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Discussion in 'Pre-Medical - MD' started by musiclink213, Mar 14, 2004.

  1. musiclink213

    musiclink213 My room is a mess
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    so won't ya please help me?? i have to do my quant lab, and for some reason, i'm trying to do the practice problem before i do the actual lab results, but i can't get the right answer. if anyone cares to take time out of their busy day to help me, it would be incredibly gratefulyl appreciated.

    this ist he problem:
    a 25.00 mL aliquot of unknown CaCO3 is titrated with EDTA to give a 22.50 mL equivalence point. Find the % CaO if the sample weigfhed .2200 g and was diluted to 250 mL in a vol. flask. The molarrity of EDTA is 0.0090 M, the FW of CaO is 56.072, FW of CaCO3 is 100.1

    any help would be appreciated, even a general formula would be useful. thanks again.
     
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  2. OP
    OP
    musiclink213

    musiclink213 My room is a mess
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    ok wait, can someone tell me if this is right?

    .02250 x .0090 = 2..25x10^-4 = mol EDTA=mol CaCO3=mol CaO

    10 x 2.025x10^-4= .002025 = mol Ca in aliquot

    .002025(56.072) = .1135 g CaO

    .1135/.2200 = 51.59%

    does that look right to anyone?
     
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  3. DrPharaohX

    DrPharaohX Free...your...mind...
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    It's 3 AM right now (east coast) and I should be asleep, but I couldn't resist:

    Here we go: (my final answer was 0.4430 % - notice it's NOT 44.30 %)

    CaCO3: 0.22 g / 100.1 g x (25/250) = 2.198 x 10^-4

    EDTA: 0.009 x 0.0225 = 2.025 x 10^-4

    [This next step assumes that reaction is one to one ratio between CaCO3 and EDTA]:

    mol CaO produced = (2.198 x 10^-4) - (2.025 x 10^-4) = 0.1738 x 10^-4 (which is 1.738 x 10^-5)

    So....

    1.738 x 10^-5 x 56.072 g = 9.74 x 10^-4 g of CaO produced

    And...

    9.74 x 10^-4 / 0.22 = 0.004430 which = 0.4430 %

    From what I remember from that lab, the yield I got for this problem seems reasonable because the yields I remember getting for our problems were often < 1%
     
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