The balanced chemical reaction of the HCl reacting with `BaOH_2` is:
`2HCl + BaOH_2 -> BaCl_2 + 2H_2O`
In reactions involving titration first determine the number of H+ and OH- ions liberated by a molecule of the acid and base respectively. Here, one molecule of HCl ionizes to give one ion of H+ and one molecule of BaOH2 forms 2 ions of OH-
142 mL of 0.25 HCl releases 0.142*0.25 = 0.0355 moles of H+. These react with the OH- ions released by the `BaOH_2` to produce water. 0.0355 moles of OH- ions are released by 0.01775 moles of `BaOH_2` . If the molarity of `BaOH_2` is M, M*0.240 = 0.01775
=> M = 0.0739
The molarity of `BaOH_2` is approximately 0.0739
You must first recognize that this is a titration problem between a strong acid and a strong base. Strong acids/bases are strong electrolytes and dissociate completely. The molarity of the solutions equals the molarity of the ions. Thus, the pH is 7 and the solution is neutralized when the number of moles of H+ equals the number of moles of OH-. The equation can be seen as:
`["H"^+]("volume"_"A") = ["OH"^-]("volume"_"B")`
Plug in your known variables...
Note that you do not need to convert mL to L because the units are the same and are proportional. Now, you can solve for the molarity of the base.
`["OH"^-] = 0.148 "M"`
Because Ba(OH)2 dissociates into one Ba2+ ion and two OH- ions, you must divide the molarity of OH- in half. Therefore, the molarity of Ba(OH)2 is:
`["Ba(OH)_2"] = 0.74"M"`
Note that there are two sig figs!