# Can someone check my work? Neutralization reaction.

1. Aug 24, 2007

### LakeMountD

1. The problem statement, all variables and given/known data

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.

2. Aug 24, 2007

### 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...

3. Aug 24, 2007

### Gokul43201

Staff Emeritus
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.