Asked by saddee

25.00 mL aliquots of the solution from Part 1 are titrated with EDTA to the Eriochrome Black T end point. A blank containing a small measured amount of Mg2+ requires 2.60 mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg2+ is added requires 28.55 mL of the EDTA to reach the end point.

How many milliliters of EDTA are needed to titrate the Ca+2 ion in the aliquot?

How many moles of EDTA are there in the volume obtained in the previous question?

What is the molarity of the EDTA solution?

Answers

Answered by DrBob222
<b>Without all of the question it is impossible to answer any but the first part.</b>
25.00 mL aliquots of the solution from Part 1 are titrated with EDTA to the Eriochrome Black T end point. A blank containing a small measured amount of Mg2+ requires 2.60 mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg2+ is added requires 28.55 mL of the EDTA to reach the end point.

How many milliliters of EDTA are needed to titrate the Ca+2 ion in the aliquot?
<b>28.55-2.60 = ??</b>

How many moles of EDTA are there in the volume obtained in the previous question?
<b>moles EDTA in 100 mL x 10 = moles EDTA in 1 L.

What is the molarity of the EDTA solution? </b>
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