What is the pH of a solution prepared by mixing 100.00 mL of 0.025 M Ca(OH)₂ with 50.00 mL of 0.080 M LiOH? Assume that the volumes are additive. A)12.67 B)12.78 C)12.95 D)13.25

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The solution involves calculating the total moles of OH- ions provided by both Ca(OH)2 and LiOH, then finding the concentration of OH- in the final solution, and finally calculating the pH from the pOH.1. Calculate moles of OH- from Ca(OH)2:$$ \text{Moles of Ca(OH)2} = 0.025 \, \text{M} \times 0.100 \, \text{L} = 0.0025 \, \text{moles} $$Since each unit of Ca(OH)2 provides 2 OH- ions:$$ \text{Moles of OH- from Ca(OH)2} = 0.0025 \, \text{moles} \times 2 = 0.005 \, \text{moles} $$2. Calculate moles of OH- from LiOH:$$ \text{Moles of LiOH} = 0.080 \, \text{M} \times 0.050 \, \text{L} = 0.004 \, \text{moles} $$LiOH provides 1 OH- ion per molecule, so:$$ \text{Moles of OH- from LiOH} = 0.004 \, \text{moles} $$3. Total moles of OH-:$$ \text{Total moles of OH-} = 0.005 \, \text{moles} + 0.004 \, \text{moles} = 0.009 \, \text{moles} $$4. Total volume of the solution:$$ \text{Total volume} = 100.00 \, \text{mL} + 50.00 \, \text{mL} = 150.00 \, \text{mL} = 0.150 \, \text{L} $$5. Concentration of OH- in the solution:$$ \text{[OH-]} = \frac{0.009 \, \text{moles}}{0.150 \, \text{L}} = 0.060 \, \text{M} $$6. Calculate pOH and then pH:$$ \text{pOH} = -\log(0.060) \approx 1.22 $$$$ \text{pH} = 14 - \text{pOH} = 14 - 1.22 = 12.78 $$Therefore, the pH of the solution is approximately 12.78.

What Is the PH of a Solution Prepared by Mixing