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?
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.
THANKS
DrBob222,
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?
To calculate the mass of Mg2+ and Ca2+ in the patient's urine, we need to use the information provided about the EDTA titration. EDTA forms a 1:1 complex with both Mg2+ and Ca2+, which means that for every mole of Mg2+ or Ca2+ present, one mole of EDTA is required for complete titration.
Let's start by calculating the number of moles of EDTA used in both titrations:
For the first titration, 26.81 mL of 0.003474 M EDTA was used. We can calculate the moles of EDTA as follows:
Moles of EDTA = Volume of EDTA (L) × Concentration of EDTA (mol/L)
= 0.02681 L × 0.003474 mol/L
= 9.31 × 10^-5 mol
For the second titration, 11.63 mL of the same EDTA solution was used. Again, we can calculate the moles of EDTA:
Moles of EDTA = Volume of EDTA (L) × Concentration of EDTA (mol/L)
= 0.01163 L × 0.003474 mol/L
= 4.04 × 10^-5 mol
Since the EDTA forms a 1:1 complex with both Mg2+ and Ca2+, the moles of Mg2+ and Ca2+ in the urine are equal to the moles of EDTA used in each titration.
Now, to calculate the mass of Mg2+ and Ca2+, we need to know the molar masses of Mg2+ and Ca2+. The molar mass of Mg2+ is approximately 24.31 g/mol, and the molar mass of Ca2+ is approximately 40.08 g/mol.
Mass of Mg2+ = Moles of Mg2+ × Molar mass of Mg2+
Mass of Mg2+ = 9.31 × 10^-5 mol × 24.31 g/mol
Similarly,
Mass of Ca2+ = Moles of Ca2+ × Molar mass of Ca2+
Mass of Ca2+ = 4.04 × 10^-5 mol × 40.08 g/mol
Now, substitute the values and calculate the masses of Mg2+ and Ca2+ in the patient's urine.