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Chem! EDTA and Molarity Question. Need help

  1. Sep 27, 2012 #1
    1. We had a titration lab where we titrated tap water with EDTA. Through our results we should be able to figure out the molarity of Ca and Mg ions. I think I'm doing this right but it seems to easy in comparison with everything else we are doing in lab and lecture so I just want to know if I'm missing something.

    2. M=mol/L
    MM of EDTA is 292.24g/mol
    EDTA solution was prepared by dissolving 0.6005g of EDTA with 0.400L.
    0.6005(g/mol)/292.24g=0.00205mol/.400L=0.005137M of EDTA soultion (I think)


    3. 0.005137M*0.0290L (the amount it took to titrate the water)=1.474x10^-4 moles. There was 100.00mL of water and 0.50mL of a buffer added but I don't think the buffer volume is counted in the molarity. So, molarity would be
    1.474x10^-4mol/100.00mL=1.474x10^-4M. This is for Mg and Ca

    For just Ca, 0.005137*0.02175L (the amount it took to titrate the water) =1.117x10^-4 mol. There was 100.00mL of water and 0.50mL of a buffer added but I don't think the buffer volume is counted in the molarity. So, molarity would be
    1.117x10^-4 mol/100.00mL=1.117x10^-4 M for Ca solution.

    For just Mg: (1.474x10^-4M) - (1.117x10^-4M)= 3.57x10^-5M

    Is that all there really is to this or am missing some extra step I should be doing because of the use EDTA?
     
  2. jcsd
  3. Sep 28, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    Concentration is OK (assuming you used a pure, anhydrous EDTA - possible, but not the only option here). Just note you can't write moles=concentration, there should be no equal sign there.

    Sum of Mg & Ca, OK.

    That's where you have lost me. Just a moment ago you wrote it took 29.0 mL to titrate the water, so apparently you did here something different. I suppose this is result of titration at much higher pH, where Mg2+ is precipitated as Mg(OH)2. If so, your calculations look OK, just your description is wrong.

    Disclaimer: I have not checked math, just the logic.
     
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