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EDTA titration lab help

  1. Oct 24, 2007 #1
    My question is:
    Using the average [EDTA], find the weight% of Ca2+ in the unknown

    My attempt:

    so, the[EDTA] i found from standardizing was 0.005 M;
    I used 29.85mL to of EDTA to titrate my 25.00mL unknown
    and the weight of my unknown used was 0.2505 g

    this is what i did:
    [EDTA] = 0.005M * 0.02985L = 1.489E-4 mol EDTA = mol Ca2+

    so that is the moles of Ca2+ , my question is do I have to convert this moles to moles in 250mL, because the unknown was prepared in a 250mL volumetric flask, and only 25.00mL was taken out from that flask for titration. Does that mean 1.489E-4 moles (calculated above) is only the amount of moles in 25.00mL???????

    any help appreciated :P
     
  2. jcsd
  3. Oct 24, 2007 #2

    chemisttree

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    Of course!
     
  4. Oct 24, 2007 #3
    so do i have to use m1v1=m2v2 to find the moles that is in 250mL

    can i just use the number in moles for m1.... or do i have to change it to concentration first

    (1.489E-4 mol)(25.00mL) = m2 (250mL)
    m2 = 1.489 E-5 moles in 250mL ???

    this is the moles of Ca2+ in the unknown... i divide by the total weight of the unknown to find the % Ca2+
     
  5. Oct 24, 2007 #4

    chemisttree

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    You are given the weight of the unknown as 0.2505. You dissolved this into 250 mL and analyzed a 25 mL aliquot of this unknown using 29.85 mL of 0.005 M EDTA.

    You used 25 mL out of a solution of 250 mL. What fraction of 250 mL is 25 mL? That math is pretty simple.
     
  6. Oct 24, 2007 #5
    so in 250mL i have 1.48E-4 mol

    25 ml /250 ml = 0.1

    0.1*1.48E-4 mol = 1.48E-5 mol.... which will be the same as what i got using mv=mv
     
  7. Oct 24, 2007 #6

    chemisttree

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    Why are you doing this????? Why did you DIVIDE by 10! Is there 1/10 as much in 250 mL as there is in 25 mL?
     
  8. Oct 24, 2007 #7
    oh oops.. should be times 10... 1.48E-4 moles * 250/25 = 1.48E-3 moles
     
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