Question

Determine the magnesium content (in ppm) of a 100.00 mL sample of drinking water if the following data were obtained. Titrant is 0.0674 M EDTA, 14.81 mL were used when an aliquot of the sample was at pH 13.0, 18.51 mL were used for an aliquot at pH 10.5.

Answer 1

at pH 13

millimoles of EDTA = 0.0674 x 14.81   = 0.998 moles

millimoles of Mg = M x 100

moles of EDTA = moles of Mg

0.998 = M x 100

Molarity of Mg = 9.982 x 10^-3 M

molarity = 9.982 x 10^-3 mol / L

            = 9.982 x 10^-3 x 24 x 1000

            = 239.57 ppm

magnesium content = 239.6 ppm

At pH = 10.5 :

millimoles of EDTA = 0.0674 x 18.51 = 1.25 moles

millimoles of Mg = M x 100

moles of EDTA = moles of Mg

1.25 = M x 100

Molarity of Mg = 0.0125 M

molarity = 0.0125 mol / L

            = 0.0125 x 24 x 1000

            = 299.4 ppm

magnesium content = 299 ppm