A 0.5215-g sample of CaCO_3, is dissolved in 12 M HCl and the resulting solution

Question

A 0.5215-g sample of CaCO_3, is dissolved in 12 M HCl and the resulting solution is diluted to 250.0 mL, in a volumetric flask. How many moles of CaCO_3, are 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 25.0-mL aliquot of the solution in Ib? 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 2.60 mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg^2+ 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 Part a? What is the molarity of the EDTA solution? A 100-mL sample of hard water is titrated with the EDTA solution in Problem 2. The same amount of Mg^2+ is added as previously, and the volume of EDTA required is 22.44 mL. What volume of EDTA is used in titrating the Ca^2+ in the hard water? How many moles of EDTA are there in that volume? How many moles of Ca^2+ are there in the 100 mL of water? If the Ca^2+ comes from CaCO_3 how many moles of CaCO_3 are there in one liter of the water? How many grams CaCO_3 per liter? If 1 ppm CaCO_3 = 1 mg per liter, what is the water hardness in ppm CaCO_3 per liter?

Explanation / Answer

Q1.

m = 0.5215 g of CaCO3

M = 12 M of HCl

dilution to V = 250 mL

a)

find moles of CaCO3:

mol = mass/MW = 0.5215/100.1 = 0.0052097902 mol of CaCO3

b)

molarity of Ca+2 in V = 250 mL

[Ca+2] = mol of Ca+2 / Total V = 0.0052097902 / (0.250) = 0.0208391608 mol of Ca+2 per liter

c)

moles in V = 25 mL

so..

we have 0.0052097902 mol of Ca+2 in 250 mL

so in 25 mL --> 25/250 * 0.0052097902 = 0.00052097902 moles

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