Question

4. A 0.3634 g sample of ore containing calcium is dissolved and the calcium ions are precipitated as CaC2O4. The CaC2O4 precipitate is filtered, washed and redissolved in acid. The pH of the resulting Ca2+ solution is adjusted. A 25.00 mL volume of 0.0376 M EDTA is added. The excess EDTA is back-titrated to the end point, requiring 23.08 mL of 0.0188 M Mg2+ solution.

a) Write a balanced chemical equation for the EDTA reactions that occur as part of the back-titration used in this analysis. (3 pts)

b) Determine the %Ca in the original ore sample. (5 pts)

Answer 1

Number of moles of Mg2+ is , n = Molarity x volume in L

= 0.0188 M x 23.08 mL x 10^-3 L/mL

= 4.34x10^-4 mol

So number of moles of EDTA titrated by Mg2+ ions is , n' = 4.34x10^-4 mol Mg2+ x ( 1 mole EDTA / 1 mole Mg2+ )

= 4.34x10^-4 mol EDTA

Actual number of moles of EDTA , N = 0.0376 mol x 25.00 mL x 10^-3L/mL

= 9.4x10^-4 mol

So number of moles of EDTA reacted with Ca2+ is , N' = N - n' = 5.06x10^-4 mol

Number of moles of Ca2+ reacted with EDTA , N" = 5.06x10^-4 mol EDTA x ( 1 mole Ca2+ / 1 mole EDTA )

= 5.06x10^-4 mol Ca2+

Mass of CaC2O4 , m = number of moles x molar mass

= 5.06x10^-4 mol x 128 g/mol

= 0.065 g

So % Ca = ( mass of CaC2O4 / mass of sample ) x 100

= ( 0.065 g / 0.3634 g ) x 100

= 17.8 %