What volume of the 0.200mole dm-3hcl is required to neutralize 25.0cm3 of 0.200 moldm-3 BaOH2?

Answer 1

To neutralize the Ba(OH)2, we need an equal number of moles of HCl. The molarity of Ba(OH)2 is 0.200 mol/dm^3, so the number of moles present in 25.0 cm^3 is (0.200 mol/dm^3) x (25.0 cm^3 / 1000 cm^3) = 0.005 mol. Since HCl and Ba(OH)2 neutralize each other in a 1:1 ratio, we need the same number of moles of HCl. Therefore, the volume of 0.200 mol/dm^3 HCl needed is (0.005 mol) / (0.200 mol/dm^3) = 0.025 dm^3, which is equivalent to 25.0 cm^3.