The urine excreted by an adult patient over 24 hours was collected and diluted to give 2.000L sample.
at ph 10 EDTA FORMS 1:1 COMPLEX WITH BOTH Mg2+ and Ca2+ and after buffering at this pH a 10mL aliquot of the sample required 26.81mL of 0.003474M EDTA for titration
The Calcium in another 10.00mL aliquot was precipitated as Calcium Oxalate CaC2O4 (s) which was isolated ,redissolved in acid ,buffered at pH 10 and titrated with 11.63mL of the EDTA solution.
Calculate mass of Mg2+ and Ca2+ in the patients urine?
6 answers
DrBob22 PLEASE hELP!!!!!!!!!!!!!!!!!
Since the same molarity EDTA was used, you can simplify this by
26.81 mL = Ca + Mg
-11.63 mL = Ca
---------
15.18 mL = Mg.
0.0158L x 0.003474M x atomic mass Ca x (2000 mL/10 mL) = g Ca.
0.01163 x 0.003474M x atomic mass Mg x (2000 mL/10 mL) = g Mg.
26.81 mL = Ca + Mg
-11.63 mL = Ca
---------
15.18 mL = Mg.
0.0158L x 0.003474M x atomic mass Ca x (2000 mL/10 mL) = g Ca.
0.01163 x 0.003474M x atomic mass Mg x (2000 mL/10 mL) = g Mg.
THANKS
DrBob222,
aren't the volumes the wrong way round?
aren't the volumes the wrong way round?
Yeah volumes at the bottom of the page are switched
Is that the correct solution of the exercise?