The following information is to be used for the next 2 questions. In order to an
ID: 491006 • Letter: T
Question
The following information is to be used for the next 2 questions. 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 31.48 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 18.94 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?
Explanation / Answer
EDTA titration with urine sample:
sample of urine: 10mL of 24hr urine sample diluted to 2L
EDTA consumed = 31.48 mL of 0.003474 M
1 M has 1 mol or 292.24 g in 1000mL
so, 0.003474M would have 0.003474 moles or 1.015 g in 1000mL
or 0.0001093 moles or 0.03195 g in 31.48 mL of EDTA
One EDTA molecule can complex only one ion.
Therefore, total number of Mg and Ca present in 10mL aliquot is 0.0001093 moles.
EDTA titration with Ca of urine sample
sample of urine: only calcium of 10mL of 24hr urine sample (diluted to 2L)
EDTA consumed = 18.94 mL of 0.003474 M
1 M has 1 mol or 292.24 g in 1000mL
so, 0.003474M would have 0.003474 moles or 1.015 g in 1000mL
and therefore 0.00006578 moles or 0.0192 g in 18.94 mL of EDTA
One EDTA molecule can complex only one ion.
Therefore, Ca present in 10mL aliquot is 0.00006578 moles.
Mg and Ca present in 2L sample
For 10mL aliquot,
Total number of Mg and Ca present is 0.0001093 moles.
Ca present is 0.00006578 moles.
For 2L,
Total number of Mg and Ca present is 0.02186 moles
Ca present is 0.0132 moles or 0.528 g or 528 mg.
Mg is therefore would be (0.02186-0.0132) = 0.0086 moles or 0.2064 g or 206.4 mg
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