Question

A 25.00 mL Limestone sample containing Ca2 and Mg2 required 15.52 mL of 0.0106 M EDTA to reach the end point. A second 25.00 mL sample of unknown was then treated with KOH to precipitate out the Mg2 , and the remaining Ca2 only solutin was titrated with the EDTA. The Ca2 only solution requred 10.68 mL of EDTA to reach the end point. Determine the amount of Ca2 and Mg2 in the unknown sample.

Answer 1

The concentration of Mg^2+ and Ca^2+ ions in 25mL of solution can be found using the equation

Ca^2+ + Mg^2+ + EDTA -----> Ca(EDTA) + Mg(EDTA)

1mole Ca^2+ = 1mole EDTA and 1mole Mg^2+ = 1mole EDTA

Thu we have M1V1 = M2V2

M1 = Molarity of ions =? and V1 = VOlume of limestone solution = 25mL

M2 = Molarity of EDTA = 0.0106 M and V2 = volume of EDTA required = 15.52 mL

M1 = M2V2/V1 = 0.0106*15.52/25 = 0.00658 M

The total concentration of ions in solution = 0.00658 M

In second experiment, only Ca^2+ ions were titrated using EDTA

M1 = ?, V1 = 25mL , M2 = 0.0106 M and V2 = 10.68 mL

Molarity of Ca^2+ ions = M2V2/v1 = 0.0106*10.68/25 = 0.00452 M

Amount of Ca^2+ ions = M* mol.wt of Ca = 0.00452*40 = 0.1808g

Molarity of Mg^2+ ions = Total conc. - conc. of Ca^2+ = 0.00658 - 0.00452 = 0.00206 M

Amount of Mg^2+ ions = M* mol.wt of Mg = 0.00206*24 = 0.0494g