EDTA Lab Question/Clarification

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To determine the molarity of the EDTA stock solution, the correct calculation is (0.2416 g / 292.25 g/mol) / 0.250 L, resulting in 0.00331 M, as the total volume of the prepared solution is used. The moles of metal ions in the water sample can be calculated using the formula (M EDTA x V EDTA) = (M ions x V ions), where the titration volume is essential for accurate results. For total hardness, the calculation involves (moles of metal ions x MW of CaCO3 mg) / 0.050 L to express it in mg/L (ppm). Clarification on the preparation method is crucial, as the description of diluting EDTA must align with the volumetric flask method used. Accurate preparation and measurement are key to successful titration and analysis.
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I weighed out 0.2416 g disodium EDTA and diluted it with about 250 mL of H2O and stuck it in a plastic bottle to use for titration. My question is, to determine the molarity of my stock EDTA standard sol'n, is it just (0.2416g / 292.25 g) / 0.250 L = 0.00331 M? Or do I have to use the VOLUME of EDTA solution used to titrate to the endpoint? (which in my case is 27.54 mL, 27.80 mL, and 27.26 mL respectively). I'm really leaning towards the first equation I wrote, but I'm not 100% sure.

Also, how would I go about calculating the moles of metal ions that I have in my water sample (50mL). EDTA has a 1:1 stoich ratio with them so would I just (M EDTA x V EDTA) = (M ions x V ions)? So then it would be something like (M EDTA x titration volume) = (M ions x 0.050 L)?

So, once I figure out the moles of metal ions in solution, to figure out the total hardness of the water sample do I just do (moles of metal ions x MW of CaCO3 mg) / (0.050L) to get it in mg/L (ppm)?

Thanks in advance for your help.
 
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What was the exact volume of EDTA titrant you prepared?

Moles of metal titrated equals moles of EDTA titrant used, at the titration's endpoint. Did you use a standard Ca+2 solution?
 
I prepped it with 250 mL. I put the disodium EDTA in a 250 mL volumetric flask and diluted it to the mark, so I assume it's that. We used water from the faucet in our lab.
 
Ponder very carefully the difference between your preparation descriptions which you gave in these two quotes:
anthonynewegn said:
I weighed out 0.2416 g disodium EDTA and diluted it with about 250 mL of H2O and stuck it in a plastic bottle to use for titration.
And
I prepped it with 250 mL. I put the disodium EDTA in a 250 mL volumetric flask and diluted it to the mark, so I assume it's that. We used water from the faucet in our lab..

So you seem to have made a standard solution of your EDTA titrant, if what you really did was the second described process. Your two descriptions are not the same. Next, your equation for how the moles are related is correct.
 
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Ah sorry, I should've clarified it better. I weighed out the 0.2416 g disodium EDTA and put it in the 250 mL volumetric flask so I could dilute it to the mark. After I did that, I poured the solution into a plastic bottle because its safer that way. (No metal ions in plastic bottle). So in my bottle, there was 250 mL of EDTA solution. Sorry for the mixup.
 
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