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EDTA titration lab help 
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#1
Oct2407, 11:04 AM

P: 129

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 = 1.489E4 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 1.489E4 moles (calculated above) is only the amount of moles in 25.00mL??????? any help appreciated :P 


#3
Oct2407, 02:21 PM

P: 129

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 (1.489E4 mol)(25.00mL) = m2 (250mL) m2 = 1.489 E5 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+ 


#4
Oct2407, 02:46 PM

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EDTA titration lab help
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. 


#5
Oct2407, 03:11 PM

P: 129

so in 250mL i have 1.48E4 mol
25 ml /250 ml = 0.1 0.1*1.48E4 mol = 1.48E5 mol.... which will be the same as what i got using mv=mv 


#6
Oct2407, 05:00 PM

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#7
Oct2407, 07:33 PM

P: 129

oh oops.. should be times 10... 1.48E4 moles * 250/25 = 1.48E3 moles



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