Posted by jenny on Monday, March 12, 2012 at 4:43pm.
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?
chemistry - jenny, Monday, March 12, 2012 at 5:15pm
DrBob22 PLEASE hELP!!!!!!!!!!!!!!!!!
chemistry - DrBob222, Monday, March 12, 2012 at 5:20pm
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.
chemistry - JENNY, Monday, March 12, 2012 at 5:34pm
chemistry - jonathan, Tuesday, April 28, 2015 at 9:08am
aren't the volumes the wrong way round?
chemistry - D, Wednesday, April 29, 2015 at 7:08am
Yeah volumes at the bottom of the page are switched
chemistry - Efro, Wednesday, March 30, 2016 at 2:56pm
Is that the correct solution of the exercise?
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