A 0.3946 g sample of CaCO, is dissolved in 12 M MCI and the resulting solution i

Question

A 0.3946 g sample of CaCO, is dissolved in 12 M MCI and the resulting solution is diluted to 250.0 ml. in a volumetric flask. How many moles of CaCO, arc used (formula mass = 100.1)? What is the molarity of the Ca^2+ in the 250 ml. of solution? How many moles of Ca^2+ are in a 25.00-ml. aliquot of the solution in lb? 25.00-mL aliquots of the solution from Problem 1 are titrated with EDTA to the Eriochrome Black T end point. A blank containing a small measured amount of Mg^2+ requires 2.77 ml- of the EDTA to reach the end point. An aliquot to which the same amount of Mg^2+ is added requires 39.04 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 ? What is the molarity of the EDTA solution?

Explanation / Answer

2) concentration of Ca+2 in the above problem is 0.0158 M

a) The volume needed to titrate Mg+2 = 2.77 mL

The volume needed to titrate Mg+2 and Ca+2 = 39.04

so volume needed to titrate only Ca+2 = 39.04 - 2.77 = 36.27mL

b) the moles of EDTA must be equal to moles of Ca+2 [as they each mole reacts with one mole of Ca+2]
Moles of CA+2 = Molarity X volume = 0.0158 X 25 = 0.395 millimoles

So moles of EDTa = 0.395 millimoles

c) Molarity of EDTA = moles / volume = 0.395 / 36.27 = 0.0109 M

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