H3PO4 and Ba(OH)2 neutralization stoichiometry

[LakeMountD]

Homework Statement

2 H3PO4 + 3 Ba(OH)2 ---> Ba3(PO4)2 + 6H2O

What volume of 0.2M H3PO4 will be required to completely neutralize 100 mL of 0.6M Ba(OH)2.

What I did was used normality and said that there are 2*3 = 6N H+ ions and 3*2 = 6N OH ions. Therefore, I multiplied the normals by molarity and got:

(6N * 0.2M) * x = (6N * 0.6M) * 100mL
1.2M * x = 3.6M * 100mL
x = (3.6M * 100mL) / 1.2M = 300 mL

Book says 200mL but there are A LOT of errors in this book and I don't think they accounted for the extra protons.

The main thing I am wondering here is why the book doesn't account for the fact that there are 3 protons for every 2 hydroxyls and since they are at a 3:2 ratio, they are are in even quantities. 3 moles of H+ * 2 moles of compound and 2 moles of OH- and 3 moles of compound.

 

[Leong]

According to the equation:
2 mols of H3PO4 will neutralize 3 mols of Ba(OH)2

0.2(? many ml of H3PO4) will neutralize 0.6(100) ml of Ba(OH)2...

 

[Gokul43201]

The textbook answer is correct. You do not want to find the moles of protons per hydroxyl, but the moles of H3PO4 for the given moles of base.