What volume of 0.2 M Ba(OH)2 must be added to 300 mL of a 0.08 M HCl solution to get a solution in which the molarity of hydroxyl OH- ions is 0.08 M ?

Let volume of Ba(OH)2 solution taken be V liters.∴nBaOH2=0.2×VBaOH2+2HCl→BaCl2+2H2OnHCl=0.3×0.08= 0.024(By stoichimetry)∴nBaOH2 left =0.2V−120.024=0.2V−0.012 molesTotal volume of solution =0.3+V liters Molarity of  OH−ion=0.2V−0.012×20.3+V= 0.08On solving, ∴V=0.15 litre      =150​ mL