Question

Determine the magnesium content (in ppm) of a 100.00 mL sample of drinking water if the following data were obtained. Titrant is 0.0677 M EDTA, 14.68 mL were used when an aliquot of the sample was at pH 13.0, 18.02 mL were used for an aliquot at pH 10.5.

Answer 1

EDTA binds Mg in a 1:1 ratio.....ppm = mg/L
EDTA in a base has lower ability to bind the metals because metal hydroxides will precipitate out of solution with the high [OH-]. lowering pH to ~11 allows for the formation of the EDTA -4 molecule which is more effective in chelating metals.

at pH 13
moles EDTA: 0.0677 * 0.01468 = 9.94x10^-4 moles
mass Mg = 9.94x10^-4 * 24.3 = 0.0239 g of Mg or 23.9 mg
Total volume = 100 + 14.68 = 114.68 mL
ppm = 23.9 / 0.11468 = 208.41 mg/L

at pH 10.5
moles EDTA = 0.0677 * 0.01802 = 1.22x10^-3 moles
mass Mg = 1.22x10^-3 * 24 = 0.02928 g or 29.28 mg
ppm = 29.28 / 0.11802 = 248.09 mg/L

Now taking the average we have:
248.09 + 208.41 / 2 = 228.25 ppm

Hope this helps