text{mol}}{0.120 \, \text{L}} = 0.07842 \, \text{M}
\]

### Step 3: Calculate the pOH.
Potassium hydroxide (KOH) is a strong base, which dissociates completely in water:
\[
\text{KOH} \rightarrow \text{K}^+ + \text{OH}^-
\]
Thus, the concentration of \(\text{OH}^-\) ions is equal to the molarity of the KOH solution, which is \(0.07842 \, \text{M}\).

The pOH is calculated as:
\[
\text{pOH} = -\log[\text{OH}^-] = -\log[0.07842] = 1.105
\]

### Step 4: Calculate the pH.
Using the relationship between pH and pOH:
\[
\text{pH} + \text{pOH} = 14
\]
\[
\text{pH} = 14 - 1.105 = 12.895
\]

Thus, the pH of the solution is approximately **12.9** (rounded to three significant digits).

---

Would you like any further clarification? Here are some related questions to deepen your understanding:

1. How does the strength of KOH as a base affect the pH of the solution?
2. What happens to the pH if more KOH is dissolved in the same volume?
3. How does temperature impact the dissociation of KOH in water?
4. Can the pH of this solution be acidic under any conditions?
5. What are the practical applications of calculating pH for industrial or laboratory purposes?

**Tip:** When dealing with strong acids or bases, remember that they fully dissociate in solution, simplifying calculations!

A chemist dissolves 528 mg of pure potassium hydroxide in enough water to make up 120 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 containing 528 mg of potassium hydroxide dissolved in 120 mL of water. This problem covers key concepts such as molarity, pH, and pOH calculation, suitable for high school chemistry students.

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