Question

A 24-hr urine specimen was diluted to exactly 2.000 L in a volumetric flask. The Ca^2+ and Mg^2+ in a 100.0-mL aliquot of the diluted solution required 20.81 mL of 0.00830 M EDTA, after being buffered to a pH of 10.0. The Ca^2+ second 100.0-mL aliquot was precipitated as CaC_2O_4(s), which was filtered and washed. The filtrate and washings were titrated with 5.98 mL of the EDTA solution. Calculate the milligrams of Ca^2+ and Mg^2+ in the 24-hr sample.

Answer 1

Moles of Mgn 100.0mL diluted sample 0, 00830M × 0.00598し = 4.96×10-5 mol Total moles of Ca and Mg* -0. 00830M 0.0208 1L 1.73x10mol Moles of Ca1.73x104 mol-496x10mol = 1.234 × 10-4mol Mass of Ca2+ in 2.0 L sample 01234 x1оґ mol x 100mL 2000mL 40.08g 1000mg 1msol 1g 98.9 mg Ca" Mass of Mg* in 20L sample = 4.96× 10-MO/X 2000mL 24.32g 1000mg 100mLm 1g 24.1mg Mg