Question

A 0.4505 g sample of CaCO3 was dissolved in the HCl and the resulting solution diluted...

A 0.4505 g sample of CaCO3 was dissolved in the HCl and the resulting solution diluted to 25.00 mL in a volumetric flask. A 25.00 mL aliquot of the solution required 29.25 mL of EDTA solution for titration to the Eriochrom Black T end point.

a. How many moles of CaCO3 were present in the solid sample?

b. What is the molar concentration of Ca^2+ are contained in a 250.00 mL aliquot of the CaCl2 solution?

c. How many moles Ca^2+ are contained in the 25.00 mL aliquot of CaCl2 solution?

d. How many moles EDTA are contained in the 29.25 mL used for the titration?

e. What is the molar concentration of the EDTA solution? f. From the molarity of the calcium solution you found in part (b), calculate the concentration in ppm of calcium ion and the concentration in ppm of CaCO3

a.moles CaCO3 = 0.4505 g/ 100.1 g/mol=0.0045

b.[Ca2+]= 0.0045/ 0.250 L=0.018M
c.moles Ca2+ = 0.018 x 0.02500 L=0.00045

d.moles of EDTA can be found from its concentration using the formula = M1V1=M2V2

[Ca2+]0.018M*25 = M2*29.25 [EDTA]

[EDTA=]M2=0.0153M

moles of EDTA = 0.0153M*0.02925 = 0.000447

e) Molarity of EDTA = 0.0153M

f) [Ca2+]= 0.0045/ 0.250 L=0.018M

0.018 mol/L times 40.08 g/mol = 0.721g/L

0.721 g / 1000 g of solution times 1000/1000 = 400 g / 1,000,000 g of solution