Chem EDTA and Molarity Question. Need help

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SUMMARY

The discussion centers on calculating the molarity of calcium (Ca) and magnesium (Mg) ions in tap water using EDTA titration. The user prepared an EDTA solution with a molarity of 0.005137 M by dissolving 0.6005 g of EDTA in 0.400 L of water. The calculations for the molarity of Ca and Mg ions were derived from the titration volumes used, resulting in 1.117x10^-4 M for Ca and 3.57x10^-5 M for Mg. The user sought confirmation on the accuracy of their calculations and whether any additional steps were necessary due to the use of EDTA.

PREREQUISITES
  • Understanding of molarity calculations (M = mol/L)
  • Familiarity with EDTA and its role in complexometric titrations
  • Knowledge of stoichiometry in chemical reactions
  • Basic skills in laboratory titration techniques
NEXT STEPS
  • Review the principles of complexometric titration using EDTA
  • Learn about the impact of pH on the precipitation of metal ions, specifically Mg(OH)2
  • Explore the concept of buffer solutions and their influence on titration results
  • Investigate common errors in molarity calculations and how to avoid them
USEFUL FOR

Chemistry students, laboratory technicians, and educators involved in analytical chemistry and titration experiments will benefit from this discussion.

AMSAMS
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1. We had a titration lab where we titrated tap water with EDTA. Through our results we should be able to figure out the molarity of Ca and Mg ions. I think I'm doing this right but it seems to easy in comparison with everything else we are doing in lab and lecture so I just want to know if I'm missing something.

2. M=mol/L
MM of EDTA is 292.24g/mol
EDTA solution was prepared by dissolving 0.6005g of EDTA with 0.400L.
0.6005(g/mol)/292.24g=0.00205mol/.400L=0.005137M of EDTA soultion (I think)


3. 0.005137M*0.0290L (the amount it took to titrate the water)=1.474x10^-4 moles. There was 100.00mL of water and 0.50mL of a buffer added but I don't think the buffer volume is counted in the molarity. So, molarity would be
1.474x10^-4mol/100.00mL=1.474x10^-4M. This is for Mg and Ca

For just Ca, 0.005137*0.02175L (the amount it took to titrate the water) =1.117x10^-4 mol. There was 100.00mL of water and 0.50mL of a buffer added but I don't think the buffer volume is counted in the molarity. So, molarity would be
1.117x10^-4 mol/100.00mL=1.117x10^-4 M for Ca solution.

For just Mg: (1.474x10^-4M) - (1.117x10^-4M)= 3.57x10^-5M

Is that all there really is to this or am missing some extra step I should be doing because of the use EDTA?
 
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AMSAMS said:
0.6005(g/mol)/292.24g=0.00205mol/.400L=0.005137M of EDTA soultion (I think)

Concentration is OK (assuming you used a pure, anhydrous EDTA - possible, but not the only option here). Just note you can't write moles=concentration, there should be no equal sign there.

0.0290L (the amount it took to titrate the water)

molarity would be
1.474x10^-4mol/100.00mL=1.474x10^-4M. This is for Mg and Ca

Sum of Mg & Ca, OK.

0.02175L (the amount it took to titrate the water)

That's where you have lost me. Just a moment ago you wrote it took 29.0 mL to titrate the water, so apparently you did here something different. I suppose this is result of titration at much higher pH, where Mg2+ is precipitated as Mg(OH)2. If so, your calculations look OK, just your description is wrong.

Disclaimer: I have not checked math, just the logic.
 

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