It is often necessary for hospitals to measure the amount of ions in patient bod

Question

It is often necessary for hospitals to measure the amount of ions in patient body fluids. Assume a patient has their urine collected over a 24-hour period. The urine specimen is diluted to 2.000 L and buffered to pH=10. A 10.00 mL sub-sample was titrated with 26.81 mL of 0.003474 M EDTA. A second 10.00 mL sub-sample was treated with oxalate to precipitate CaC2O4 solid. The solid was then re-dissolved and titrated with 8.63 mL of EDTA solution. Assuming that a healthy human being will excrete 75 to 150 mg of magnesium and 50 to 300 mg of calcium per day in their urine, what is your diagnosis for this patient? (The EDTA titration measures the combined moles of Mg ion and Ca ion, while the oxalate isolation and titration measures the moles of Ca ion only).

Explanation / Answer

Mg2+ + EDTA4- => Mg(EDTA)2-

Ca2+ + EDTA34- => Ca(EDTA)2-


Moles of (Mg2+ + Ca2+) in 10 mL = moles of EDTA in first titration

= volume x concentration of EDTA

= 26.81/1000 x 0.003474 = 9.314 x 10^(-5) mol


Moles of Ca2+ in 10 mL = moles of EDTA in second titration

= volume x concentration of EDTA

= 8.63/1000 x 0.003474 = 2.998 x 10^(-5) mol


Moles of Ca2+ in urine = moles of Ca2+ in 2 L (2000 mL) sample

= 2000/10 x 2.998 x 10^(-5) = 0.005996 mol


Mass of Ca2+ in urine = moles x molar mass of Ca

= 0.005996 x 40.078

= 0.240 g = 240 mg


Moles of Mg2+ in 10 mL = 9.314 x 10^(-5) - 2.998 x 10^(-5)

= 6.316 x 10^(-5) mol


Moles of Mg2+ in urine = moles of Mg2+ in 2 L (2000 mL) sample

= 2000/10 x 6.316 x 10^(-5) = 0.012632 mol


Mass of Mg2+ in urine = moles x molar mass of Mg

= 0.012632 x 24.305

= 0.307 g = 307 mg


Thus the Mg2+ content = 307 mg is too high (>150 mg) while the Ca2+ content = 240 mg is within the normal range (50-300 mg)

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