What volume of 0.500 M HCl is required to neutralize 25.0 mL of 0.200 M Ca(OH)₂?
Q1. What volume of 0.500 M HCl is required to neutralize 25.0 mL of 0.200 M Ca(OH)₂?
Answer: 40.0 mL
Explanation: 2×0.200×0.025 /0.500 = 0.04 L = 40.0 mL. B ignores 2:1 molar ratio.