A urine specimen collected over a 24 hour period is diluted to 2.000 L and buffe

Question

A urine specimen collected over a 24 hour period is diluted to 2.000 L and buffered to pH=10. A 10.00 ml sample is then titrated with 26.81ml of 0.003474 M EDTA. This titrates both Ca2+ and Mg 2+ ion. A second 10.00ml sample is treated with oxalate to precipitate CaC2O4 solid. The solid is then dissolved and titrated with 8.63ml of the same EDTA solution. This step isolates the Ca2+ ion. How do the patient levels compare to the normal amounts of 75 to 150mg of magnesium and 50 to 300mg of calcium per day?

Explanation / Answer

Ca 2++Mg2+=26.81 ml

Ca2+ = 8.63 ml

Mg2+=18.18 ml

Gm of Ca in 2 Liter

gm of Ca=0.00863 L x 0.003474 (Mol/lit) x atomic mass Ca x (2000 mL/10 mL)

gm of Ca=0.00863*0.003474*40*2000/10=0.2398 gm=239.8 mg
gm of Mg=0.01818 x 0.003474M x atomic mass Mg x (2000 mL/10 mL)

gm of Mg=0.01818*0.003474*24*200=0.30315 =303.15 mg

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