Calculating Ca2+ Weight Percentage Using EDTA Titration Method | Lab Help

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To calculate the weight percentage of Ca2+ in the unknown solution using the EDTA titration method, the moles of Ca2+ must be determined from the titration data. The user calculated 1.489E-4 moles of EDTA used, which corresponds to the same amount of moles of Ca2+. Since only 25.00 mL was titrated from a 250 mL solution, the total moles of Ca2+ in the entire solution is 1.489E-4 moles multiplied by 10, resulting in 1.489E-3 moles. Finally, to find the weight percentage, this total moles of Ca2+ should be divided by the weight of the unknown (0.2505 g) and multiplied by 100. This process ensures an accurate calculation of the weight percentage of Ca2+ in the unknown.
higherme
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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
 
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higherme said:
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

Of course!
 
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+
 
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.
 
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
 
higherme said:
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

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?
 
oh oops.. should be times 10... 1.48E-4 moles * 250/25 = 1.48E-3 moles
 
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