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.