1. A 0.5188g sample of CaCO3 is dissolved in 6M HCl and the resulting solution i
ID: 533098 • Letter: 1
Question
1. A 0.5188g sample of CaCO3 is dissolved in 6M HCl and the resulting solution is diluted to 250.0mL in a volumetric flask. (a)How many moles of CaCO3 are in the sample? (formula mass = 100.1) (b)What is the molarity of the Ca2+ ion in the 250mL of solution? (c) How Many moled of Ca2+ ion are in a 250mL aliquot of the solution?
2. 25.00mL aliquots of the solution from problem 1 are titrated with EDTA to the Calmagite end-point. An aliquot is found to require 28.55 mL of the EDTA before the solution becomes sky blue. (a) How many moles of EDTA are there in the volume that is used? (b) What is the molarity of the EDTA solution?
3. A 100mL sample of hard water is titrated with the EDTA solution in problem 2. The volume of EDTA required to reach the end point is 22.44 mL. (a) How many moles of EDTA are used in the titration? (b) How many moles of Ca2+ ion are there in the 100mL water sample? (c) If the Ca2+ ion comes from CaCO3, how many moles of CaCO3 are there in 1L of the hard water? How many grams CaCO3 per liter? (d) If 1 ppm CaCO3=1mg/L, what is the water hardness in ppm CaCO3?
Explanation / Answer
1. mass of CaCO3 = 0.5188 g
molar mass of CaCO3 = 100.1 g/mol
no of moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3 = 0.5188/100.1 = 0.00518 mol
no of moles of Ca2+ = 0.00518 mol
molarity of Ca2+ ion in sample = no of moles / volume of solution = 0.00518/ 0.250 = 0.02072 M
No of moles of Ca2+ ion in 250 mL sample = 0.00518 mol
2. no of moles of EDTA = no of moles of Ca2+
(MV)EDTA = (MV) Ca2+
MEDTA = (0.02072 * 25) / 28.55 = 0.0181 M
(a) no of moles of EDTA = 0.02072 * 0.025 = 0.000518 moles
(b) molarity of EDTA = 0.0181 M
3. no of moles of EDTA = no of moles of Ca2+
(MV)EDTA = (MV) Ca2+
MCa2+ = (0.0181 * 22.44) / 100 = 0.00406 M
moles of Ca2+ = 0.0181 * 0.02244 = 0.000406 mol
molarity of Ca2+ solution = 0.00406 M
no of moles of Ca2+ in 1L solution = 0.00406 mol
no of moles of CaCO3 = 0.00406 mol
mass of CaCO3 = no of moles of CaCO3 * molar mass of CaCO3 = 0.00406 * 100.1 = 0.0000406 g
(d) now 0.0000406 g of CaCO3 present in 1 L of solution
0.0406 mg of CaCO3 present in 1 L of solution
thus 0.0406 ppm of CaCO3 is the hardness of water.
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