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The problem in the image asks to calculate the pH of a solution where 233 mg of pure barium hydroxide \((Ba(OH)_2)\) is dissolved in enough water to make up 90 mL of solution, at 25°C.

### Step-by-step solution:

#### 1. **Molar mass of Barium Hydroxide \((Ba(OH)_2)\):**
   The molar mass of \(Ba(OH)_2\) is the sum of the atomic masses:
   - \(Ba\) = 137.33 g/mol
   - \(O\) = 16.00 g/mol (two oxygen atoms, so \(16.00 \times 2 = 32.00\))
   - \(H\) = 1.008 g/mol (two hydrogen atoms, so \(1.008 \times 2 = 2.016\))

   \[
   Molar \ mass \ of \ Ba(OH)_2 = 137.33 + 32.00 + 2.016 = 171.346 \ g/mol
   \]

#### 2. **Convert 233 mg to grams:**
   \[
   233 \ mg = 0.233 \ g
   \]

#### 3. **Moles of \(Ba(OH)_2\):**
   The number of moles of \(Ba(OH)_2\) is given by:

   \[
   \text{Moles of } Ba(OH)_2 = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{0.233 \ g}{171.346 \ g/mol} \approx 0.00136 \ mol
   \]

#### 4. **Concentration of \(Ba(OH)_2\) in the solution:**
   The volume of the solution is 90 mL, which is 0.090 L. The concentration (molarity) of \(Ba(OH)_2\) is:

   \[
   \text{Concentration of } Ba(OH)_2 = \frac{\text{moles}}{\text{volume (L)}} = \frac{0.00136 \ mol}{0.090 \ L} \approx 0.0151 \ M
   \]

#### 5. **Dissociation of Barium Hydroxide:**
   Barium hydroxide is a strong base and dissociates completely in water:
   \[
   Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-
   \]
   For each mole of \(Ba(OH)_2\), 2 moles of \(OH^-\) ions are produced. Therefore, the concentration of \(OH^-\) ions is:

   \[
   [OH^-] = 2 \times [Ba(OH)_2] = 2 \times 0.0151 \ M = 0.0302 \ M
   \]

#### 6. **Calculate pOH:**
   The pOH is calculated using the formula:

   \[
   \text{pOH} = -\log[OH^-] = -\log(0.0302) \approx 1.52
   \]

#### 7. **Calculate pH:**
   Since the temperature is 25°C, we can use the relation between pH and pOH:

   \[
   pH + pOH = 14
   \]
   \[
   pH = 14 - pOH = 14 - 1.52 = 12.48
   \]

Thus, the pH of the solution is **12.48**.

Would you like further details on any of the steps?

### Relative questions:
1. How does the dissociation of a strong base like \(Ba(OH)_2\) affect the pH?
2. What are the key differences in calculating the pH for weak versus strong bases?
3. How would temperature changes affect the dissociation constant and pH of the solution?
4. What is the relationship between pH, pOH, and the ion product of water (\(K_w\)) at different temperatures?
5. How does the molarity of a solution influence its pH and pOH?

### Tip:
Always ensure your significant figures match the precision given in the problem to avoid calculation errors.

A chemist dissolves 233 mg of pure barium hydroxide in enough water to make up 90 mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25°C.)

This problem involves calculating the pH of a solution with 233 mg of barium hydroxide dissolved in 90 mL of water at 25°C. The solution covers molarity, strong base dissociation, and logarithmic calculations for pH.

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