Question

Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10. A 50.0 mL sample of water containing both Ca? and Mg is titrated in another 50.0 mL sample, the Mg* was precipitated as Mg(OH)2 and then, Ca2* was titrated at pH 13 with 9.25 ml of the same EDTA solution. Calculate ppm CaCo, (FW-100.09) and MgCO, (FW-84.31) in the sample. FWca: 40.08; FWm: 24.30

Answer 1

Cat2 and Mg+2 forms 1:1 complex with EDTA Cat2+ EDTA-4 4> [Ca(EDTA)]-2 -4 > [Mg(EDTA)]-2 Mgt2+ EDTA-4 Number of moles of EDTA initially added - M*V 0.0114 M * 16.55 mL 0.18867 mmol Number of moles of Ca+2 and Mgt2-0.18867 mmol Number of moles of EDTA added for titration of Ca+2-M *V= 0.0114 M * 9.25 mL = 0.10545 mmol Number of moles of Ca+2 0.10545 mmol Number of moles of Mg+2-0.18867-0.10545 = 0.08322 mmol Mass of CaCO3 moles* mol.wt. 0.10545 mmol * 100.09 g/mol 10.55449 mg 0.010554 g Mass of MgCO,-moles * mol.wt. = 0.08322 mmol * 84.31 g/m。 7.016278 mg = 0.007016 g mass of CaCO volume of solution 0.010554 10 211 3ー×10 ppm CaCO ppm CaCO, ppm MgCO ppm MgCO 50.0 ×10 gco,--mass of volume of solution ppn Mgco,-0 970104 50.0