OK, I suppose I need to fully explain myself here
Barium Hydroxide is a strong base. So the ionic equation is
Ba2+ + 2OH- + 2H+ + 2Cl ---> Ba2+ + 2Cl- +2 H20
The net ionic equation is then
2OH- + 2H+ ---> 2H2O
Which since all of the coeficiencts are the same is
H+ + OH- ---> H2O
Which gives you the correct stoichiometry here
Now, the moles of H+ used is the same as the moles of HCl used in your tittration, which is, as calculated by stewed_ant is 0.0035mol. This is equal to the moles of OH- neutralized. So the molarity of the OH- is 0.14M, which agrees with the book.
Now here is the thing. That is the molarity of the OH- in solution, not molarity of Ba(OH)2. Using stoichiometry, the molarity of Barium hydroxide is half of the molarity of OH-, which is, as Stewed_Ant calculated is 0.07M. So what you should do konichiwa2x is check what the book is asking for. Is it molarity of hydroxide or molarity of barium hydroxide, which are two different values. The answer in the book is for molarity of hydroxide ions not barium hydroxide. So it is the book's mistake not mine. I merely calculated the molarity of hydroxide ions. So please read your question again to se what is being asked exactly. It is not unusual for the book to mix things up
Stewed_ant, you were right for the question that konichiwa2x posted. I was merely trying to diagnose the desceptancies in answers here. My equation is the well used dilution equation that can be used for 1:1 stoichiometry instances (like a lot of acid-base neutraliztions). I know, after five years of chemistry (I am a grad. student) I have used it a million times, I am surprised you never seen it.
Cheers to you both