Question

Determine the calcium content (in ppm) of a 100.00 mL sample of drinking water if the following data were obtained. Titrant is 0.0505 M EDTA, 13.49 mL were used when an aliquot of the sample was at pH 13.0, 16.99 mL were used for an aliquot at pH 10.5.

Answer 1

EDTA binds Ca 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
0.0505 M EDTA x 0.01349 L = 6.8 x10^-4 moles EDTA used

so, 6.8 x10^-4moles Ca+2 chelated

6.8 x10^-4moles Ca+2 x 40.0 g/mole = 0.0272 g
ppm = mg/L...0.0272 g = 27.2 mg
113.49 ml = 0.11349 L
Mg = 27.2 mg / 0.11349 L = 240 ppm

at pH 10.5
0.0505M EDTA x 0.01699 L = 8.57 x10^-4 moles EDTA
so, 8.57 x10^-4 moles Ca+2 chelated
8.57 x10^-4 moles Ca x 40.0 g/mole = 0.0343 g
ppm = mg/LL...0.0343 g = 34.3 mg
116.99 ml = 0.11699 L
Mg = 34.3 mg / 0.11699L = 293 ppm