How many ml of 0.200 M HCl are required to completely neutralize 50.0 ml of 0.150 M Ba(OH)2 solution?

To neutralize Ba(OH)₂, which dissociates into Ba²⁺ and 2 OH⁻ ions, we need to account for the 2:1 ratio of HCl to Ba(OH)₂. The moles of Ba(OH)₂ in 50.0 mL of 0.150 M solution are (0.150 , \text{M} \times 0.050 , \text{L} = 0.0075 , \text{mol}). Thus, (0.0075 , \text{mol} \times 2 = 0.015 , \text{mol}) of HCl is required. The volume of 0.200 M HCl needed is ( \frac{0.015 , \text{mol}}{0.200 , \text{M}} = 0.075 , \text{L} = 75.0 , \text{mL}).