Question

In order to analyze for Mg and Ca, a 24-hour urine sample was diluted to 2.000 L. After the solution was buffered to pH 10, a 10.00 mL aliquot was titrated with 44.96 mL of 0.003474 M EDTA. The calcium in a second 10.00 mL aliquot was isolated as CaC2O4, redissolved in acid, and titrated with 17.35 mL of the EDTA solution. (Note: Normal levels for magnesium are 15 to 300 mg per day and for calcium are 50 to 400 mg per day.)

How many mg of Ca were in the original sample?

How many mg of Mg were in the original sample?

Answer 1

The EDTA bonds to the metal ions in a ratio 1:1.

mol of EDTA second aliquot= 0.003474M x 17.35x10^-3L= 6.02739x10^-5 mol = mol of Ca in the aliquot

The second aliquot is taken because in the first titration you titrate both metal ions, in the second aliquot they transform all Ca to CaC2O4, an isoluble compound in order to separate Ca from Mg.

Now for the first aliquot:

mol EDTA= 44.96x10^-3L x 0.003474M= 1.5619104x10^-4 mol = total moles of Mg + Ca

Now, we know that in the second aliquot we have  6.02739x10^-5 mol of Ca.

1.5619104x10^-4 mol = mol of Ca + mol of Mg =  6.02739x10^-5 mol + mol Mg

mol Mg= 9.591714x10^-5 mol

This amount of moles are for an aliquot of 10 mL, so, we want to know the moles in 2L (original sample):

6.02739x10^-5 mol of Ca ------------- 10mL

x= 0.012 mol of Ca------------2000mL

9.591714x10^-5 mol of Mg ------------ 10mL

x= 0.0192 mol of Mg ------------2000mL

mass of Ca= 0.012mol x 40.078g/mol= 0.481g = 481 mg

mass Mg= 0.0192 mol x 24.3050g/mol= 0.467g= 467mg