The analysis of Ca 2+ and Mg 2+ in urine samples could be performed in a way sim

Question

The analysis of Ca2+ and Mg2+ in urine samples could be performed in a way similar to the EDTA analysis of the calcium content in water. For example, a 25.0 mL aliquot of a urine sample (buffered to pH = 10.0) required 17.2 mL of 0.0776 M EDTA to react with all of the Ca2+ and Mg2+ ions in the sample. Then another identical 25.0 mL urine sample was analyzed, except this one was treated with NaOH to precipitate all of the Mg2+ from the solution. This second solution required 13.2 mL of the 0.0776 M EDTA to react with the Ca2+ ions. Report the moles of Ca2+ and Mg2+ in the urine.

Explanation / Answer

1) moles of Ca2+ ions = molarity of EDTA x volume of EDTA

= 0.0776 M x 13.2 ml

= 1.024 mmol

2) moles of Ca2+ and Mg2+ ions = 17.2 ml x 0.0776 M

= 1.335 mmol

therefore

moles of Mg2+ ions = 1.335 mmol - 1.024 mol

= 0.311 mmol

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