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Posted by on Sunday, February 3, 2013 at 8:26am.

An EDTA solution is standardized against a solution of primary standard CaCO3 (0.5622 g dissolve in 1000 ml of solution) and titrating aliquots of it with the EDTA. If a 25.00 ml aliquot required 21.88 ml of the EDTA, what is the concentration of the EDTA?

  • analytical chemistry - , Sunday, February 3, 2013 at 12:51pm

    mols CaCO3 = grams/molar mass
    M CaCO3 = mols/L
    mol = 0.5622/100 = 0.005622 and that is in 1L; therefore M = 0.005622M.
    Then I would use
    MEDTA x 21.88mL EDTA = MCaCO3 x 25.00 mL CaCO3. Solve for MEDTA

  • analytical chemistry - , Monday, March 11, 2013 at 12:01pm

    first get for the moles of CaCO3
    mols CaCO3= 0.5662g*(1mol CaCO3/100gCaCO3)
    Molarity of CaCO3= mols CaCO3/L of solution
    =0.005622mols/1L
    M EDTA= [(molarity CaCO3)(mL aliquot)]/
    mL of EDTA
    M EDTA= [(0.005622)(25)]/21.88
    = 0.00642 M

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