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Ca(H2PO4)2 + NaOH(limited) → CaHPO4 + ?

  1. Oct 8, 2009 #1
    This is a problem I was trying to help some chem students with. This is all that was given.

    Ca(H2PO4)2 + NaOH(limited) → CaHPO4 +

    Will one hydrogen be neutralized on each dihydrogen phosphate to yield Na2HPO4 and water? This is my guess so that the two phosphate ions stay the same/get the same neutralization. It's hard to say what those four hydrogens and limited NaOH would actually make...

    Ca(H2PO4)2 + 2NaOH(limited) → CaHPO4 + Na2HPO4 + 2H2O

    They had another problem that was exactly the same but had Mg in place of Ca. Would that reaction be the same?
  2. jcsd
  3. Oct 9, 2009 #2


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    Staff: Mentor

    Seems logical to me. CaHPO4 is weakly soluble, so it can easily precipitate. Ditto for MgHPO4.

  4. Oct 9, 2009 #3
    Just remembered that the problem said the Ca(H2PO4)2 and CaHPO4 were solid, and Mg(H2PO4)2 and MgHPO4 were aqueous. I suppose that doesn't affect the first one, but what about the second one with the Mg?
  5. Oct 10, 2009 #4


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    Staff: Mentor

    Ksp for CaHPO4 is 6.58 and for MgHPO4 5.82 - this is not a large difference, so I don't think they should behave differently. Unless we are talking about some fancy geochemical reactions.

  6. Oct 10, 2009 #5
    This is just introductory chemistry, so nothing too complicated.
    Thanks for your response.
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