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Can someone check my work? Neutralization reaction.

  1. Aug 24, 2007 #1
    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. jcsd
  3. Aug 24, 2007 #2
    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...
  4. Aug 24, 2007 #3


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    Staff Emeritus
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    Gold Member

    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.
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