A 0.5100-g sample of is dissolved in 12 M HC1 and the resulting solution is dilu
ID: 1053766 • Letter: A
Question
A 0.5100-g sample of is dissolved in 12 M HC1 and the resulting solution is diluted to 250.0 mL in a volumetric flask. How many moles of are used (M.W. = 100.1 g/mol)? 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 soln in b.? 25.00-mL aliquots of the solution in problem 1 are titrated with EDTA to the calmagite end point. A blank containing a small measured amount of Mg^2+ requires 2.50 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 mL 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 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, and the volume of EDTA required is 22.51 mL. What is the volume of EDTA used in titrating the Ca^2+? How many moles of EDTA are there in that volume? How many moles of Ca^2+ are there in 100 mL of water?
Explanation / Answer
1) Mass of calcium carbonate = 0.5100 g
molarity of HCl = 12 M
Final volume of solution = 250 ml
molar mass of calcium carbonate = 100.1 g /mol
a) moles of calcium carbonate = 0.5100/100.1 = 0.0051 mol
b) Molarity of calcium ion = 0.0051 mol /0.250 L = 0.0204 M
c) If we take 25 ml of the solution then the moles of calcium ion will be = 0.0204 M x 0.025 = 0.00051 mol
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.