Let's update the calculations based on the correct concentration of \( \text{HCl} \), which is 0.15 mol/L. We'll follow the same steps as before:

### 1. Calculate the moles of \( \text{OH}^- \) ions from \( \text{Ca(OH)}_2 \):
Given:
- Volume of \( \text{Ca(OH)}_2 \) = 15 mL = 0.015 L
- Concentration of \( \text{Ca(OH)}_2 \) = 0.1 mol/L

Since \( \text{Ca(OH)}_2 \) dissociates into 2 \( \text{OH}^- \) ions:
\[
\text{Moles of } \text{OH}^- = 0.015 \, \text{L} \times 0.1 \, \text{mol/L} \times 2 = 0.003 \, \text{mol}
\]

### 2. Calculate the moles of \( \text{H}^+ \) ions from \( \text{HCl} \):
Given:
- Volume of \( \text{HCl} \) = 12 mL = 0.012 L
- Concentration of \( \text{HCl} \) = 0.15 mol/L

\[
\text{Moles of } \text{H}^+ = 0.012 \, \text{L} \times 0.15 \, \text{mol/L} = 0.0018 \, \text{mol}
\]

### 3. Determine the limiting reactant:
We compare the moles of \( \text{H}^+ \) and \( \text{OH}^- \). 

- \( \text{Moles of OH}^- = 0.003 \, \text{mol} \)
- \( \text{Moles of H}^+ = 0.0018 \, \text{mol} \)

Since \( \text{H}^+ \) is the limiting reactant, it will react completely with \( \text{OH}^- \).

Remaining \( \text{OH}^- \) after the reaction:
\[
\text{Remaining } \text{OH}^- = 0.003 \, \text{mol} - 0.0018 \, \text{mol} = 0.0012 \, \text{mol}
\]

### 4. Calculate the concentration of \( \text{OH}^- \) in the final solution:
Total volume of the solution:
\[
\text{Total Volume} = 15 \, \text{mL} + 12 \, \text{mL} = 27 \, \text{mL} = 0.027 \, \text{L}
\]
\[
\text{Concentration of } \text{OH}^- = \frac{0.0012 \, \text{mol}}{0.027 \, \text{L}} \approx 0.0444 \, \text{mol/L}
\]

### 5. Find the pOH and then the pH:
\[
\text{pOH} = -\log[\text{OH}^-] = -\log[0.0444] \approx 1.352
\]
\[
\text{pH} = 14 - \text{pOH} = 14 - 1.352 \approx 12.648
\]

The pH of the solution is approximately **12.648**.

Would you like more details on any of the steps? Here are some related questions:

1. How do the concentrations of reactants affect the final pH of the solution?
2. Why does \( \text{Ca(OH)}_2 \) provide twice the amount of \( \text{OH}^- \) ions compared to its concentration?
3. How would the pH change if the volumes of \( \text{HCl} \) and \( \text{Ca(OH)}_2 \) were swapped?
4. What is the importance of using a strong acid and base in this reaction?
5. How does the neutralization process impact the pH of a solution?

**Tip:** Always double-check the concentrations and volumes of your reactants before beginning a calculation, as these values are crucial for determining the correct pH.

Learn how to calculate pH from the concentration of HCl (0.15 mol/L) using detailed step-by-step calculations. This problem involves chemical equilibrium, acid-base reactions, and pH calculations, suitable for advanced high school and above.

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